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pochemuha
3 years ago
15

Need help with the question below

Mathematics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

Incorrect

Step-by-step explanation:

The formula for slope is [ y2-y2/x2-x1 ] and Emma calculated the slope using the opposite of the formula [ x2-x1/y2-y1 ]. That was her mistake.

300-900/180-165

-600/15

-40

Best of Luck!

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What is the area of the figure Enter in the box
just olya [345]

Answer:

Total Area = 532 m²

Step-by-step explanation:

Triangle = ½bh.

Rectangle = lw.

Total Area = sum of individual areas =

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+ (18 m × 18 m) =

(64 m²) + (144 m²) + (324 m²) =

532 m²

8 0
3 years ago
For f(x), evaluate the following:<br> a. f(0)<br> b. f(6)
Mashutka [201]
F(0) = 0 + 4 = 4 Because 0 < 5
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6 0
4 years ago
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As a bowl of soup cools, the temperature of the soup is given by the twice-differentiable function H for 0<img src="https://tex.
Aleks04 [339]

The rate of change of temperature with time at a point in time is given by

the derivative of the function for the temperature of the soup.

The correct responses are;

  • a) H'(5) is approximately<u> -2.6 degrees Celsius per minute</u>.
  • b) Yes
  • c) The equation for the line tangent is<u> y = -3.6·t + 90.8</u>
  • The approximate value of C(5) is <u>72.8 °C</u>

  • d) The rate of change of the temperature of the soup at t = 3 minutes is <u>-3.6 degrees Celsius per minute</u>.

Reasons:

a) From the data in the table, we have;

The approximate value of H'(5) is given by the average value of the rate of

change of the temperature with time between points, t = 3, and t = 8

Therefore;

\displaystyle H'(5) = \mathbf{\frac{H(8) - H(3)}{8 - 3}}

Which gives;

\displaystyle H'(5) =  \frac{80 - 67}{8 - 3} = \mathbf{2.6}

Therefore, H'(5) = <u>-2.6°C per minute</u>

b) Given that the function is twice differentiable over the interval, 0 ≤ t ≤ 12, the function for the change in temperature is continuous in the interval 0 ≤ t ≤ 12

At t = 0, H(0) = 90 °C

At t = 12, H(12) = 58 °C

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Therefore, there exist a temperature, of 60 °C between 90° C and 58 °C

c) The given derivative of <em>C</em> is, C'(t) = \mathbf{-3.6 \cdot e^{-0.05 \cdot t}}

At t = 3, we have;

The \ slope \ at \ t = 3 \ is \ C'(3) = -3.6 \cdot e^{-0.05 \times 3} \approx -3.1

Therefore, we have;

y - 80 ≈ -3.1 × (x - 3)

The equation for the tangent is; y = -3.6 × (x - 3) + 80

y = -3.6·x + 10.8 + 80 = -3.6·x + 90.8

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The  value of C(5) is approximately, C(5) ≈ -3.6 × 5 + 90.8 = 72.8

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<u />

d) Based on the the model above, the rate at which the temperature of the

soup is changing at t = 3 minutes is <u>-3.6 degrees per minute</u>.

Learn more about calculus and concepts here:

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8 0
3 years ago
Help Me please!!!!! Rn
Sergio039 [100]

Answer:

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Step-by-step explanation:

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x = 20

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do you want me to explain what the distributive property is????

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