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ozzi
2 years ago
6

What are the solutions?8x^2-12x-23=-5​

Mathematics
1 answer:
erastova [34]2 years ago
4 0

Answer:

x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

Step-by-step explanation:

Solve for x over the real numbers:

8 x^2 - 12 x - 23 = -5

Divide both sides by 8:

x^2 - (3 x)/2 - 23/8 = -5/8

Add 23/8 to both sides:

x^2 - (3 x)/2 = 9/4

Add 9/16 to both sides:

x^2 - (3 x)/2 + 9/16 = 45/16

Write the left hand side as a square:

(x - 3/4)^2 = 45/16

Take the square root of both sides:

x - 3/4 = (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

x = 3/4 + (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

Answer:  x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

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Solve the right triangle. round your answers to the nearest tenth. help.
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Answer:

19.2

Step-by-step explanation:

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2 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.
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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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Step-by-step explanation:

We are given the following information:

Confidence level = 90%

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Variance = 1.96

\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{1.96} = 1.4

Formula:

\text{Error} = z_{critical}\dfrac{\sigma}{\sqrt{n}}

Putting all the values,

z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64

0.09 = 1.64\times \dfrac{1.4}{\sqrt{n}}\\\\n = (\dfrac{1.64\times 1.4}{0.09})^2 = 650.81 \approx 651

651 is the minimum number of people over age 30 they must include in their sample.

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