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sweet [91]
4 years ago
13

The points (-4,1) and (2,-4) are the endpoints of the line segment . Which coordinate pair is the midpoint of the line segment?

Mathematics
1 answer:
nata0808 [166]4 years ago
4 0
The midpoint is (-1, -3/2)
to find the midpoint of something you must use the formula (x2 + x1/2, y2+y1/2)
So our formula becomes (2-4/2, -4+1/2)
which then equals (-1, -3/2)
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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

  • 58 °C < 60 < 90 °C

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

  • The equation for the tangent is; <u>y = -3.6·x + 90.8</u>

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

  • <u>C(5) ≈ 72.8°</u>

<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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Answer:


Step-by-step explanation:

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