$\dfrac{3xyz^2}{6y^4}=\dfrac{3}{6}\cdot\dfrac{xyz^2}{y^4}=\dfrac{1}{2}\cdot\dfrac{xz}{y^3}=\dfrac{xz}{2y^3}\\\\\\\dfrac{3xyz^2}{6y^4}\cdot\dfrac{2y}{xz^4}=\dfrac{xz^2}{2y^3}\cdot\dfrac{2y}{xz^4}=\dfrac{2}{2}\cdot\dfrac{xyz^2}{xy^3z^4}=1\cdot\dfrac{1}{y^2z^2}=\dfrac{1}{y^2z^2}$

8 0

3 years ago

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Nine years less than 5 times the age a

hichkok12 [17]

5a-9
I think this is the answer I am a little unsure of what the question is.

4 0

3 years ago

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Consider a diving board that is 10 feet above a pool. If the ladder is 5 feet away from the base of the diving board, approximat

OleMash [197]

Answer: 15 feet

Step-by-step explanation:

From the question, we are informed that a diving board is 10 feet above a pool. Since the ladder is 5 feet away from the base of the diving board, to calculate how tall is the ladder is, we subtract -5 from 10. This will be:

= 10 - (-5)

= 10 + 5

= 15 feet

The answer is option C.

5 0

3 years ago