Question

The variable z varies jointly as the square root of x and the cube root of y. If z=108.0 when x=100 and y=64, find z when x=9 and y=27.

Answer 1

How do I set this up and do it?

Answer 2

@ganeshie8
@jim_thompson5910
@phi
@thomaster

Answer 3

@hartnn
@Luigi0210
@Preetha
@bibby

Answer 4

@Destinymasha
@kropot72
@ranga
@beccaboo333
@sourwing
@someonepls

Answer 5

So, I've personally never heard it said this way, but "Z varies jointly with..." implies:
\[Z = C \sqrt{x}\sqrt[3]{y}\]
where C is some constant.

We are told that when x = 100 and y = 64, that z = 108. We plug these values in and solve for C.

We can then find Z for the given x and y.

Answer 6

So
\[108=C \sqrt{100}\sqrt[3]{64}\]

Answer 7

Right

Answer 8

So that's 100=C(40)
Which is C=2.5, amirite?

Answer 9

108 = 40 C

C = 2.7

Answer 10

Oh my bad, I wrote it incorrectly

Answer 11

So
\[z=2.7\sqrt{9}\sqrt[3]{27}\]

Answer 12

Right

Answer 13

z=24.3?

Answer 14

Yep :)

Answer 15

Thanks! Can you help me with another one?

Answer 16

Sure

Answer 17

Give me a bit to write it down

Answer 18

The kinetic energy K of a moving object varies jointly with its mass m and the square of the velocity v. If an object weighing 17 kilograms and moving with a velocity of 10 meters per second has a kinetic energy of 850 Joules, find its kinetic energy when the velocity is 22 meters per second.

Answer 19

There we go. Let me try to set it up real quick to see if I learned anything :D

Answer 20

\[K=Cm \sqrt{v}\]

Answer 21

Yes?

Answer 22

Close. It varies with the SQUARE of the velocity. Not the square root.

Answer 23

Ohh
\[K=Cmv ^{2}\]

Answer 24

Yep!

Answer 25

:)
So
\[850=C(17)(10^{2})\]

Answer 26

Yep

Answer 27

850=C(170)
850/170=C
C=5

Answer 28

10 * 10 = 100

100* 17 = 1700

Answer 29

My god, it's always the silly little mistakes that get me

Answer 30

Haha, I know the feeling :)

Answer 31

850/1700=0.5

Answer 32

Soooooo

\[K=0.5(17)(22^{2})\]

Answer 33

K=0.5(8228)

Answer 34

K=4114 is my final answer

Answer 35

That's it :)

Answer 36

Thank you so much! I got these last two questions right

Answer 37

Outstanding!

Any time :)