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2 years ago

Please help me answer these questions if you give an answer you are a lifesaver

Mathematics

1 answer:

il63 [147K]2 years ago

4 0

On Part A you have to subtract $2.39-$1.99. Then in Part B you have to multiply $4.50 by 2. And then I think you have to add $3.79 when u are done multiplying.

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PLEASE HELP! GIVEE ME THE 14 ANSWERS FOR THIS FOR BRAINLIEST

gulaghasi [49]

Answer:

Step-by-step explanation:

oh boy

1) 9 < 16

2) -4 < 5

3) 18 > -19

4) -38 < -16

5) 6 > 4

6) -14 > -25

7) -45 > -55

8) -2 > -10

9) -4, -1, 1, 3, 6

10) -15, -13, -5, 8

11) -9, -6, -5, -1, 0

12) -4, -2, 1, 3

13) -10, -9, 7, 8, 11

14) -33, -25, -14, -4, 7

hope this helps <3

6 0

3 years ago

A large bag of marbles contains an equal amount of red, orange, yellow, blue, and green marbles. Quinn randomly selects 1 marble

Leni [432]

The answer will be A.

8 0

2 years ago

Consider the inequality -5(x+7)<-10. write an inequality representing the solution for x

Virty [35]

Answer:

$x > 3$

Step-by-step explanation:

$- 5(x + 7) < - 10$

Dividing by - 5

$x + 7 > 10 \\ x > 3$

6 0

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.

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

8 0

2 years ago

A videographer earns $485 for every wedding he records and $18 for every extra copy of the video his costumers order . How much

kupik [55]

All you have to do is start by multiplying 485 and 34 (485*34) to find how much he made by just weddings. That should be equal to 16,490. Then you have to multiply 18 by 87 because he charged 18 per copy and there was 87 copies so (18*87). This should now be equal to 1,566. Since you want the total you now should add them which should be 16,490+1,566 and that is equal to 18,056.

6 0

2 years ago