430,110 views
40 votes
40 votes
What values of c and d make the equation true? Assume x>0 and 20.

3
50x6y³ 5y √√2y
=
O c = 1, d = 3
O c = 1, d = 9
O c = 2, d = 8
O c = 2, d = 9

User Vijay Madhavapeddi
by
2.8k points

1 Answer

17 votes
17 votes

Answer:

c=1 and c=3

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}.

9x

8

50x

6

y

3

=

9x

2

25⋅2y

2

⋅y

=

3x

5y

2y

.

If

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

3x

5y

2y

is equal to

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

dx

5y

c

2y

,

then

\begin{gathered}y=y^c\Rightarrow c=1,\\ \\3x=dx\Rightarrow d=3.\end{gathered}

y=y

c

⇒c=1,

3x=dx⇒d=3.

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.