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3 years ago

If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?

Mathematics

2 answers:

Anestetic [448]3 years ago

6 0

$(k\circ h)(x)=k(h(x))$

$\displaystyle k(h(x))= \frac{1}{h(x)}= \frac{1}{5x}$

galina1969 [7]3 years ago

6 0

For those who may still need it in the future, the correct answer is actually: B

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400/2 = 200
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3 years ago

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3 years ago

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 -2.6 degrees Celsius per minute.

b) Yes

c) The equation for the line tangent is y = -3.6·t + 90.8
The approximate value of C(5) is 72.8 °C

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

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) = -2.6°C per minute

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 C 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; y = -3.6·x + 90.8

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

C(5) ≈ 72.8°

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

soup is changing at t = 3 minutes is -3.6 degrees per minute.

Learn more about calculus and concepts here:

brainly.com/question/20336420

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2 years ago