$\bf \begin{cases}
3=a\cdot b^0\\
75=a\cdot b^2
\end{cases}\qquad b^0=1\implies 
\begin{cases}
3=a\cdot 1\implies 3=a\\
----------\\
75=a\cdot b^2\implies 75=3\cdot b^2\\\\
\frac{75}{3}=b^2\implies 25=b^2\\\\
\pm \sqrt{25}=b\implies \pm 5 = b
\end{cases}$

recall, anything raised to the 0, zero, power, is 1, with the sole exception of 0

8 0

4 years ago

CAN SOMEONE PLEASE HELP ME!!

QveST [7]

<h2>A) Wrong</h2><h2>B) Wrong</h2><h2>C) Right </h2><h2>D) Right</h2><h2>C and D is right because when you solve the problem it ends up as - 12 - 6f. </h2><h2>But to me it is mostly C. </h2><h2>I hope this helps. </h2><h2>I tried my best </h2><h2><u>Have a great dayyyy!!! :) </u></h2>

3 0

3 years ago

A mountain climber started at 150 feet below sea level. What would he

Ber [7]

Answer:

3150 feet for both

Step-by-step explanation:

correct?

4 0

3 years ago

Please help me 13 points​

Please help me 13 points