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daser333 [38]
4 years ago
14

How can i get the answer

Mathematics
1 answer:
weqwewe [10]4 years ago
6 0
6 miles is the entire diameter, divide that by 2 to get your RADIUS.
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Which polygon has perpendicular sides?
Pavlova-9 [17]

Answer:

c

Step-by-step explanation:

6 0
3 years ago
Solve for x and write your answer in simplest form.<br><br> -1/2 -1/2(4/5x+1) = -2 -3x
dimaraw [331]

The solution of the equation which is as described in the task content when expressed in its simplest form is; -5/13.

<h3>What is the solution of the equation as described in the task content?</h3>

It follows from the task content that the solution to the equation is to be determined and expressed in its simplest form.

Given;

-1/2 -1/2(4/5x+1) = -2 -3x

By distributing the coefficients: we have;

-1/2 - 2/5x - 1/2 = -2 -3x

Hence, by collecting like terms; we have;

-1/2 - 1/2 + 2 = -3x + (2/5)x

1 = -13x/5

Therefore; by cross-multiplication; we have;

5 = -13x

Divide both sides by; -13.

x = -5/13.

Ultimately, the solution of the equation which is as described in the task content when expressed in its simplest form is; -5/13.

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4 0
1 year ago
Find coordinates for (-3, -6) when translated 5 units up
Leni [432]
The coordinates would be (-3,-1)

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4 years ago
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How do you write 57 and 62/100 in expanded form
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57 divided by 100 and 62 divided by 100
7 0
3 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

  • 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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8 0
3 years ago
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