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Andrews [41]
3 years ago
10

Solve and show work. 4-(6x+5)=7-2(3x+4)

Mathematics
1 answer:
luda_lava [24]3 years ago
8 0
4-(6x + 5) = 7 - 2(3x + 4)
Distribute: 4 - 6x - 5 = 7 - 6x - 8
Combine like terms (constants): -1 - 6x = -1 - 6x
Add 1 to both sides: -6x = -6x
Divide both sides by -6: x = x

Hope this helps!
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Answer: Solution

Step-by-step explanation:  I believe the answer is solution

8 0
3 years ago
What is the average rate of change for this function for the interval from x=3 to x=5
sveta [45]

Answer:

A. 16

Step-by-step explanation:

We have to find rate of change of function from x=3 to x=5

The average rate of change for the interval a≤x≤b is given by:

Rate of change=  (f(b)-f(a))/(b-a)

In our question,

a=3

and

b=5

We can see from the table that at x = 3, the value of function is 18

So,

f(a)=18

and

At x=5, the value of function is 50

f(b)=50

Rate of change=(50-18)/(5-3)

=32/2

=16

So the rate of change of function between x=3 and x=5 is 16 ..  

Option A is the correct answer..

3 0
3 years ago
What is the answer of 5.21 + 0.6-0.27/3
ruslelena [56]
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3 0
4 years ago
Read 2 more answers
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:

brainly.com/question/20336420

8 0
3 years ago
What is the slope of a sliding board tgat rises 2 feet for every horizontal change of 36 inches?
n200080 [17]
To find slope, it's rise over run, so the answer would be 2/36, simplified to 1/18.
7 0
3 years ago
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