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kenny6666 [7]
1 year ago
10

What values of c and d make the equation true? RootIndex 3 StartRoot 162 x Superscript c Baseline y Superscript 5 Baseline EndRo

ot = 3 x squared y (RootIndex 3 StartRoot 6 y Superscript d Baseline EndRoot) c = 2, d = 2 c = 2, d = 4 c = 6, d = 2 c = 6, d = 4
Mathematics
2 answers:
bonufazy [111]1 year ago
7 0

Answer:

its c

Step-by-step explanation:

Reil [10]1 year ago
3 0

Answer:

<em>c=6, d=2</em>

Step-by-step explanation:

<em>Equations </em>

We must find the values of c and d that make the below equation be true

\sqrt[3]{162x^cy^5}=3x^2y \sqrt[3]{6y^d}

Let's cube both sides of the equation:

\left (\sqrt[3]{162x^cy^5}\right )^3=\left (3x^2y \sqrt[3]{6y^d}\right)^3

The left side just simplifies the cubic root with the cube:

162x^cy^5=\left (3x^2y \sqrt[3]{6y^d}\right)^3

On the right side, we'll simplify the cubic root where possible and power what's outside of the root:

162x^cy^5=3^3x^6y^3 (6y^d)

Simplifying

x^cy^5=x^6y^{3+d}

Equating the powers of x and y separately we find

c=6

5=3+d

d=2

The values are

\boxed{c=6,d=2}

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