Question

SOMEONE HELP PLEASEEEE

Suppose the volume of a given object varies directly as the cube of the radius of its base. If one example of the object with radius 4.5 feet has a volume of 127.2 ft3, what is the volume of a second example with radius 2.5 feet?

Answer 1

use proportions

Answer 2

I think it's 57.24

Answer 3

A. 70.67 feet3
B. 42.27 feet3
C. 39.25 feet3
D. 21.81 feet3

Answer 4

oh

Answer 5

>.< mhm im stuck

Answer 6

volume varies directly as the cube of the radius
that means
V= k*r^3 (k is a number)
they give us V and r^3 for one of the objects, so we can find k

127.2= k*(4.5)^3

can you find k?

Answer 7

no lol im stuck. >.<

Answer 8

do you have a calculator?
can you find 4.5^3 (4.5 cubed, which means 4.5*4.5*4.5)

Answer 9

91.125

Answer 10

127.2= k*(4.5)^3

can you rewrite this equation, replacing 4.5^3 with its value (the number you found)

Answer 11

127.2=k*(91.125) ?

Answer 12

yes. now to "solve for k".

divide both sides by the number multiplying k:
127.2/91.125= k*91.125/91.125

on the right side, any number divided by itself is 1. Can you rewrite the equation
(do the division on both sides)

Answer 13

1.39 and a whole bunch more number ?

Answer 14

yes, but write it as an equation. start with
127.2/91.125= k*91.125/91.125

what do we get ?

Answer 15

1.39=k ?

Answer 16

yes.
so k = 1.39 (roughly, because we dropped some numbers)
but let's see what we get when we use it.

V= k*r^3 (volume varies directly as the radius cubed is how you interpret this)
using k = 1.39 we have

V= 1.33*r^3

second example with radius 2.5 feet, replace r in the equation with the number 2.5
what do we get? write down the equation

Answer 17

*1.39 not 1.33

Answer 18

V=21.71

Answer 19

which should be close to one of your choices. If you recompute k with say 4 decimals, the answer will match

Answer 20

21.81?

Answer 21

yes.
if you used k= 1.3959 (rounded)
you would get
2.81 (rounded to the nearest 100th)

Answer 22

okay thanks hun <3

Answer 23

if you want more info on how to do the algebra, try khan's site:
http://www.khanacademy.org/math/algebra/solving-linear-equations/v/simple-equations

it is not too hard to learn this ...