Question

What would the radius be of a soccer ball that has a volume of 523.6 in3?


4 in


6 in


5 in


4.5 in

Answer 1

use Volume of Sphere formula...\[V=\frac{4}{3}\pi r^3\]

Answer 2

ik that, but i dont know how i would set that up because it says the volume is 523.6 so can you show me how i would set that up?

Answer 3

yes i can

Answer 4

Volume V = 523.6 in^3 is given... then you can solve the radius r...

Answer 5

oops so its 523.6 = 4/3 pi r ^3

Answer 6

thats what i did before, and i didnt get any of those answers up there.

Answer 7

there is an answer...

Answer 8

Okay so lets go through step for step

Answer 9

@fionacndg you can continue...

Answer 10

so what we need to do is isolate r^2

Answer 11

okay so what I would do first is to change 4/3 to a regular number

Answer 12

like?

Answer 13

wait this is the wrong formula.... try this
r = cuberoot(3V / 4π)

Answer 14

Okay the answer is 5 I have no idea why my computer is spazing so much sorry.the keys arent working sorry_))))__)))..

Answer 15

its fine thank you :)

Answer 16

do u think u could help me with this problem??

Answer 17

If point A is located at the coordinate (1, 2). Find the coordinate for point B if the midpoint of the two coordinates is located at (5, -1).


(3, 0.5)


(-3, 5)


(8, 3)


(9, -4)

Answer 18

the line joining A and B... includes the midpoint (5, -1)... they have a common slope... the slope m for A and midpoint is...\[m=\frac{rise~(change~ in~ y)}{run~(change~in~x)}=\frac{y_2-y_1}{x_2-x_1}=\frac{-1-2}{5-1}=\frac{-3}{4}=-\frac{3}{4}\]Using Point-Slope form,\[(y-y_1)=m(x-x_1)\]using A (1,2), we have \[(y-2)=-\frac{3}{4}(x-1)=\frac{-3x+3}{4}\]\[4(y-2)=-3x+3\]\[4y-8=-3x+3\]\[3x+4y=11~~~~(1)\]Another formula is the Distance Formula between two points...\[d=\sqrt{(x_m-x_A)^2+(y_m-y_A)^2}=\sqrt{(5-1)^2+(-1-2)^2}\]\[d=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5~units\]This is the same distance between the midpoint (5,-1) and the unkown point B... using the same formula as distance formula...\[5=\sqrt{(x-5)^2+(y-(-1))^2}\]\[5^2=(x-5)^2+(y+1)^2=25~~~~(2)\]denotes equation of a circle with center at midpoint (5,-1) and radius r=5... the segment AB is the diameter of the said circle, and we can say that both points A and B lie in the circle of equation 2, or line of equation 1 intersect circle of equation 2 at points A, B and center point of the circle.
We can substitute equation 1 to equation 2... then we can solve point A (1,2) again and the other point B (x,y)...