Question

What values of c and d make the equation true? Assume x>0 and y >=0.

square root of 50x^6y^3/9x^8 = 5y^c square root of 2y/dx.


A.c = 1, d = 3

B.c = 1, d = 32

C.c = 2, d = 8

D.c = 2, d = 32

Answer 1

Answer:

A. c=1, d=3

Step-by-step explanation:

If x>0 and y>0, then

$\sqrt{\dfrac{50x^6y^3}{9x^8}}=\sqrt{\dfrac{25\cdot 2y^2\cdot y}{9x^2}}=\dfrac{5y\sqrt{2y}}{3x}.$

If

$\dfrac{5y\sqrt{2y}}{3x}$

is equal to

$\dfrac{5y^c\sqrt{2y}}{dx},$

then

$y=y^c\Rightarrow c=1,\\ \\3x=dx\Rightarrow d=3.$