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Gekata [30.6K]
3 years ago
10

the temperature in an experiment is increased at a constant rate over a period of time until the temperature reaches 0 degrees c

elsius. the equation y=5/2x-70 gives the temperature y in degrees celsius x hours after the experiment begins. Find and interpret the x- and y- intercepts.
Mathematics
2 answers:
dsp733 years ago
7 0

Answer:

x = 28 (So it takes 28 hours for the temperature to reach 0°C.)

y = -70 (So the temperature when the experiment begins is −70°C.)

Step-by-step explanation:

To find the the x value, you remove the y value or replace it with 0.

0 = 5/2x - 70

Next you move the variable.

70 (2/5) = x           (multiply 70 by 2 to get 140 and divide that by 5)

And now you simplify to get:

28

To find the y value, you remove the x value or replace it with 0.

y =  0 - 70

And now you simplify to get:

-70

x = 28

y = -70

Karo-lina-s [1.5K]3 years ago
5 0

x = 28 (So it takes 28 hours for the temperature to reach 0°C.)

y = -70 (So the temperature when the experiment begins is −70°C.)

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Answer:

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Step-by-step explanation:

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= (1/9) [½ + ⅓]²

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3 years ago
A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on t
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Answer:

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Let L represent side parallel to the river and W represent width of fence.

The required fencing (F) would be F=L+2W.

We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.

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We are told that the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot, so total cost (C) of fencing would be C=20L+5(2W)\Rightarrow 20L+10W.

From equation (1), we will get:

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Upon substituting this value in cost equation, we will get:

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To minimize the cost, we need to find critical points of the the derivative of cost function as:

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C''=3200,000W^{-3}

C''=\frac{3200,000}{W^3}

Now, we will substitute W=400 in 2nd derivative as:

C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05

Since 2nd derivative is positive at W=400, therefore, width of 400 ft of the fencing will minimize the cost.

Upon substituting W=400 in L=\frac{80,000}{W}, we will get:

L=\frac{80,000}{400}\\\\L=200

Therefore, the side parallel to the river will be 200 feet.

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