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julsineya [31]
3 years ago
15

I don't know how to do this problem

Mathematics
1 answer:
Colt1911 [192]3 years ago
3 0
A. 180
b. 230
c. I'm not sure what that on is
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The answer is 3/4..............
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Please help with my school
worty [1.4K]

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You didn't send the pool dimensions ill answer it when u do

Step-by-step explanation:

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Month 1: 93 in
Elodia [21]

Answer:

9.5 inches per month

Step-by-step explanation:

Given the information about the depth of the pond:

Month 1:     93 in

Month 2:    98 in

Month 3:    112 in

From month 1 to month 3, the average rate of change of the depth of the pond, in inches per month can be calculated using formula

\dfrac{\text{Depth}_3-\text{Depth}_1}{3-1}

Hence, from month 1 to month 3, what is the average rate of change of the depth of the pond, in inches per month is

\dfrac{112-93}{3-1}=\dfrac{19}{2}=9.5\ in/month

4 0
3 years ago
1. Ruth Lockwood earns $2,917 a month as a supervisor for Acme
soldi70 [24.7K]

Answer:

a. 25%

b. $1560

c. $130

Step-by-step explanation:

a.

If 75% is paid by company, the rest is paid by Ruth.

Out of 100%, 75% is paid by company, so Ruth pays 100 - 75 = 25%

b.

Per year, the total amount paid is 6240, but Ruth pas 25% of this. So,

25% = 0.25

To get the answer, we multiply 0.25 with 6240.

0.25 * 6240 = 1560

So, $1560 is paid by Ruth annually.

c.

Since 1560 is paid annually (in 12 months), each month Ruth has to pay:

1560/12 = $130

Hence, $130 is deducted each month from her paycheck

6 0
3 years ago
Express the complex number in trigonometric form. -6 + 6 square root of 3 i ?
bagirrra123 [75]
\bf \begin{array}{clclll}
-6&+&6\sqrt{3}\ i\\
\uparrow &&\uparrow \\
a&&b
\end{array}\qquad 
\begin{cases}
r=\sqrt{a^2+b^2}\\
\theta =tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\
-------------------------------\\\\

\bf r=\sqrt{(-6)^2+(6\sqrt{3})^2}\implies r=\sqrt{36+(6^2\cdot 3)}\implies r=\sqrt{144}
\\\\\\
r=12\leftarrow 
\\\\\\
\theta =tan^{-1}\left( \frac{6\sqrt{3}}{-6} \right)\implies \theta =tan^{-1}\left( -\sqrt{3}\right)\implies \theta =
\begin{cases}
\frac{2\pi }{3}\leftarrow \\
\frac{5\pi }{3}
\end{cases}

now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.

thus         \bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]
5 0
3 years ago
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