Cylinder Volumes

9.1: The Same But Different (5 minutes)

CCSS Standards

Building On

Building Towards

  • HSG-GMD.A.1

Routines and Materials

Instructional Routines

  • Think Pair Share

In previous grades, students learned the formula for the volume of a cylinder. In this task, students compare and contrast methods for finding volumes of cylinders and prisms. Monitor for vocabulary such as diameter and radius , and for expressions like \(Bh\) , \(\pi r^2\) , and \(\ell w h\) .

Arrange students in groups of 2. Tell students there are many possible answers for the questions. After quiet work time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Follow with a whole-class discussion.

Student Facing

Here are two solids.

Expand image

  • What information would you need to calculate the volume of each solid?
  • What is the same and different about how you would find the volume of each solid?

Student Response

For access, consult one of our IM Certified Partners .

Activity Synthesis

In the discussion, be sure the expressions \(Bh\) and \(\pi r^2\) come up. Ask students if they’ve seen the expression \(\ell w h\) , and to describe how this relates to \(Bh\) . Make sure they can describe what the formula \(V=Bh\) means in wordsβ€”to find the volume of a prism or cylinder, we multiply the area of the solid’s base by the height of the solid.

9.2: Water Transfer (15 minutes)

  • Anticipate, Monitor, Select, Sequence, Connect
  • MLR5: Co-Craft Questions

In previous grades, students learned how to calculate the volume of a cylinder. In this activity, they revisit that process. Students are prompted to think about volume in 1-unit layers. This will be helpful in upcoming lessons when students learn about Cavalieri’s Principle, or the idea that if two solids have equal-area cross sections at all heights, the solids have equal volumes. Cavalieri’s Principle will be used to derive a formula for the volume of a pyramid.

Note that in this activity, β€œvolume of a container” is used as shorthand for β€œthe volume of the region enclosed by the container.”

Monitor for students who calculate the volume of the water in the prism then work backwardΒ to find the height of the water in the cylinder, and for those who use deductive logic (for example,Β reasoning that because the two bases have equal area, the height of the water must be equal in both solids) to arrive at the same answer.

Students reason abstractly and quantitatively (MP2) when they compare the prism and cylinder volumes.

Here are two containers. All measurements are in centimeters.

  • Draw a representation of this situation.
  • The water is poured from the prism into the cylinder. What is the height of the water in the cylinder? Explain your reasoning.
  • Suppose the prism contained water that reached a height of 3 cm instead of 1 cm. If the water were poured into the empty cylinder, what would the height of the water in the cylinder be?

Anticipated Misconceptions

Students may be unsure how to work with the prism measurement \(4\pi\) cm. Invite them to write out the area calculation as \(9\boldcdot4\pi\) . Alternatively, they can use an approximation of \(\pi\) to get rounded answers.

If students struggle to find the height of the water in the cylinder, ask them what they already know about the cylinder, and to consider what additional information they would need to find the height. Prompt them to use their responses to the first question to help them answer this one.

The goal is to make connections between methods of calculating volumes for prisms and cylinders. Ask previously identified students to share their reasoning, starting with those who did volume calculations and worked backward, and ending with those who used deductive reasoning. Connect these methods by asking these questions:

  • β€œDescribe what volume formulas you used. Do they apply to just one of the containers, or to both?” (Sample responses: The expression \(\ell wh\) applies to the prism. The expression \(\pi r^2 h\) applies to the cylinder. The expression \(BH\) applies to both.)
  • β€œConsider the volume formula \(V=Bh\) . How does this formula help explain why the heights were the same for the same volume of water in each container?” (The area of the base, \(B\) , is the same for each container: \(36\pi\) square centimeters. The volume formula can be rewritten \(h=V\div B\) , so if the containers have the same volume of water in each and have bases with the same area, the height will be the same.)

Invite students to describe how the volumes of the two containers compare (they are equal), and challenge them to describe another solid that has the same volume as these two.

9.3: Revisiting Rotation (15 minutes)

  • HSG-GMD.A.3
  • HSG-GMD.B.4
  • MLR8: Discussion Supports

Students combine their experience with solids of rotation and their understanding of cylinder volume.

No specific directions are given in regard to specifying how students should express their volume answer. Monitor for students who leave their answers in terms of \(\pi\) and those who found an approximate decimal answer.

Suppose each two-dimensional figure is rotated around the vertical axis shown. Each small square in the grid represents 1 square centimeter.

Description: <p>A grid. A vertical line is drawn 3 units to the right. A six sided figure is shaded with points at 3 units right, 1 unit up. Another point at 8 units right, 1 unit up. Another point at 8 units right, 4 units up. Another point at 5 units right, 4 units up. Another point at 5 units right, 7 units up. Another point at 3 units right, 7 units up.</p>

Description: <p>A grid. A vertical line is drawn 3 units to the right. A four sided figure is shaded with points at 6 units right, 1 unit up. Another point 8 units right, 1 unit up. Another point 8 units right, 7 units up. Another point 6 units right, 7 units up.</p>

For each solid:

  • Either sketch or Β describe in words the three-dimensional solid that would form.
  • Find the solid’s volume.

Are you ready for more?

  • The height is tripled to \(3h\) and the radius remains \(r\) .
  • The radius is tripled to \(3r\) and the height remains \(h\) .
  • Given a cube of side length \(s\) , write an expression for the volume if the side length were tripled to \(3s\) .
  • Which change affected the shape’s volume the most? The least? Explain or show your reasoning.

Students may struggle to visualize the rotated figures. For the L shape, encourage them to divide it into smaller pieces and consider each independently. For the rectangle, remind them of the doughnut cross section they drew in a previous lesson.

In the discussion, explore different aspects of the calculations students made. Invite one student who left the answer in terms of \(\pi\) and another who found a decimal answer to share their results. Remind students that the former is called an exact answer, while the latter is called an approximate answer because the value of \(\pi\) needs to be approximated and rounded to find a decimal answer. Emphasize that the choice of expression depends on the situation at hand.

If time permits, ask students would would happen if the first figure were rotated around the horizontal axis formed by the bottom edge of the figure. Challenge them to decide if the volume would be the same as with the rotation around the given vertical axis. (The volume of this figure would actually be \(99\pi\) cubic centimeters.)

Lesson Synthesis

Display the images for all to see, or bring in similar objects. Ask students to identify which solids’ volumes can be found by using the processes from this lesson. Challenge them to explain their reasoning using precise geometric language. Invite them to describe the types of solids for which the formula \(V=Bh\) applies.

Sample responses: We could find the volumes of the stick of butter, the bolt and nut, and the pencil using the formula \(V=Bh\) . These shapes are all made of prisms and cylinders, so you can think of them as being built from stacked, congruent layers. The pyramid, traffic cone, and sphere can’t use the formula \(V=Bh\) . If we think about stacking layers to form these shapes, the shape and size of the layers would change as you move through the object.

Photo of the Louvre Pyramid.

Attribution: By derwiki. Public Domain. Pixabay. Source .

Photo of a nut and a bolt.

Attribution: By Alexas_Fotos. Public Domain. Pixabay. Source .

A photo of a yellow pencil

Attribution: By WikimediaImages. Public Domain. pixabay. Source .

Photo of a traffic pylon.

Attribution: By manfredrichter. Public Domain. Pixabay. Source .

Photo of a stick of butter.

Attribution: By Ahecht. Public Domain. Wikimedia Commons. Source .

Photo of a basketball

Attribution: By alles. Public Domain. Pixabay. Source .

9.4: Cool-down - Cylinder Strategies (5 minutes)

  • HSA-SSE.A.1.b

Student Lesson Summary

Cylinder and prism volumes can be found by multiplying the area of the figure’s base by its height. The formula \(V=Bh\) , where \(V\) represents volume, \(B\) is the area of the base, and \(h\) is height, captures this concept.Β Consider the solid formed by rotating this rectangle around the horizontal axis shown. The result is a hollow cylinder of height 5 units with inner radius 1 unit and outer radius 4 units.

Description: <p>A grid. A horizontal line is drawn 1 unit up. A four sided figure is shaded with points at 4 units right, 2 units up. Another point 4 units right, 5 units up. Another point 9 units right, 2 units up. Another point 9 units right and 5 units up.</p>

Description: <p>A cylinder with the middle cylinder missing. The height is 5. The distance between the outside of the cylinder and interior cylinder is 3. The radius of the interior cylinder is 1.</p>

To calculate the volume of the outer cylinder, start by finding the area of the circular base. The circle’s radius measures 4 units, so its area is \(16 \pi\) square units because \(\pi(4)^2=16\pi\) . Multiply that by the cylinder’s height of 5 units to get \(80\pi\) cubic units.

For the inner cylinder, the area of the base is \(\pi\) square units, because \(\pi (1)^2 = \pi\) . The volume is therefore \(5\pi\) cubic units.Β Now subtract the volume of the inner, hollow part from the volume of the outer cylinder to get the volume of the solid: \(75 \pi\) cubic units because \(80\pi - 5\pi = 75\pi\) .

problem solving lesson 9 5 answer key

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problem solving lesson 9 5 answer key

x = 4 3 , x = βˆ’4 3 x = 4 3 , x = βˆ’4 3

y = 3 3 , y = βˆ’3 3 y = 3 3 , y = βˆ’3 3

x = 7 , x = βˆ’7 x = 7 , x = βˆ’7

m = 4 , m = βˆ’4 m = 4 , m = βˆ’4

c = 2 3 i , c = βˆ’2 3 i c = 2 3 i , c = βˆ’2 3 i

c = 2 6 i , c = βˆ’2 6 i c = 2 6 i , c = βˆ’2 6 i

x = 2 10 , x = βˆ’2 10 x = 2 10 , x = βˆ’2 10

y = 2 7 , y = βˆ’2 7 y = 2 7 , y = βˆ’2 7

r = 6 5 5 , r = βˆ’ 6 5 5 r = 6 5 5 , r = βˆ’ 6 5 5

t = 8 3 3 , t = βˆ’ 8 3 3 t = 8 3 3 , t = βˆ’ 8 3 3

a = 3 + 3 2 , a = 3 βˆ’ 3 2 a = 3 + 3 2 , a = 3 βˆ’ 3 2

b = βˆ’2 + 2 10 , b = βˆ’2 βˆ’ 2 10 b = βˆ’2 + 2 10 , b = βˆ’2 βˆ’ 2 10

x = 1 2 + 5 2 x = 1 2 + 5 2 , x = 1 2 βˆ’ 5 2 x = 1 2 βˆ’ 5 2

y = βˆ’ 3 4 + 7 4 , y = βˆ’ 3 4 βˆ’ 7 4 y = βˆ’ 3 4 + 7 4 , y = βˆ’ 3 4 βˆ’ 7 4

a = 5 + 2 5 , a = 5 βˆ’ 2 5 a = 5 + 2 5 , a = 5 βˆ’ 2 5

b = βˆ’3 + 4 2 , b = βˆ’3 βˆ’ 4 2 b = βˆ’3 + 4 2 , b = βˆ’3 βˆ’ 4 2

r = βˆ’ 4 3 + 2 2 i 3 , r = βˆ’ 4 3 βˆ’ 2 2 i 3 r = βˆ’ 4 3 + 2 2 i 3 , r = βˆ’ 4 3 βˆ’ 2 2 i 3

t = 4 + 10 i 2 , t = 4 βˆ’ 10 i 2 t = 4 + 10 i 2 , t = 4 βˆ’ 10 i 2

m = 7 3 , m = βˆ’1 m = 7 3 , m = βˆ’1

n = βˆ’ 3 4 , n = βˆ’ 7 4 n = βˆ’ 3 4 , n = βˆ’ 7 4

ⓐ ( a βˆ’ 10 ) 2 ( a βˆ’ 10 ) 2 β“‘ ( b βˆ’ 5 2 ) 2 ( b βˆ’ 5 2 ) 2 β“’ ( p + 1 8 ) 2 ( p + 1 8 ) 2

ⓐ ( b βˆ’ 2 ) 2 ( b βˆ’ 2 ) 2 β“‘ ( n + 13 2 ) 2 ( n + 13 2 ) 2 β“’ ( q βˆ’ 1 3 ) 2 ( q βˆ’ 1 3 ) 2

x = βˆ’5 , x = βˆ’1 x = βˆ’5 , x = βˆ’1

y = 1 , y = 9 y = 1 , y = 9

y = 5 + 15 i , y = 5 βˆ’ 15 i y = 5 + 15 i , y = 5 βˆ’ 15 i

z = βˆ’4 + 3 i , z = βˆ’4 βˆ’ 3 i z = βˆ’4 + 3 i , z = βˆ’4 βˆ’ 3 i

x = 8 + 4 3 , x = 8 βˆ’ 4 3 x = 8 + 4 3 , x = 8 βˆ’ 4 3

y = βˆ’4 + 3 3 , y = βˆ’4 βˆ’ 3 3 y = βˆ’4 + 3 3 , y = βˆ’4 βˆ’ 3 3

a = βˆ’7 , a = 3 a = βˆ’7 , a = 3

b = βˆ’10 , b = 2 b = βˆ’10 , b = 2

p = 5 2 + 61 2 , p = 5 2 βˆ’ 61 2 p = 5 2 + 61 2 , p = 5 2 βˆ’ 61 2

q = 7 2 + 37 2 , q = 7 2 βˆ’ 37 2 q = 7 2 + 37 2 , q = 7 2 βˆ’ 37 2

c = βˆ’9 , c = 3 c = βˆ’9 , c = 3

d = 11 , d = βˆ’7 d = 11 , d = βˆ’7

m = βˆ’7 , m = βˆ’1 m = βˆ’7 , m = βˆ’1

n = βˆ’2 , n = 8 n = βˆ’2 , n = 8

r = βˆ’ 7 3 , r = 3 r = βˆ’ 7 3 , r = 3

t = βˆ’ 5 2 , t = 2 t = βˆ’ 5 2 , t = 2

x = βˆ’ 3 8 + 41 8 , x = βˆ’ 3 8 βˆ’ 41 8 x = βˆ’ 3 8 + 41 8 , x = βˆ’ 3 8 βˆ’ 41 8

y = 5 3 + 10 3 , y = 5 3 βˆ’ 10 3 y = 5 3 + 10 3 , y = 5 3 βˆ’ 10 3

y = 1 , y = 2 3 y = 1 , y = 2 3

z = 1 , z = βˆ’ 3 2 z = 1 , z = βˆ’ 3 2

a = βˆ’3 , a = 5 a = βˆ’3 , a = 5

b = βˆ’6 , b = βˆ’4 b = βˆ’6 , b = βˆ’4

m = βˆ’6 + 15 3 , m = βˆ’6 βˆ’ 15 3 m = βˆ’6 + 15 3 , m = βˆ’6 βˆ’ 15 3

n = βˆ’2 + 2 6 5 , n = βˆ’2 βˆ’ 2 6 5 n = βˆ’2 + 2 6 5 , n = βˆ’2 βˆ’ 2 6 5

a = 1 4 + 31 4 i , a = 1 4 βˆ’ 31 4 i a = 1 4 + 31 4 i , a = 1 4 βˆ’ 31 4 i

b = βˆ’ 1 5 + 19 5 i , b = βˆ’ 1 5 βˆ’ 19 5 i b = βˆ’ 1 5 + 19 5 i , b = βˆ’ 1 5 βˆ’ 19 5 i

x = βˆ’1 + 6 , x = βˆ’1 βˆ’ 6 x = βˆ’1 + 6 , x = βˆ’1 βˆ’ 6

y = 1 + 2 , y = 1 βˆ’ 2 y = 1 + 2 , y = 1 βˆ’ 2

c = 2 + 7 3 , c = 2 βˆ’ 7 3 c = 2 + 7 3 , c = 2 βˆ’ 7 3

d = 9 + 33 4 , d = 9 βˆ’ 33 4 d = 9 + 33 4 , d = 9 βˆ’ 33 4

r = βˆ’5 r = βˆ’5

t = 4 5 t = 4 5

ⓐ 2 complex solutions; β“‘ 2 real solutions; β“’ 1 real solution

ⓐ 2 real solutions; β“‘ 2 complex solutions; β“’ 1 real solution

ⓐ factoring; β“‘ Square Root Property; β“’ Quadratic Formula

ⓐ Quadratic Forumula; β“‘ Factoring or Square Root Property β“’ Square Root Property

x = 2 , x = βˆ’ 2 , x = 2 , x = βˆ’2 x = 2 , x = βˆ’ 2 , x = 2 , x = βˆ’2

x = 7 , x = βˆ’ 7 , x = 2 , x = βˆ’2 x = 7 , x = βˆ’ 7 , x = 2 , x = βˆ’2

x = 3 , x = 1 x = 3 , x = 1

y = βˆ’1 , y = 1 y = βˆ’1 , y = 1

x = 9 , x = 16 x = 9 , x = 16

x = 4 , x = 16 x = 4 , x = 16

x = βˆ’8 , x = 343 x = βˆ’8 , x = 343

x = 81 , x = 625 x = 81 , x = 625

x = 4 3 x = 2 x = 4 3 x = 2

x = 2 5 , x = 3 4 x = 2 5 , x = 3 4

The two consecutive odd integers whose product is 99 are 9, 11, and βˆ’9, βˆ’11

The two consecutive even integers whose product is 128 are 12, 14 and βˆ’12, βˆ’14.

The height of the triangle is 12 inches and the base is 76 inches.

The height of the triangle is 11 feet and the base is 20 feet.

The length of the garden is approximately 18 feet and the width 11 feet.

The length of the tablecloth is approximatel 11.8 feet and the width 6.8 feet.

The length of the flag pole’s shadow is approximately 6.3 feet and the height of the flag pole is 18.9 feet.

The distance between the opposite corners is approximately 7.2 feet.

The arrow will reach 180 feet on its way up after 3 seconds and again on its way down after approximately 3.8 seconds.

The ball will reach 48 feet on its way up after approximately .6 second and again on its way down after approximately 5.4 seconds.

The speed of the jet stream was 100 mph.

The speed of the jet stream was 50 mph.

Press #1 would take 12 hours, and Press #2 would take 6 hours to do the job alone.

The red hose take 6 hours and the green hose take 3 hours alone.

ⓐ up; β“‘ down

ⓐ down; β“‘ up

ⓐ x = 2 ; x = 2 ; β“‘ ( 2 , βˆ’7 ) ( 2 , βˆ’7 )

ⓐ x = 1 ; x = 1 ; β“‘ ( 1 , βˆ’5 ) ( 1 , βˆ’5 )

y -intercept: ( 0 , βˆ’8 ) ( 0 , βˆ’8 ) x -intercepts ( βˆ’4 , 0 ) , ( 2 , 0 ) ( βˆ’4 , 0 ) , ( 2 , 0 )

y -intercept: ( 0 , βˆ’12 ) ( 0 , βˆ’12 ) x -intercepts ( βˆ’2 , 0 ) , ( 6 , 0 ) ( βˆ’2 , 0 ) , ( 6 , 0 )

y -intercept: ( 0 , 4 ) ( 0 , 4 ) no x -intercept

y -intercept: ( 0 , βˆ’5 ) ( 0 , βˆ’5 ) x -intercepts ( βˆ’1 , 0 ) , ( 5 , 0 ) ( βˆ’1 , 0 ) , ( 5 , 0 )

The minimum value of the quadratic function is βˆ’4 and it occurs when x = 4.

The maximum value of the quadratic function is 5 and it occurs when x = 2.

It will take 4 seconds for the stone to reach its maximum height of 288 feet.

It will 6.5 seconds for the rocket to reach its maximum height of 676 feet.

β“‘ The graph of g ( x ) = x 2 + 1 g ( x ) = x 2 + 1 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted up 1 unit. The graph of h ( x ) = x 2 βˆ’ 1 h ( x ) = x 2 βˆ’ 1 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted down 1 unit.

β“‘ The graph of h ( x ) = x 2 + 6 h ( x ) = x 2 + 6 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted up 6 units. The graph of h ( x ) = x 2 βˆ’ 6 h ( x ) = x 2 βˆ’ 6 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted down 6 units.

β“‘ The graph of g ( x ) = ( x + 2 ) 2 g ( x ) = ( x + 2 ) 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted left 2 units. The graph of h ( x ) = ( x βˆ’ 2 ) 2 h ( x ) = ( x βˆ’ 2 ) 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shift right 2 units.

β“‘ The graph of g ( x ) = ( x + 5 ) 2 g ( x ) = ( x + 5 ) 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted left 5 units. The graph of h ( x ) = ( x βˆ’ 5 ) 2 h ( x ) = ( x βˆ’ 5 ) 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted right 5 units.

f ( x ) = βˆ’4 ( x + 1 ) 2 + 5 f ( x ) = βˆ’4 ( x + 1 ) 2 + 5

f ( x ) = 2 ( x βˆ’ 2 ) 2 βˆ’ 5 f ( x ) = 2 ( x βˆ’ 2 ) 2 βˆ’ 5

ⓐ f ( x ) = 3 ( x βˆ’ 1 ) 2 + 2 f ( x ) = 3 ( x βˆ’ 1 ) 2 + 2 β“‘

ⓐ f ( x ) = βˆ’2 ( x βˆ’ 2 ) 2 + 1 f ( x ) = βˆ’2 ( x βˆ’ 2 ) 2 + 1 β“‘

f ( x ) = ( x βˆ’ 3 ) 2 βˆ’ 4 f ( x ) = ( x βˆ’ 3 ) 2 βˆ’ 4

f ( x ) = ( x + 3 ) 2 βˆ’ 1 f ( x ) = ( x + 3 ) 2 βˆ’ 1

β“‘ ( βˆ’4 , βˆ’2 ) ( βˆ’4 , βˆ’2 )

β“‘ ( βˆ’ ∞ , 2 ] βˆͺ [ 6 , ∞ ) ( βˆ’ ∞ , 2 ] βˆͺ [ 6 , ∞ )

β“‘ ( βˆ’1 , 5 ) ( βˆ’1 , 5 )

β“‘ ( βˆ’ ∞ , 2 ] βˆͺ [ 8 , ∞ ) ( βˆ’ ∞ , 2 ] βˆͺ [ 8 , ∞ )

( βˆ’ ∞ , βˆ’4 ] βˆͺ [ 2 , ∞ ) ( βˆ’ ∞ , βˆ’4 ] βˆͺ [ 2 , ∞ )

[ βˆ’3 , 5 ] [ βˆ’3 , 5 ]

[ βˆ’1 βˆ’ 2 , βˆ’1 + 2 ] [ βˆ’1 βˆ’ 2 , βˆ’1 + 2 ]

( βˆ’ ∞ , 4 βˆ’ 2 ) βˆͺ ( 4 + 2 , ∞ ) ( βˆ’ ∞ , 4 βˆ’ 2 ) βˆͺ ( 4 + 2 , ∞ )

ⓐ ( βˆ’ ∞ , ∞ ) ( βˆ’ ∞ , ∞ ) β“‘ no solution

ⓐ no solution β“‘ ( βˆ’ ∞ , ∞ ) ( βˆ’ ∞ , ∞ )

Section 9.1 Exercises

a = Β± 7 a = Β± 7

r = Β± 2 6 r = Β± 2 6

u = Β± 10 3 u = Β± 10 3

m = Β± 3 m = Β± 3

x = Β± 6 x = Β± 6

x = Β± 5 i x = Β± 5 i

x = Β± 3 7 i x = Β± 3 7 i

x = Β± 9 x = Β± 9

a = Β± 2 5 a = Β± 2 5

p = Β± 4 7 7 p = Β± 4 7 7

y = Β± 4 10 5 y = Β± 4 10 5

u = 14 , u = βˆ’2 u = 14 , u = βˆ’2

m = 6 Β± 2 5 m = 6 Β± 2 5

r = 1 2 Β± 3 2 r = 1 2 Β± 3 2

y = βˆ’ 2 3 Β± 2 2 9 y = βˆ’ 2 3 Β± 2 2 9

a = 7 Β± 5 2 a = 7 Β± 5 2

x = βˆ’3 Β± 2 2 x = βˆ’3 Β± 2 2

c = βˆ’ 1 5 Β± 3 3 5 i c = βˆ’ 1 5 Β± 3 3 5 i

x = 3 4 Β± 7 2 i x = 3 4 Β± 7 2 i

m = 2 Β± 2 2 m = 2 Β± 2 2

x = 3 + 2 3 , x = 3 βˆ’ 2 3 x = 3 + 2 3 , x = 3 βˆ’ 2 3

x = βˆ’ 3 5 , x = 9 5 x = βˆ’ 3 5 , x = 9 5

x = βˆ’ 7 6 , x = 11 6 x = βˆ’ 7 6 , x = 11 6

r = Β± 4 r = Β± 4

a = 4 Β± 2 7 a = 4 Β± 2 7

w = 1 , w = 5 3 w = 1 , w = 5 3

a = Β± 3 2 a = Β± 3 2

p = 1 3 Β± 7 3 p = 1 3 Β± 7 3

m = Β± 2 2 i m = Β± 2 2 i

u = 7 Β± 6 2 u = 7 Β± 6 2

m = 4 Β± 2 3 m = 4 Β± 2 3

x = βˆ’3 , x = βˆ’7 x = βˆ’3 , x = βˆ’7

c = Β± 5 6 6 c = Β± 5 6 6

x = 6 Β± 2 i x = 6 Β± 2 i

Answers will vary.

Section 9.2 Exercises

ⓐ ( m βˆ’ 12 ) 2 ( m βˆ’ 12 ) 2 β“‘ ( x βˆ’ 11 2 ) 2 ( x βˆ’ 11 2 ) 2 β“’ ( p βˆ’ 1 6 ) 2 ( p βˆ’ 1 6 ) 2

ⓐ ( p βˆ’ 11 ) 2 ( p βˆ’ 11 ) 2 β“‘ ( y + 5 2 ) 2 ( y + 5 2 ) 2 β“’ ( m + 1 5 ) 2 ( m + 1 5 ) 2

u = βˆ’3 , u = 1 u = βˆ’3 , u = 1

x = βˆ’1 , x = 21 x = βˆ’1 , x = 21

m = βˆ’2 Β± 2 10 i m = βˆ’2 Β± 2 10 i

r = βˆ’3 Β± 2 i r = βˆ’3 Β± 2 i

a = 5 Β± 2 5 a = 5 Β± 2 5

x = βˆ’ 5 2 Β± 33 2 x = βˆ’ 5 2 Β± 33 2

u = 1 , u = 13 u = 1 , u = 13

r = βˆ’2 , r = 6 r = βˆ’2 , r = 6

v = 9 2 Β± 89 2 v = 9 2 Β± 89 2

x = 5 Β± 30 x = 5 Β± 30

x = βˆ’7 , x = 3 x = βˆ’7 , x = 3

m = βˆ’11 , m = 1 m = βˆ’11 , m = 1

n = 1 Β± 14 n = 1 Β± 14

c = βˆ’2 , c = 3 2 c = βˆ’2 , c = 3 2

x = βˆ’5 , x = 3 2 x = βˆ’5 , x = 3 2

p = βˆ’ 7 4 Β± 161 4 p = βˆ’ 7 4 Β± 161 4

x = 3 10 Β± 191 10 i x = 3 10 Β± 191 10 i

Section 9.3 Exercises

m = βˆ’1 , m = 3 4 m = βˆ’1 , m = 3 4

p = 1 2 , p = 3 p = 1 2 , p = 3

p = βˆ’4 , p = βˆ’3 p = βˆ’4 , p = βˆ’3

r = βˆ’3 , r = 11 r = βˆ’3 , r = 11

u = βˆ’7 Β± 73 6 u = βˆ’7 Β± 73 6

a = 3 Β± 3 2 a = 3 Β± 3 2

x = βˆ’4 Β± 2 5 x = βˆ’4 Β± 2 5

y = βˆ’2 , y = 1 3 y = βˆ’2 , y = 1 3

x = βˆ’ 3 4 Β± 15 4 i x = βˆ’ 3 4 Β± 15 4 i

x = 3 8 Β± 7 8 i x = 3 8 Β± 7 8 i

v = 2 Β± 2 13 v = 2 Β± 2 13

y = βˆ’4 , y = 7 y = βˆ’4 , y = 7

b = βˆ’2 Β± 11 6 b = βˆ’2 Β± 11 6

c = βˆ’ 3 4 c = βˆ’ 3 4

q = βˆ’ 3 5 q = βˆ’ 3 5

ⓐ no real solutions no real solutions β“‘ 1 1 β“’ 2 2

ⓐ 1 1 β“‘ no real solutions no real solutions β“’ 2 2

ⓐ factor factor β“‘ square root square root β“’ Quadratic Formula Quadratic Formula

ⓐ Quadratic Formula Quadratic Formula β“‘ square root square root β“’ factor factor

Section 9.4 Exercises

x = Β± 3 , x = Β± 2 x = Β± 3 , x = Β± 2

x = Β± 15 , x = Β± 2 i x = Β± 15 , x = Β± 2 i

x = Β± 1 , x = Β± 6 2 x = Β± 1 , x = Β± 6 2

x = Β± 3 , x = Β± 2 2 x = Β± 3 , x = Β± 2 2

x = βˆ’1 , x = 12 x = βˆ’1 , x = 12

x = βˆ’ 5 3 , x = 0 x = βˆ’ 5 3 , x = 0

x = 0 , x = Β± 3 x = 0 , x = Β± 3

x = Β± 11 2 , x = Β± 7 x = Β± 11 2 , x = Β± 7

x = 25 x = 25

x = 4 x = 4

x = 1 4 x = 1 4

x = 1 25 , x = 9 4 x = 1 25 , x = 9 4

x = βˆ’1 , x = βˆ’512 x = βˆ’1 , x = βˆ’512

x = 8 , x = βˆ’216 x = 8 , x = βˆ’216

x = 27 8 , x = βˆ’ 64 27 x = 27 8 , x = βˆ’ 64 27

x = 27 512 , x = 125 x = 27 512 , x = 125

x = 1 , x = 49 x = 1 , x = 49

x = βˆ’2 , x = βˆ’ 3 5 x = βˆ’2 , x = βˆ’ 3 5

x = βˆ’2 , x = 4 3 x = βˆ’2 , x = 4 3

Section 9.5 Exercises

Two consecutive odd numbers whose product is 255 are 15 and 17, and βˆ’15 and βˆ’17.

The first and second consecutive odd numbers are 24 and 26, and βˆ’26 and βˆ’24.

Two consecutive odd numbers whose product is 483 are 21 and 23, and βˆ’21 and βˆ’23.

The width of the triangle is 5 inches and the height is 18 inches.

The base is 24 feet and the height of the triangle is 10 feet.

The length of the driveway is 15.0 feet and the width is 3.3 feet.

The length of table is 8 feet and the width is 3 feet.

The length of the legs of the right triangle are 3.2 and 9.6 cm.

The length of the diagonal fencing is 7.3 yards.

The ladder will reach 24.5 feet on the side of the house.

The arrow will reach 400 feet on its way up in 2.8 seconds and on the way down in 11 seconds.

The bullet will take 70 seconds to hit the ground.

The speed of the wind was 49 mph.

The speed of the current was 4.3 mph.

The less experienced painter takes 6 hours and the experienced painter takes 3 hours to do the job alone.

Machine #1 takes 3.6 hours and Machine #2 takes 4.6 hours to do the job alone.

Section 9.6 Exercises

ⓐ down β“‘ up

ⓐ x = βˆ’4 x = βˆ’4 ; β“‘ ( βˆ’4 , βˆ’17 ) ( βˆ’4 , βˆ’17 )

ⓐ x = 1 x = 1 ; β“‘ ( 1 , 2 ) ( 1 , 2 )

y -intercept: ( 0 , 6 ) ; ( 0 , 6 ) ; x -intercept ( βˆ’1 , 0 ) , ( βˆ’6 , 0 ) ( βˆ’1 , 0 ) , ( βˆ’6 , 0 )

y -intercept: ( 0 , 12 ) ; ( 0 , 12 ) ; x -intercept ( βˆ’2 , 0 ) , ( βˆ’6 , 0 ) ( βˆ’2 , 0 ) , ( βˆ’6 , 0 )

y -intercept: ( 0 , βˆ’19 ) ; ( 0 , βˆ’19 ) ; x -intercept: none

y -intercept: ( 0 , 13 ) ; ( 0 , 13 ) ; x -intercept: none

y -intercept: ( 0 , βˆ’16 ) ; ( 0 , βˆ’16 ) ; x -intercept ( 5 2 , 0 ) ( 5 2 , 0 )

y -intercept: ( 0 , 9 ) ; ( 0 , 9 ) ; x -intercept ( βˆ’3 , 0 ) ( βˆ’3 , 0 )

The minimum value is βˆ’ 9 8 βˆ’ 9 8 when x = βˆ’ 1 4 . x = βˆ’ 1 4 .

The maximum value is 6 when x = 3.

The maximum value is 16 when x = 0.

In 5.3 sec the arrow will reach maximum height of 486 ft.

In 3.4 seconds the ball will reach its maximum height of 185.6 feet.

20 computers will give the maximum of $400 in receipts.

He will be able to sell 35 pairs of boots at the maximum revenue of $1,225.

The length of the side along the river of the corral is 120 feet and the maximum area is 7,200 square feet.

The maximum area of the patio is 800 feet.

Section 9.7 Exercises

β“‘ The graph of g ( x ) = x 2 + 4 g ( x ) = x 2 + 4 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted up 4 units. The graph of h ( x ) = x 2 βˆ’ 4 h ( x ) = x 2 βˆ’ 4 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shift down 4 units.

β“‘ The graph of g ( x ) = ( x βˆ’ 3 ) 2 g ( x ) = ( x βˆ’ 3 ) 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted right 3 units. The graph of h ( x ) = ( x + 3 ) 2 h ( x ) = ( x + 3 ) 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted left 3 units.

f ( x ) = βˆ’3 ( x + 2 ) 2 + 7 f ( x ) = βˆ’3 ( x + 2 ) 2 + 7

f ( x ) = 3 ( x + 1 ) 2 βˆ’ 4 f ( x ) = 3 ( x + 1 ) 2 βˆ’ 4

ⓐ f ( x ) = ( x + 3 ) 2 βˆ’ 4 f ( x ) = ( x + 3 ) 2 βˆ’ 4 β“‘

ⓐ f ( x ) = ( x + 2 ) 2 βˆ’ 1 f ( x ) = ( x + 2 ) 2 βˆ’ 1 β“‘

ⓐ f ( x ) = ( x βˆ’ 3 ) 2 + 6 f ( x ) = ( x βˆ’ 3 ) 2 + 6 β“‘

ⓐ f ( x ) = βˆ’ ( x βˆ’ 4 ) 2 + 0 f ( x ) = βˆ’ ( x βˆ’ 4 ) 2 + 0 β“‘

ⓐ f ( x ) = βˆ’ ( x + 2 ) 2 + 6 f ( x ) = βˆ’ ( x + 2 ) 2 + 6 β“‘

ⓐ f ( x ) = 5 ( x βˆ’ 1 ) 2 + 3 f ( x ) = 5 ( x βˆ’ 1 ) 2 + 3 β“‘

ⓐ f ( x ) = 2 ( x βˆ’ 1 ) 2 βˆ’ 1 f ( x ) = 2 ( x βˆ’ 1 ) 2 βˆ’ 1 β“‘

ⓐ f ( x ) = βˆ’2 ( x βˆ’ 2 ) 2 βˆ’ 2 f ( x ) = βˆ’2 ( x βˆ’ 2 ) 2 βˆ’ 2 β“‘

ⓐ f ( x ) = 2 ( x + 1 ) 2 + 4 f ( x ) = 2 ( x + 1 ) 2 + 4 β“‘

ⓐ f ( x ) = βˆ’ ( x βˆ’ 1 ) 2 βˆ’ 3 f ( x ) = βˆ’ ( x βˆ’ 1 ) 2 βˆ’ 3 β“‘

f ( x ) = ( x + 1 ) 2 βˆ’ 5 f ( x ) = ( x + 1 ) 2 βˆ’ 5

f ( x ) = 2 ( x βˆ’ 1 ) 2 βˆ’ 3 f ( x ) = 2 ( x βˆ’ 1 ) 2 βˆ’ 3

Section 9.8 Exercises

β“‘ ( βˆ’ ∞ , βˆ’5 ) βˆͺ ( βˆ’1 , ∞ ) ( βˆ’ ∞ , βˆ’5 ) βˆͺ ( βˆ’1 , ∞ )

β“‘ [ βˆ’3 , βˆ’1 ] [ βˆ’3 , βˆ’1 ]

β“‘ ( βˆ’ ∞ , βˆ’6 ] βˆͺ [ 3 , ∞ ) ( βˆ’ ∞ , βˆ’6 ] βˆͺ [ 3 , ∞ )

β“‘ [ βˆ’3 , 4 ] [ βˆ’3 , 4 ]

( βˆ’ ∞ , βˆ’4 ] βˆͺ [ 1 , ∞ ) ( βˆ’ ∞ , βˆ’4 ] βˆͺ [ 1 , ∞ )

( 2 , 5 ) ( 2 , 5 )

( βˆ’ ∞ , βˆ’5 ) βˆͺ ( βˆ’3 , ∞ ) ( βˆ’ ∞ , βˆ’5 ) βˆͺ ( βˆ’3 , ∞ )

[ 2 βˆ’ 2 , 2 + 2 ] [ 2 βˆ’ 2 , 2 + 2 ]

( βˆ’ ∞ , 5 βˆ’ 6 ) βˆͺ ( 5 + 6 , ∞ ) ( βˆ’ ∞ , 5 βˆ’ 6 ) βˆͺ ( 5 + 6 , ∞ )

( βˆ’ ∞ , βˆ’ 5 2 ] βˆͺ [ βˆ’ 2 3 , ∞ ) ( βˆ’ ∞ , βˆ’ 5 2 ] βˆͺ [ βˆ’ 2 3 , ∞ )

[ βˆ’ 1 2 , 4 ] [ βˆ’ 1 2 , 4 ]

( βˆ’ ∞ , ∞ ) . ( βˆ’ ∞ , ∞ ) .

no solution

Review Exercises

y = Β± 12 y = Β± 12

a = Β± 5 a = Β± 5

r = Β± 4 2 i r = Β± 4 2 i

w = Β± 5 3 w = Β± 5 3

p = βˆ’1 , 9 p = βˆ’1 , 9

x = 1 4 Β± 3 4 x = 1 4 Β± 3 4

n = 4 Β± 10 2 n = 4 Β± 10 2

n = βˆ’5 Β± 2 3 n = βˆ’5 Β± 2 3

( x + 11 ) 2 ( x + 11 ) 2

( a βˆ’ 3 2 ) 2 ( a βˆ’ 3 2 ) 2

d = βˆ’13 , βˆ’1 d = βˆ’13 , βˆ’1

m = βˆ’3 Β± 10 i m = βˆ’3 Β± 10 i

v = 7 Β± 3 2 v = 7 Β± 3 2

m = βˆ’9 , βˆ’1 m = βˆ’9 , βˆ’1

a = 3 2 Β± 41 2 a = 3 2 Β± 41 2

u = βˆ’6 Β± 2 2 u = βˆ’6 Β± 2 2

p = 0 , 6 p = 0 , 6

y = βˆ’ 1 2 , 2 y = βˆ’ 1 2 , 2

c = βˆ’ 1 3 Β± 2 7 3 c = βˆ’ 1 3 Β± 2 7 3

x = 3 2 Β± 1 2 i x = 3 2 Β± 1 2 i

x = 1 4 , 1 x = 1 4 , 1

r = βˆ’6 , 7 r = βˆ’6 , 7

v = βˆ’1 Β± 21 8 v = βˆ’1 Β± 21 8

m = βˆ’4 Β± 10 3 m = βˆ’4 Β± 10 3

a = 5 12 Β± 23 12 i a = 5 12 Β± 23 12 i

u = 5 Β± 21 u = 5 Β± 21

p = 4 Β± 5 5 p = 4 Β± 5 5

c = βˆ’ 1 2 c = βˆ’ 1 2

ⓐ 1 β“‘ 2 β“’ 2 β““ 2

ⓐ factor β“‘ Quadratic Formula β“’ square root

x = Β± 2 , x = Β± 2 3 x = Β± 2 , x = Β± 2 3

x = Β± 1 , x = Β± 1 2 x = Β± 1 , x = Β± 1 2

x = 16 x = 16

x = 64 , x = 216 x = 64 , x = 216

Two consecutive even numbers whose product is 624 are 24 and 26, and βˆ’24 and βˆ’26.

The height is 14 inches and the width is 10 inches.

The length of the diagonal is 3.6 feet.

The width of the serving table is 4.7 feet and the length is 16.1 feet.

The speed of the wind was 30 mph.

One man takes 3 hours and the other man 6 hours to finish the repair alone.

ⓐ up β“‘ down

x = 2 ; ( 2 , βˆ’7 ) x = 2 ; ( 2 , βˆ’7 )

y : ( 0 , 15 ) x : ( 3 , 0 ) , ( 5 , 0 ) y : ( 0 , 15 ) x : ( 3 , 0 ) , ( 5 , 0 )

y : ( 0 , βˆ’46 ) x : none y : ( 0 , βˆ’46 ) x : none

y : ( 0 , βˆ’64 ) x : ( βˆ’8 , 0 ) y : ( 0 , βˆ’64 ) x : ( βˆ’8 , 0 )

The maximum value is 2 when x = 2.

The length adjacent to the building is 90 feet giving a maximum area of 4,050 square feet.

f ( x ) = 2 ( x βˆ’ 1 ) 2 βˆ’ 6 f ( x ) = 2 ( x βˆ’ 1 ) 2 βˆ’ 6

ⓐ f ( x ) = 3 ( x βˆ’ 1 ) 2 βˆ’ 4 f ( x ) = 3 ( x βˆ’ 1 ) 2 βˆ’ 4 β“‘

ⓐ f ( x ) = βˆ’3 ( x + 2 ) 2 + 7 f ( x ) = βˆ’3 ( x + 2 ) 2 + 7 β“‘

β“‘ ( βˆ’ ∞ , βˆ’2 ) βˆͺ ( 3 , ∞ ) ( βˆ’ ∞ , βˆ’2 ) βˆͺ ( 3 , ∞ )

[ βˆ’2 , 1 ] [ βˆ’2 , 1 ]

( 2 , 4 ) ( 2 , 4 )

[ 3 βˆ’ 5 , 3 + 5 ] [ 3 βˆ’ 5 , 3 + 5 ]

Practice Test

w = βˆ’2 , w = βˆ’8 w = βˆ’2 , w = βˆ’8

m = 1 , m = 3 2 m = 1 , m = 3 2

y = 2 3 y = 2 3

y = 1 , y = βˆ’27 y = 1 , y = βˆ’27

ⓐ down β“‘ x = βˆ’4 x = βˆ’4 β“’ ( βˆ’4 , 0 ) ( βˆ’4 , 0 ) β““ y : ( 0 , 16 ) ; x : ( βˆ’4 , 0 ) y : ( 0 , 16 ) ; x : ( βˆ’4 , 0 ) β“” minimum value of βˆ’4 βˆ’4 when x = 0 x = 0

( βˆ’ ∞ , βˆ’ 5 2 ) βˆͺ ( 2 , ∞ ) ( βˆ’ ∞ , βˆ’ 5 2 ) βˆͺ ( 2 , ∞ )

The diagonal is 3.8 units long.

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Access for free at https://openstax.org/books/intermediate-algebra/pages/1-introduction
  • Authors: Lynn Marecek
  • Publisher/website: OpenStax
  • Book title: Intermediate Algebra
  • Publication date: Mar 14, 2017
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra/pages/1-introduction
  • Section URL: https://openstax.org/books/intermediate-algebra/pages/chapter-9

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Go Math Answer Key

McGraw Hill Math Grade 5 Chapter 9 Lesson 11 Answer Key Problem Solving: Working Backwards

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 9 Lesson 11 Problem Solving: Working Backwards are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 9 Lesson 11 Problem Solving: Working Backwards

Question 1. Regina and Alicia had 4\(\frac{1}{2}\) cups of water at the end of a day of driving. They drank 8\(\frac{1}{4}\) cups before lunch and 11\(\frac{1}{4}\) cups in the afternoon. How much water did they start with? Answer: Number of cups of water they start with = 24.

Explanation: Number of cups of water Regina and Alicia had at the end of a day of driving = 4\(\frac{1}{2}\). Number of cups of water they drank before lunch = 8\(\frac{1}{4}\). Number of cups of water they drank in the afternoon = 11\(\frac{1}{4}\). Number of cups of water they start with = Number of cups of water Regina and Alicia had at the end of a day of driving + Number of cups of water they drank before lunch + Number of cups of water they drank in the afternoon = 4\(\frac{1}{2}\) + 8\(\frac{1}{4}\) + 11\(\frac{1}{4}\) = [(8 + 1) Γ· 2] + [(32 + 1) Γ· 4] + [(44 + 1) Γ· 4] = \(\frac{9}{2}\)Β  + \(\frac{33}{4}\) + \(\frac{45}{4}\) = [(9 Γ— 2) Γ· (2 Γ— 2)]Β  + \(\frac{33}{4}\) + \(\frac{45}{4}\) = \(\frac{18}{4}\)Β  + \(\frac{33}{4}\) + \(\frac{45}{4}\) = (18 + 33 + 45) Γ· 4 = 96 Γ· 4 = 24.

Question 2. A small shop has 5.9 pounds of lunch meat at the end of Friday. The shop sold 4.9 pounds on Friday morning and 9.35 pounds on Friday afternoon. How many pounds of lunchmeat did the store start with? Answer: Number of pounds of lunchmeat the store start with = 20.15.

Explanation: Number of pounds of lunch meat a small shop has at the end of Friday = 5.9. Number of pounds of lunch meat a small shop on Friday morning = 4.9. Number of pounds of lunch meat a small shop on Friday afternoon = 9.35. Number of pounds of lunchmeat the store start with = Number of pounds of lunch meat a small shop has at the end of Friday + Number of pounds of lunch meat a small shop on Friday morning + Number of pounds of lunch meat a small shop on Friday afternoon = 5.9 + 4.9 + 9.35 = 10.8 + 9.35 = 20.15.

Question 3. Use the digits 7, 3, 6, 5, 1, and 9 to write three numbers less than 600,000 but greater than 500,000. Use each digit only once for each number. Answer: Three numbers less than 600,000 but greater than 500,000. => 513679. => 531679. => 561379.

Question 4. Ms. Torres is on a road trip. She drives a total of 2,731.82 miles in three weeks. She drives 791.38 miles the second week. She drives 1,086.14 miles the third week. How many miles did she drive the first week? Answer: Number of miles She drives in the first week = 854.3.

Explanation: Number of miles She drives in three weeks = 2,731.82. Number of miles She drives in the second week = 791.38. Number of miles She drives in the third week = 1,086.14. Number of miles She drives in the first week = Number of miles She drives in three weeks – (Number of miles She drives in the second week + Number of miles She drives in the third week) = 2731.82 – (791.38 + 1086.14) = 2731.82 -1877.52 = 854.3.

Question 5. Mr. Kerry has 4 boxes of cocoa mix. Each box has 20 bags. He uses two cups of water and one bag of cocoa mix to make a mug of cocoa. How many cups of water will Mr Kerry use if he uses all the bags of cocoa mix? Answer: Number of cups of water if he uses all the bags of cocoa mix = 160.

Explanation: Number of boxes of cocoa mix Mr. Kerry has = 4. Number of bags each box has = 20. Number of cups of water he uses to make a mug of cocoa = 2. Number of bags he uses to make a mug of cocoa = 1. Total number of bags of cocoa Mr. Kerry has = Number of boxes of cocoa mix Mr. Kerry has Γ— Number of bags each box has = 4 Γ— 20 = 80. Number of cups of water if he uses all the bags of cocoa mix = Number of cups of water he uses to make a mug of cocoa Γ— Total number of bags of cocoa Mr. Kerry has = 2 Γ— 80 = 160.

Question 6. Ling is 56\(\frac{3}{4}\) in. tall. Lauren is 49\(\frac{1}{3}\) in. tall. How much taller is Ling than Lauren? Answer: 7.42 inches taller is Ling than Lauren.

Explanation: Number of inches height is Ling = 56\(\frac{3}{4}\) Number of inches height is Lauren = 49\(\frac{1}{3}\) Difference: Number of inches height is Ling – Number of inches height is Lauren = 56\(\frac{3}{4}\) – 49\(\frac{1}{3}\) = [(224 + 3) Γ· 4] – [(147 + 1) Γ· 3] = (227 Γ· 4) – (148 Γ· 3) = 56.75 – 49.33 = 7.42 inches.

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CCSS Math Answers

Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing

Redefine your true self with the Go Math Answer Key for Grade 5 curated by subject experts. Score higher grades in your exams and refer to Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing to have strong command over fundamentals. Download the HMH Go Math 5th Grade Solution Key Chapter 9 free of cost and kick start your preparation immediately.

You will get the necessary skillset needed to draw the line plots and graphs from 5th Grade Go Math Answer Key Ch 9. Access Detailed Solutions to all the problems and learn how to solve related problems when you encounter them during your exams. Seek Homework Help needed by accessing the Go Math Grade 5 Solution Key Chapter 9 Patterns and Graphing. Cross Check the Solutions from our Go Math Grade 5 Answer Key Algebra: Patterns and Graphing and understand the areas you are facing difficulty.

Lesson 1: Line Plot

Share and Show – Page No. 371

Unlock the problem – page no. 372.

Lesson 2: Ordered Pairs

Share and Show – Page No. 375

Problem solving – page no. 376.

Lesson 3: Investigate β€’ Graph Data

Share and Show – Page No. 379

Problem solving – page no. 380.

Lesson 4: Line Graphs

Share and Show – Page No. 383

Connect to science – page no. 384.

Mid-Chapter Checkpoint

Mid-Chapter Checkpoint – Vocabulary – Page No. 385

Mid-chapter checkpoint – page no. 386.

Lesson 5: Numerical Patterns

Share and Show – Page No. 389

Problem solving – page no. 390.

Lesson 6: Problem Solving β€’ Find a Rule

Share and Show – Page No. 393

On your own – page no. 394.

Lesson 7: Graph and Analyze Relationships

Share and Show – Page No. 397

Problem solving – page no. 398.

Chapter 9 Review/Test

  • Chapter Review/Test – Page No. 399

Chapter Review/Test – Page No. 400

Chapter review/test – page no. 401, chapter review/test – page no. 402.

Use the data to complete the line plot. Then answer the questions.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 1

Question 1. What is the combined mass of the beads with a mass of 1/5 gram? \(\frac{β–‘}{β–‘}\) grams

Answer: \(\frac{3}{5}\) grams

Explanation: For first we will count the number of \(\frac{1}{5}\) grams for each amount. Draw an x for the number of times each amount is recorded to complete the line plot. There are 3 xs above \(\frac{1}{5}\) on the line plot, so the combined mass of the beads is 3 fifths 3 Γ— \(\frac{1}{5}\) = 3/5 gram.

Question 2. What is the combined mass of all the beads with a mass of \(\frac{2}{5}\) gram? _____ grams

Explanation: For first we will count the number of \(\frac{2}{5}\) grams for each amount. Draw an x for the number of times each amount is recorded to complete the line plot. There are 5 xs above \(\frac{2}{5}\) on the line plot, so the combined mass of the beads is 5 two fifths. 5 Γ— \(\frac{2}{5}\) = 2 grams

Question 3. What is the combined mass of all the beads on the necklace? _____ grams

Explanation: Total mass of all the beads on the necklace is \(\frac{3}{5}\) + 2 + \(\frac{8}{5}\) + \(\frac{9}{5}\) = \(\frac{30}{5}\) = 6 Therefore the combined mass of all the beads on the necklace is 6.

Question 4. What is the average weight of the beads on the necklace? \(\frac{β–‘}{β–‘}\) grams

Answer: \(\frac{3}{7}\) grams

Explanation: Divide the sum by the number of beads to find the average. The number of beads = 3 + 5 + 3 + 2 = 14 Divide by 6. 6 Γ· 14 = 3/7 So, the average mass of the beads on the necklace is 3/7 gram.

On Your Own

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 2

Question 5. How much milk combined is used in \(\frac{1}{4}\)-cup amounts? \(\frac{β–‘}{β–‘}\) cup

Answer: \(\frac{1}{2}\) cup

Explanation: For first we will count the number of \(\frac{1}{4}\) cups for each amount. 2 Γ— \(\frac{1}{4}\) = \(\frac{1}{2}\)

Question 6. How much milk combined is used in \(\frac{1}{2}\)-cup amounts? ______ cups

Answer: 3 cups

Explanation: For first we will count the number of \(\frac{1}{2}\) cups for each amount. There are 6 \(\frac{1}{2}\) cups 6 Γ— \(\frac{1}{2}\) = 3 cups

Question 7. How much milk combined is used in \(\frac{3}{4}\)-cup amounts? _____ \(\frac{β–‘}{β–‘}\) cups

Answer: 1 \(\frac{1}{2}\) cups

Explanation: For first we will count the number of \(\frac{3}{4}\) cups for each amount. There are 2 \(\frac{3}{4}\) cups of milk. 2 Γ— \(\frac{3}{4}\) = \(\frac{3}{2}\) Convert from improper fraction to the mixed fraction. \(\frac{3}{2}\) = 1 \(\frac{1}{2}\) cups

Question 8. How much milk is used in all the orders of pancakes? _____ cups

Answer: 5 cups

Explanation: \(\frac{1}{2} c\) + [/latex]\frac{1}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] +[/latex]\frac{1}{4} c[/latex] + [/latex]\frac{1}{2} c[/latex] + [/latex]\frac{1}{2} c[/latex] = 3 + [/latex]\frac{1}{4} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{3}{4} c[/latex] + [/latex]\frac{1}{4} c[/latex] = 3 + 1 + 1 = 5cups Therefore 5 cups of milk is used in all the orders of pancakes.

Question 9. What is the average amount of milk used for an order of pancakes? \(\frac{β–‘}{β–‘}\) cup of milk

Answer: \(\frac{1}{2}\) cup of milk

Explanation: There are 6 \(\frac{1}{2}\) cups of milk. The average amount of milk used for an order of pancakes is \(\frac{1}{2}\) cup.

Question 10. Describe an amount you could add to the data that would make the average increase. Type below: _________

Answer: \(\frac{3}{4}\) cup We can add \(\frac{3}{4}\) to the data to increase the average amount of milk.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 3

Answer: I need to know the average amount of cat food that Dewey ate daily.

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Question 11. c. What steps could you use to find the average amount of food that Dewey ate daily? Type below: _________

Explanation: Number of days = 10 1/4 + 1/4 + 3/8 + 1/2 + 1/2 + 1/2 + 5/8 + 5/8 + 5/8 + 3/4 = 1 + 1 + 1/4 + 3/8 + 1/2 + 15/8 2 + 18/8 + 3/4 = 2 + 3 = 5 The average amount of food is 5 Γ· 10 = 5/10 = \(\frac{1}{2}\) cup

Question 11. d. Fill in the blanks for the totals of each amount measured. \(\frac{1}{4}\) cup: __________ \(\frac{3}{8}\) cup: __________ \(\frac{1}{2}\) cup: __________ \(\frac{5}{8}\) cup: __________ \(\frac{3}{4}\) cup: __________ Type below: _________

Answer: There are 2 xs above \(\frac{1}{4}\) cup: 2 There is 1 x above \(\frac{3}{8}\) cup: 1 There are 3 xs above \(\frac{1}{2}\) cup: 3 There are 3 xs above \(\frac{5}{8}\) cup: 3 There is 1 x above \(\frac{3}{4}\) cup: 1

Question 11. e. Find the total amount of cat food eaten over 10 days. _____ + _____ + _____ + _____ + _____ = _____ So, the average amount of food Dewey ate daily was ______. Type below: _________

Answer: Number of days = 10 1/4 + 1/4 + 3/8 + 1/2 + 1/2 + 1/2 + 5/8 + 5/8 + 5/8 + 3/4 = 1 + 1 + 1/4 + 3/8 + 1/2 + 15/8 2 + 18/8 + 3/4 = 2 + 3 = 5 cups

Question 12. Test Prep How many days did Dewey eat the least amount of cat food? Options: a. 1 day b. 2 day c. 3 day d. 4 day

Answer: 1 day By seeing the above line plot we can say that Dewey eats the least amount of cat food on day 1. Thus the correct answer is option A.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 5

Question 1. C( _____ , _____ )

Answer: 6, 3

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for C is (6, 3).

Question 2. D( _____ , _____ )

Answer: 3, 0

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. Thus the ordered pair for D is (3, 0)

Question 3. E( _____ , _____ )

Answer: 9, 9

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. Thus the ordered pair for E (9, 9)

Question 4. F( _____ , _____ )

Answer: 10, 5

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. Thus the ordered pair for F is (10, 5)

Plot and label the points on Coordinate Grid A.

Question 5. M (0, 9) Type below: _________

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Question 6. H (8, 6) Type below: _________

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Question 7. K (10, 4) Type below: _________

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Question 8. T (4, 5) Type below: _________

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Question 9. W (5, 10) Type below: _________

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Question 10. R (1, 3) Type below: _________

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Question 11. G( _____ , _____ )

Answer: 6, 4

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for G is (6, 4)

Question 12. H( _____ , _____ )

Answer: 4, 9

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for H is (4, 9)

Question 13. I( _____ , _____ )

Answer: 0, 7

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for I is (0, 7)

Question 14. J( _____ , _____ )

Answer: 9, 5

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for J is (9, 5)

Question 15. K( _____ , _____ )

Answer: 3, 3

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for K is (3, 3)

Question 16. L( _____ , _____ )

Answer: 5, 2

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for L is (5, 2)

Question 17. M( _____ , _____ )

Answer: 1, 1

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for M is (1, 1)

Question 18. N( _____ , _____ )

Answer: 2, 5

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for N is (2, 5)

Question 19. O( _____ , _____ )

Answer: 7, 8

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for O is (7, 8)

Question 20. P( _____ , _____ )

Answer: 10, 10

Explanation: Locate the point for which you want to write an ordered pair. Look below at the x-axis to identify the points horizontal distance from 0, which is its x-coordinate. Look to the left at the y-axis to identify the points vertical distance from 0, which is it’s y-coordinate. So, the ordered pair for P is (10, 10)

Plot and label the points on Coordinate Grid B.

Question 21. W (8, 2)

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Question 22. E (0, 4)

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Question 23. X (2, 9)

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Question 24. B (3, 4)

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Question 25. R (4, 0)

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Question 26. F (7, 6)

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Question 27. T (5, 7)

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Question 28. A (7, 1)

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Question 29. S (10, 8)

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Question 30. Y (1, 6)

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Question 31. Q (3, 8)

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Question 32. V (3, 1)

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Question 33. What ordered pair gives the location of Bryant Park? ( _____ , _____ )

Answer: 4, 8

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Question 34. What’s the Error? Nathan says that Madison Square Garden is located at (0, 3) on the map. Is his ordered pair correct? Explain. Type below: __________

Answer: He needs to put point 3 on Y-axis but he placed on X-Axis.

Question 35. The Empire State Building is located 5 blocks right and 1 block up from (0, 0). Write the ordered pair for this location. Plot and label a point for the Empire State Building. Type below: __________

Answer: 5, 1

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Question 36. Paulo walks from point B to Bryant Park. Raul walks from point B to Madison Square Garden. If they only walk along the grid lines, who walks farther? Explain. __________

Answer: Paulo By seeing the above graph we can say that Paulo walks farther along the grid lines.

Question 37. Explain how to find the distance between Bryant Park and a hot dog stand at the point (4, 2). _____ city blocks

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Question 38. Test Prep Use the map above. Suppose a pizzeria is located at point B. What ordered pair describes this point? Options: a. (4,2) b. (3,4) c. (2,4) d. (4,4)

Answer: (2,4)

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Question 1. a. Write the ordered pairs for each point. Type below: __________

Answer: A(1, 30), B (2, 35), C (3, 38), D (4, 41), E (5, 44)

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Question 1. b. What does the ordered pair (3, 38) tell you about Ryan’s age and height? Type below: __________

Answer: The ordered pair tells that the age of Ryan is 3 and height is 38 inches.

Question 1. c. Why would the point (6, 42) be nonsense? Type below: __________

Answer: The point (6, 42) be nonsense because the height will be increased. In the above-ordered pair the height is decreased. So, the statement is nonsense.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 10

Question 2. a. Write the ordered pairs for each point. Type below: __________

Answer: We can write the ordered pairs by using the above table Day is the x-axis and height is the y-axis. The coordinates are A (5,1), B (10,3), C (15, 8), D (20,12), E (25,16), F(30,19).

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Question 2. b. How would the ordered pairs be different if the heights of the plants were measured every 6 days for 30 days instead of every 5 days?

Answer: If the heights of the plants were measured every 6 days for 30 days instead of every 5 days the coordinates will be A (6,1), B (12,3), C (18, 8), D (24,12), E (30,16)

What’s the Error?

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 12

Question 1. What scale and intervals would be appropriate to make a graph of the data? Type below: __________

Answer: Scale is 1 cm = 10Β°F Months will be on the x-axis. The temperature will be on the y-axis.

Question 2. Write the related pairs as ordered pairs. Type below: __________

Answer: The related pairs are A (Jan, 40), B (Feb, 44), C (Mar, 54), D (Apr, 62), E (May, 70)

Question 3. Make a line graph of the data. Type below: __________

Question 4. Use the graph to determine between which two months the least change in average temperature occurs. Type below: __________

Answer: By seeing the above graph we can say that Jan and Feb has the least change in the average temperature.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 17

Question 5. Write the related number pairs for the plant height as ordered pairs. Type below: __________

Answer: The related number pairs of the above table are A (1, 20), B(2, 25), C (3, 29), D (4, 32)

Question 6. What scale and intervals would be appropriate to make a graph of the data? Type below: __________

Answer: The above table says that the X-Axis is Month and Y-Axis is Height in inches. Scale is 1 cm = 5 inches.

Explanation: The horizontal axis could represent months from 1 to 4. In this case, the scale interval is one month. The vertical axis could represent height from 20 inches to 32 inches but we can show a break in the scale between 1 inch and 16 inches since there are no heights between 0 inches and 20 inches, the scale interval is 1 inch.

Question 7. Make a line graph of the data. Type below: __________

Question 8. Use the graph to find the difference in height between Month 1 and Month 2. Type below: __________

Answer: By observing the above graph we can say that the difference between months 1 and 2 is 5 inches. 25 – 20 = 5 inches From the graph we can see that the plant grew the most between 1 and 2 months (about 5 inches), the least change is between 3 and 4 months (about 3 inches).

Question 9. Use the graph to estimate the height at 1 \(\frac{1}{2}\) months. _____ in.

Answer: The estimated height at 1 \(\frac{1}{2}\) months is 22.5 inches. The average of month 1 and month 2 is (20 + 25) Γ· 2 = 45/2 = 22.5 inches.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 18

Use the graph for 10–13.

Question 10. About how much precipitation falls in Redding, California, in February? _____ inches

Answer: From the graph, we can see that the precipitation in February is 4.2 inches.

Question 11. What is the average temperature for Redding, California, in February? _____ Β°F

Answer: From the graph, we can see that the temperature in February is 50Β°F.

Question 12. Explain how the overlay graph helps you relate precipitation and temperature for each month. Type below: __________

Answer: The average temperature for each month is plotted on the graph with the blue line and the red bar graph represents the precipitation. As the temperature increases the precipitation decreases.

Question 13. Describe how the average temperature changes in the first 5 months of the year. Type below: __________

Answer: From the graph, we can see that the temperature in the first 5 months of the year but the amount of precipitation is decreasing. It’s logical because when the temperature is increasing the amount of precipitation is decreasing.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 20

Answer: Day 5

Explanation: By seeing the above graph we can say that the snow level has increased 3 feet from day 4 to Day 5. Thus the correct answer is option C.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 21

Question 1. The ______ is the horizontal number line on the coordinate grid. __________

Answer: X-Axis The X-Axis is the horizontal number line on the coordinate grid.

Question 2. A ______ is a graph that uses line segments to show how data changes over time. __________

Answer: Line graph A Line graph is a graph that uses line segments to show how data changes over time.

Concepts and Skills

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 22

Question 3. How many kittens weigh at least \(\frac{3}{8}\) of a pound? ______ kittens

Explanation: The line plot shows that there are 4 xs above \(\frac{3}{8}\), 3 xs above \(\frac{1}{2}\) and 2 xs on \(\frac{5}{8}\). To find the kittens weigh at least \(\frac{3}{8}\) we need to add all above \(\frac{3}{8}\) = 4 + 3 + 2 = 9

Question 4. What is the combined weight of all the kittens? ______ lb

Explanation: There are 3 xs above 1/4 on the line plot, so the combined weight of Kitten in the animal shelter is 3 fourths 3 Γ— 1/4 = 3/4. There are 4 xs above 3/8 on the line plot, so the combined weight of kittens in the animal shelter is 4 three eights or 4 Γ— 3/8 = 12/8 = 3/2 There are 3 xs above 1/2 on the line plot, so the combined weight of kittens in the animal shelter is 3 halves = 3/2 There are 2 xs above 5/8 on the line plot, so the combined weight of kittens in the animal shelter is 10/8 3/4 + 1 4/8 + 1 1/2 + 1 2/8 = 3/4 + 12/8 + 3/2 + 10/8 = 6/8 + 12/8 + 12/8 + 10/8 = 40/8 = 5 lb

Question 5. What is the average weight of the kittens in the shelter? ______ lb

Answer: 5/12

Explanation: Divide the sum you found in example 4. 5 lb by the number of the kittens to find the average. The number of kittens is 12 so we will divide 5 lb by 12. 5 Γ· 12 = 5/12. Thus the average weight of the kittens in the shelter as 5/12 lb.

Use the coordinate grid at the right for 6–13.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 23

Question 6. A( ______ , ______ )

Answer: 1, 6 The ordered pair for A is (1,6)

Question 7. B( ______ , ______ )

Answer: 2, 2 The ordered pair for B is (2, 2)

Question 8. C( ______ , ______ )

Answer: 4, 4 The ordered pair for C is (4, 4)

Question 9. D( ______ , ______ )

Answer: 0, 3 The ordered pair for D is (0, 3)

Plot and label the point on the coordinate grid.

Question 10. E(6, 2) Type below: __________

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Question 11. F(5, 0) Type below: __________

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Question 12. G(3, 4) Type below: __________

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Question 13. H(3, 1) Type below: __________

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Question 14. Jane drew a point that was 1 unit to the right of the y-axis and 7 units above the x-axis. What is the ordered pair for this location? ( ______ , ______ )

Answer: (1, 7) The ordered pair for the location is (1, 7).

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Mid-Chapter Checkpoint img 24

Answer: 2 \(\frac{1}{6}\) miles

Explanation: There are 3 xs above \(\frac{1}{3}\) = 3 Γ— \(\frac{1}{3}\) = 1 There are 1 x above \(\frac{1}{2}\) = 1 Γ— \(\frac{1}{2}\) = \(\frac{1}{2}\) There is 1 x above \(\frac{2}{3}\) = 1 Γ— \(\frac{2}{3}\) = \(\frac{2}{3}\) 1 + \(\frac{2}{3}\) + \(\frac{1}{2}\) = (6 + 3 + 4)/6 = 13/6 The mixed fraction of 13/6 is 2 \(\frac{1}{6}\) miles Thus she walked 2 \(\frac{1}{6}\) miles in 5 days.

Use the given rules to complete each sequence. Then, complete the rule that describes how nickels are related to dimes.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 26

Answer: The number of Dimes is 2 times the number of Nickels. We need to add 5 to Nickels = 5 + 5 + 5 + 5 + 5 = 25 We need to add 10 to Dimes = 10 + 10 + 10 + 10 + 10 = 50

Complete the rule that describes how one sequence is related to the other. Use the rule to find the unknown term.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 27

Answer: The amount spent is 4 times the number of books so we multiply the number of books by 4 to find the amount spent. Multiply 4 to the amount spent = 24 Γ— 4 = 96

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 28

Answer: The weight of Bag is 3 times the number of marbles So, we divide the weight of Bag by 3 to find the number of marbles. Divide 360 by 3 360/3 = 120

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 29

Answer: The muffins is 6 times the number of eggs so we multiply the number of eggs by 6 to find the muffins. The unknown term in the table we will find when multiply 18 by 6. 18 Γ— 6 = 108 The unknown term is 108.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 30

Answer: The number of meters is 400 times the number of laps so we divide the number of meters by 400 to find the number of laps. The unknown term in the table we will find when divide 6400 by 400. 6400 Γ· 400 = 16 The unknown term is 16.

Question 6. Suppose the number of eggs used in Exercise 4 is changed to 3 eggs for each batch of 12 muffins, and 48 eggs are used. How many batches and how many muffins will be made? ______ batches ______ muffins

Answer: 16 batches 192 muffins will be made.

Explanation: If we change to 3 eggs for each batch of 12 muffins and 48 eggs are used we will have 16 batches. 16 Γ— 3 = 48 The muffins are 4 times the number of eggs so we multiply the number of eggs by 4 to fins the number of muffins. If the number of batches is 16 and there are 48 eggs to find the number of muffins we will multiply the number of eggs 48 with 4: 48 Γ— 4 = 192 192 muffins will be made.

Question 7. Emily has a road map with a key that shows an inch on the map equals 5 miles of actual distance. If a distance measured on the map is 12 inches, what is the actual distance? Write the rule you used to find the actual distance. ______ miles

Answer: 60 miles

Explanation: For first the total length of roads is 5 inches + 7 inches = 12 inches 1 inch on the map represents 5 miles of actual distance so to find what actual distance corresponding to 12 inches we will find with using proportion. 1 inch : 5 inches = 12 inches: x inches 1 Γ— x = 5 Γ— 12 x = 60 miles The actual distance which Emily will drive is 60 miles. The rule which we used to find the actual distance is multiplied by 5 which is a mark in solution.

Question 8. To make a shade of lavender paint, Jon mixes 4 ounces of red tint and 28 ounces of blue tint into one gallon of white paint. If 20 gallons of white paint and 80 ounces of red tint are used, how much blue tint should be added? Write a rule that you can use to find the amount of blue tint needed. ______ oz

Answer: 560 oz

Explanation: For one gallon of white paint, Jon mixed 28 ounces of blue tint so to find how much tint corresponding to 20 gallons of white paint we will find with using proportion. 1 gallon of white paint:28 ounces of blue tint = 20 gallons of white paint:x ounces of the blue tint. 1 Γ— x = 28 Γ— 20 x = 560 560 ounces of blue tint should be added. The rule which we used to find the amount of blue tint needed is to multiply by 28 which is the mark in solution.

Question 9. In the cafeteria, tables are arranged in groups of 4, with each table seating 8 students. How many students can sit at 10 groups of tables? Write the rule you used to find the number of students. ______ students

Answer: 320 students

Explanation: Tables are arranged in groups of 4, with each table seating 8 students, so in one group sit 4 Γ— 8 = 32 students To find how many students can sit at 10 groups of tables, we will find when multiplying 32 students with 10. 32 Γ— 10 = 320 Finally, 320 students can sit at 10 groups of tables. The rule which we used to find the number of students is to multiply by 32 which is marked is a solution.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 31

Answer: 105; Multiply by 3.

Explanation: The unknown number in Sequence number 7 we will get when multiply 35 with 3 because the rule that releases the number of miles to the number of runners is multiplying by 3. The unknown number is: 35 Γ— 3 = 105 Thus the correct answer is option C.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 32

Answer: Possible rule for posts: 27 Possible rule for rails: 54

Explanation: The number of posts is 3 times the number of sections. So, we multiply the number of posts by 3. With using the rule the unknown number is 9 Γ— 3 = 27 Thus the possible rule for posts is 27. Now multiply the number of rails by 2. With using the rule the unknown number is 27 Γ— 2 = 54 Thus the possible rule for rails is 54.

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Answer: 108 rails

Explanation: The number of posts is 3 times the number of sections. So, we multiply the number of posts by 3. With using the rule the unknown number is 9 Γ— 3 = 27 Thus the possible rule for posts is 27. Now multiply the number of rails by 4. With using the rule the unknown number is 27 Γ— 4 = 108 Thus the possible rule for rails is 108.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 37

Answer: $15

Explanation: Leslie is buying a coat on layaway for $135. She will pay $15 each week until the coat is paid for. Multiply the number of weeks by 15. 15 Γ— 8 = $120 Now subtract $120 from $135 = $135 – $120 = $15

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 38

Answer: 1000

Explanation: The possible rule for Hour Worked: We can see that the difference between terms is 2. So, the rule which describes this is Add 2. The possible rule for Jane’s Pay: We can see that the difference between terms is 50. So, the rule which describes this is Add 50. Jane’s Pay is 25 times the hours worked so, we will multiply the hours worked by 25 to find Jane’s Pay. The unknown number Jane’s Pay we will find when multiplying 40 with 25: 40 Γ— 25 = 1000 She earns 1000 dollars.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 39

Answer: 120

Explanation: Possible rule for Tokens: We can see that the difference between terms is 8. So, the rule which describes this is Add 8. Possible rule for Games: We can see that the difference between terms is 4. So, the rule which describes this is Add 4. Tokens are 2 times the games so, we will divide the tokens by 2 to find how many games can she3 play. The unknown number of games we will find when dividing 120 with 2: 120 Γ· 2 = 60 She can play 60 games for 120 tokens.

Question 6. Paul is taking a taxicab to a museum. The taxi driver charges a $3 fee plus $2 for each mile traveled. How much does the ride to the museum cost if it is 8 miles away?

Answer: $40

Explanation: Paul is taking a taxicab to a museum. The taxi driver charges a $3 fee plus $2 for each mile traveled. That means the driver charged $5 per mile. For 8 miles = 8 Γ— $5 = $40.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 40

Answer: 2 + 4 + 4 + 4

Explanation: We can see that the difference between two consecutive figures is 4 squares. So, the rule which describes this is Add 4. Thus figure 4 has 14 squares. Thus the correct answer is option C.

Graph and label the related number pairs as ordered pairs. Then complete and use the rule to find the unknown term.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 41

Sense or Nonsense?

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 49

Elsa said that George’s chili was hotter than Lou’s because the graph showed that the amount of hot sauce in George’s chili was always 3 times as great as the amount of hot sauce in Lou’s chili.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing img 51

Answer: Elsa’s Answer makes sense.

Explanation: Elsa’s answer makes sense because the amount of hot sauce in George’s chili was always 3 times as great as the amount of hot sauce in Lou’s chili. To prove this we will take two points from the graph which has an equal amount of cups of chili and compares the amount of hot sauce in George’s chili with the amount of hot sauce in Lou’s chili. If we take 4 cups of George’s chili and Lou’s chili the amount of hot sauce in George’s chili is 12 teaspoons and the amount of hot sauce in Lou’s chili is 4 teaspoons. 12 is 3 times greater than 4 so Elsa’s answer makes sense.

Chapter Review/Test – Vocabulary – Page No. 399

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 52

Question 1. The __________ is the point where the x-axis and y-axis meet. Its __________ is 0, and its __________ is 0. The ________ is the point where the x-axis and y-axis meet. Its ________ is 0, and its ________ is 0.

Answer: The Origin is the point where the x-axis and y-axis meet. Its x-coordinate is 0, and its y-coordinate is 0.

Question 2. A __________ uses line segments to show how data changes over time.

Answer: A line graph uses line segments to show how data changes over time.

Check Concepts

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 53

Question 3. Write related number pairs of data as ordered pairs. Type below: __________

Answer: The ordered pair for week 1 is (1, 2) The ordered pair for week 2 is (2, 6) The ordered pair for week 3 is (3, 14) The ordered pair for week 4 is (4, 16)

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 54

The ordered pair for week 1 is (1, 2) The ordered pair for week 2 is (2, 6) The ordered pair for week 3 is (3, 14) The ordered pair for week 4 is (4, 16)

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 55

Answer: Multiply the number of eggs by 6 to find the number of cupcakes. The unknown number in batches 6 we will get when multiply 18 with 6 because the rule that releases the number of eggs to the number of cupcakes is multiplying by 6. The number of eggs is multiple of 3 and the number of cupcakes is multiple of 6.

Fill in the bubble completely to show your answer.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 56

Answer: (0, 7) By seeing the above graph we can find the location of the hole label T i.e., (0, 7)

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 57

Question 7. What is the average of the data in the line plot? Options: a. \(\frac{1}{2}\) pound b. 1 pound c. 6 pounds d. 6 \(\frac{3}{4}\) pounds

Answer: 6 pounds

Explanation: There are 3 xs above \(\frac{1}{2}\) pound = 3 Γ— \(\frac{1}{2}\) = 3/2 There are 4 xs above \(\frac{2}{3}\) pound = 4 Γ— \(\frac{2}{3}\) = 8/3 There is 1 x above \(\frac{5}{6}\) pound = 5/6 There are 2 xs above \(\frac{1}{6}\) = 2/6 There are 2 xs above \(\frac{1}{3}\) = 2/3 3/2 + 8/3 + 5/6 + 2/6 + 2/3 = 6 pounds Thus the correct answer is option C.

Question 8. How many bags of rice weigh at least \(\frac{1}{2}\) pound? Options: a. 2 b. 3 c. 5 d. 8

Explanation: By seeing the above line plot we can find the number of bags of rice weigh at least \(\frac{1}{2}\) pound There are 3 xs above \(\frac{1}{2}\) pound = 3 Γ— \(\frac{1}{2}\) = 3/2 There are 4 xs above \(\frac{2}{3}\) pound = 4 Γ— \(\frac{2}{3}\) = 8/3 There is 1 x above \(\frac{5}{6}\) pound = 5/6 Total number of bags of rice weigh at least \(\frac{1}{2}\) pound = 3 + 4 + 1 = 8 Thus the correct answer is option D.

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 58

Question 9. Compare Tori’s and Martin’s savings. Which of the following statements is true? Options: a. Tori saves 4 times as much per week as Martin. b. Tori will always have exactly $15 more in savings than Martin has. c. Tori will save 15 times as much as Martin will. d. On week 5, Martin will have $30 and Tori will have $90.

Answer: Tori saves 4 times as much per week as Martin.

Explanation: By seeing the above table we can say that Tori saves 4 times as much per week as Martin. Thus the correct answer is option A.

Question 10. What rule could you use to find Tori’s savings after 10 weeks? Options: a. Add 10 from one week to the next. b. Multiply the week by 2. c. Multiply Martin’s savings by 4. d. Divide Martin’s savings by 4.

Answer: Multiply Martin’s savings by 4.

Explanation: We can find the savings of Tori by multiplying the savings of Martins by 4. Thus the suitable statement is Multiply Martin’s savings by 4. Therefore the correct answer is option C.

Question 11. In an ordered pair, the x-coordinate represents the number of hexagons and the y-coordinate represents the total number of sides. If the x-coordinate is 7, what is the y-coordinate? Options: a. 6 b. 7 c. 13 d. 42

Explanation: Given that x-coordinate represents the number of hexagons. Thus x-coordinate is 6. And also given that the y-coordinate represents the number of sides. The figure hexagon contains 6 sides. So, the y-coordinate is 6. Thus the ordered pair is (7, 6) Therefore the correct answer is option A.

Question 12. Point A is 2 units to the right and 4 units up from the origin. What ordered pair describes point A? Options: a. (2, 0) b. (2, 4) c. (4, 2) d. (0, 4)

Answer: (2, 4)

Explanation: Point A is 2 units to the right and 4 units up from the origin. 2 units will be located on the x-axis and 4 units will be on the y-axis. Thus the ordered pair for point A is (2, 4) Therefore the correct answer is option B.

Constructed Response

Question 13. Mr. Stevens drives 110 miles in 2 hours, 165 miles in 3 hours, and 220 miles in 4 hours. How many miles will he drive in 5 hours? Explain how the number of hours he drives is related to the number of miles he drives. _____ miles

Answer: 275 miles

Explanation: Given that, Mr. Stevens drives 110 miles in 2 hours, 165 miles in 3 hours, and 220 miles in 4 hours. We have to divide the number of miles by number of hours That means, 110/2, 165/3, 220/4 the distance gone in 5 hours can be found with this equation 110/2 x ?/5 multiply 110 by 5 then divide the product by 2 110 Γ— 5= 550 550/2 =275 Thus the answer is Mr. Stevens goes 275 miles in 5 hr.

Performance Task

Go Math Grade 5 Answer Key Chapter 9 Algebra Patterns and Graphing Chapter Review/Test img 59

Question 14. B). Use the graph to estimate the temperature at 7 minutes. Estimate: _____ Β°F

Answer: By seeing the above graph we can say that the estimated temperature at 7 minutes is 15Β°F.

Question 14. C). Write a question that can be answered by making a prediction. Then answer your question and explain how you made your prediction. Type below: __________

Question: Estimate the temperature at 5 minutes by using the graph. Answer: By seeing the above table we can say that the estimated temperature at 5 minutes is 13Β°F

Fall in love with Maths by utilizing the Go Math 5th Standard 5 Answer Key. Make use of the Go Math Grade 5 Answer Key Chapter 9 Algebra: Patterns and Graphing as a reference for all your queries. Keep in touch with our site to avail updates on Class Specific Go Math Answer Key at your fingertips.

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