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Neporo4naja [7]
4 years ago
11

A manufacturer uses a tank to supply water to his machines while they are working. When the machines are on the tank loses 1.5 g

allons of water an hour at the same time 1 gallon of water is pumped into the tank from a well. Write an equation that models the amount of water (w) in the tank for any hour (h) that the machines are on. The tank starts full with 150 gallons of water in it.
Mathematics
1 answer:
shepuryov [24]4 years ago
3 0
OK. The first step is to find the amount the tank is losing per hour by balancing its gains and losses. Since it is gaining one gallon per hour but losing 1.5 gallons per hour, it is overall losing .5 gallons an hour. If there was originally 150 gallons in the tank, you can use the equation w=-.5h+150 to find the amount of water after h hours. 
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Answer:

D) 31 child tickets and 49 adult tickets

Step-by-step explanation:

6 0
3 years ago
What are the steps I need to do to get the answer?
Margaret [11]

Answer:

B. 10.9

Step-by-step explanation:

Area of triangle= 1/2*24*h

∠N=90° ⇒ Area of triangle= 1/2*MN*ON

MN²= h²+(24-7)²

ON²= h²+7²

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7 0
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5 0
3 years ago
Read 2 more answers
I need help with finding the answer to a) and b). Thank you!
shtirl [24]

Answer:

\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }

\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}

\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}

Step-by-step explanation:

We are given that:

\displaystyle \sin(x)=-\frac{21}{29}

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

a=\sqrt{29^2-21^2}=20

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

180

So:

\displaystyle 90 < \frac{x}{2} < 135

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 +  20/29}{2}

Simplify:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}

2)

Likewise:

\displaystyle  \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

\displaystyle  \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }

Simplify:

\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}

3)

Finally:

\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}

Therefore:

\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}

5 0
3 years ago
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