Asked by cal
I am also having trouble with this one, thanks!
If y varies jointly as x and the cube root of z, and y=120 when x=3 and z=8, find y when x=4 and z=27.
If y varies jointly as x and the cube root of z, and y=120 when x=3 and z=8, find y when x=4 and z=27.
Answers
Answered by
bobpursley
y=k*x*cubroot z
120=k*3*2 find k, then put that k in to
y=k*x*cubroot z to solve the last.
120=k*3*2 find k, then put that k in to
y=k*x*cubroot z to solve the last.
Answered by
Anonymous
Y = K * X * ∛Z
120= K (3)(2)...CUBE ROOT OF 8 IS 2
120= 6 K
120/6 =K
20= K
Y= (20)(4)(3)...CUBE ROOT OF 27 IS 3
Y= 240
120= K (3)(2)...CUBE ROOT OF 8 IS 2
120= 6 K
120/6 =K
20= K
Y= (20)(4)(3)...CUBE ROOT OF 27 IS 3
Y= 240
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.