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4vir4ik [10]
3 years ago
11

The coldest temperature in the city today was 14.8 degrees (Fahrenheit) lower than the hottest temperature. Let H represent the

high
temperature and let L represent the low temperature.
i) What is a formula that correctly relates H and L?
ii) If the low temperature was 68.9 degrees, what was the high temperature?
i) L = H + 14.8
ii) 54.1 degree

i) H = L + 14.8
ii) 83.7 degrees

i) H = L + 14.8
ii) 54.1 degrees

i) L = H + 14.8
ii) 83.7 degrees
Mathematics
1 answer:
devlian [24]3 years ago
6 0
<h3>The formula that correctly relates H and L is H = L + 14.8</h3><h3>The high temperature is 83.7 degrees</h3>

<em><u>Solution:</u></em>

Let H represent the high  temperature

Let L represent the low temperature

Given that,

<em><u>The coldest temperature in the city today was 14.8 degrees (Fahrenheit) lower than the hottest temperature</u></em>

L = H - 14.8

Rearrange,

H = L + 14.8

<em><u>If the low temperature was 68.9 degrees, what was the high temperature?</u></em>

L = 68.9

Then,

H = 68.9 + 14.8

H = 83.7

Thus the high temperature is 83.7 degrees

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Answer:

The answer is

<h2>y =  -  \frac{2}{3} x + 5</h2>

Step-by-step explanation:

Equation of a line is y = mx + c

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To find the equation we must first find the slope of the line.

So the slope of the line using points (-3, 7) and (9,-1) is

<h3>m =  \frac{ - 1 - 7}{9 + 3}  =  -  \frac{8}{12}  =  -  \frac{2}{3}</h3>

Now we use the formula

<h3>y - y1 = m(x - x1)</h3>

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So the equation of the line using point

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<h3>y - 7 =  -  \frac{2}{3} (x + 3) \\ y - 7 =  -  \frac{2}{3} x - 2 \\ y =  -  \frac{2}{3} x - 2 + 7</h3>

We have the final answer as

<h3>y =  -  \frac{2}{3} x + 5</h3>

Hope this helps you

5 0
3 years ago
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Answer:

C

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. y'=(3\sin x)'=3\cos x. Differentiate again to get y''=(3\cos x)'=-3\sin x, then y''+y=-3\sin x+3\sin x=0\neq 3\sin x so this choice of y doesn't solve the equation.

B. y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x and y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x, then y''+y=10\sin x+6\cos x\neq 3\sin x so y is not a solution

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D.y'=-3\sin x and y''=-3\cos x, then y''+y=0 thus y isn't a solution

E. y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x hence y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x. Then y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x then y is not a solution.

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