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2 years ago

The radii of two similar cones are 1 and 2.5. What is the ratio of their volumes?

Mathematics

1 answer:

algol132 years ago

6 0

Answer:

Assuming the cones are the same height, then

V1 = pi * r^2 * h/3 = 3.14159*1^2*h/3

V2 = 3.14159*2.5^2*h/3 = 3.14159*6.25*h/3

V1/V2=(3.14159*h/3)/(3.14159*6.25)*h/3

V1/V2 = 1/6.25 = .16

V2/V1=(3.14159*6.25)*h/3/(3.14159*h/3)

V2/V1=6.25/1 = 6.25

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A store is having a sale on walnuts and chocolate chips. For 7 pounds of walnuts and 9 pounds of chocolate chips, the total cost

Jlenok [28]

Answer:

Walnuts cost $1.75, chocolate chips cost $2.75

Step-by-step explanation:

<h3>Let the costs be:</h3>

Walnuts - x
Chocolate chips - y

<h3>Set equations as per question</h3>

For 7 pounds of walnuts and 9 pounds of chocolate chips, the total cost is $37:

7x + 9y = 37

For 5 pounds of walnuts and 3 pounds of chocolate chips, the total cost is $17:

5x + 3y = 17

<h3>Solve the system by elimination</h3>

Multiply the second equation by 3 and subtract the first equation, solve for x:

3(5x + 3y) - (7x + 9y) = 3(17) - 37
15x + 9y - 7x - 9y = 51 - 37
8x = 14
x = 14/8
x = 1.75

Find the value of y:

7*1.75 + 9y = 37
12.25 + 9y = 37
9y = 37 - 12.25
9y = 24.75
y = 24.75/9
y = 2.75

6 0

2 years ago

PLEASE HELP ME!! 1.02: Triangle Similarity 1<br><br> What is the value of X?

hichkok12 [17]

Answer:

Step-by-step explanation:

ΔARP and ΔCRD are similar. Therefore the sides are in proportion:

$\dfrac{AR}{CR}=\dfrac{PR}{DR}$

We have:

AR = 10 + x

CR = x

PR = 15 + 42 = 57

DR = 42

Substitute:

$\dfrac{10+x}{x}=\dfrac{57}{42}$ <em>cross multiply</em>

$42(10+x)=57x$ <em>use distributive property</em>

$(42)(10)+(42)(x)=57x$

$420+42x=57x$ <em>subtract 42x from both sides</em>

$420=15x$ <em>divide both sides by 15</em>

$x=28$

6 0

3 years ago

PLEASE HELP TEST IS TIMED

DaniilM [7]

45.5

hope it help thanks

6 0

3 years ago

10 POINTS!! PLEASE HELP ASAP!!

Ber [7]

If the polygons are similar, then we can set up a similarity ratio using sides AB and EF and any other 2 sides where both side lengths are given. That appears to be true for sides BC and FG. Our similarity ratio would be this then, going from the bigger polygon stuff on top and the smaller polygon stuff on bottom:

$\frac{8}{6}=\frac{6}{y}$ .