Question

Find the volume of the cone. (Round to the nearest tenth.)
Answer

1250.7 mm

2014.1 mm3

289.4 mm3

1229.4 mm3

Answer 1

The formula you should use is \[V_{cone}=\frac{1}{3}\pi r^2 h\]

Answer 2

radius is not clear

Answer 3

i think it's 38

Answer 4

There's a curvy arrow beside the 38, maybe it's circumference of the base?

Answer 5

is 38 the radius ?

Answer 6

yes

Answer 7

So \(C=2 \pi r\) first, and then the formula above, assuming 38 is the circumference. That would make the drawing more to scale as well.

Answer 8

so whats the answer

Answer 9

What part did you get stuck on?

Answer 10

finding the answer

Answer 11

Your first task is to find the radius. If you're given a circumference of 38, then you can find the radius with \(C=2 \pi r\).

Answer 12

238.64

Answer 13

but thats not an answer

Answer 14

Hrm, my radius was much smaller than that. Could you post your calculation?

Answer 15

2x3.14x38

Answer 16

2*3.14*r = 38

Answer 17

r = 38/6.28

Answer 18

Where did that 14 come from?

Perhaps we should rearrange the equation first \[C=2 \pi r <=> r=\frac{C}{2\pi}\]

Answer 19

?

Answer 20

If \[r=\frac{C}{2\pi}\] you just need to put 38 in for C, and find the value of the fraction.

Answer 21

59.66?

Answer 22

So you took 38, and divided it by 2\(\pi\), and got 59.66?

Answer 23

yes

Answer 24

but its wrong

Answer 25

im getting really confused

Answer 26

The value of 2\(\pi\) is about 6.28, so if you divide 38 by 6.28, the number you get out should be smaller than 38.

Answer 27

19

Answer 28

nope

Answer 29

That's 38 divided by 2. Divide that by \(\pi\)

Answer 30

6.05

Answer 31

yup

Answer 32

Alright, now we're getting somewhere! Now you need to use the equation for the volume of a cone: \[V=\frac{1}{3}\pi r^2 h\]

Answer 33

344.79

Answer 34

yea

Answer 35

so how do we get the answer

Answer 36

1/3*3.14*(6.05)^2*9

Answer 37

but thats nott the answer

Answer 38

344.79 should be the answer, but for some reason it's not on the list you gave. I'm quite stumped.