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3 years ago

If a student knows a car traveled 300 kilometers in three hours, what can the student find?

Mathematics

2 answers:

sveta [45]3 years ago

6 0

Its speed 100 km per hour. 300/3

maksim [4K]3 years ago

6 0

Answer:

I believe the answer is its ACCELERATION.

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NetSell, a the TV remote control supplier for Lumyn Electronics, has a weekly production cost of q TV remote controls that is gi

natali 33 [55]

Answer:

a. $\frac{dC(q)}{dq} = 0.000012q^2 -0.06q + 100$

b. $\frac{dR(q)}{dq}=-0.01q+200$

$U'(2000)=-0.000012(2000)^2+0.05(2000)+100 = 152$

$U'(7000)=-0.000012(7000)^2+0.05(7000)+100 = -138$

Step-by-step explanation:

a) The marginal cost function is given by the derivative of the total cost function, in this way the marginal cost function for this company is:

$\frac{dC(q)}{dq} = \frac{dC(q)}{dq} (0.000004q^ 3 - 0.03q ^ 2 + 100q + 75000) = 0.000012q^2 -0.06q + 100$

b) The income function is given by the relation $R (q) = P (q) q = -0.005q^2 + 200q$ .

The marginal revenue function for the company is given by the derivative of the revenue function, in this way the marginal revenue function is:

$\frac{dR(q)}{dq}= -0.01q+200$

(c) The profit function of the company is given by the relation $U (q) = R (q) - C (q)$ , and the marginal utility function is given by the derivative of the utility function, in this way , the marginal utility function is:

$\dfrac{dU(q)}{dq}=R'(q) - C'(q) = -0.01q+200 - (0.000012q^2-0.06q+100) = -0.000012q^2+0.05q+100$

When q = 2000, the marginal utility is:

$U'(2000)=-0.000012(2000)^2+0.05(2000)+100 = 152$

When q = 7000, the marginal utility is:

$U'(7000)=-0.000012(7000)^2+0.05(7000)+100 = -138$

6 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

Will mark you as brainliest. If your good with polynomials.

kobusy [5.1K]

To solve/simplify this all you have to do is group like terms (the x^2's with each other, the x's with each other, and the normal numbers, -8)

14x^2-8+5x-6x^2+2x

group the x^2 (add 14x^2 to -6x^2)

8x^2-8+5x+2x

group the x's together (add 5x and 2x together)

8x^2+7x-8

Your answer will be d) 8x^2+7x-8

4 0

3 years ago

What is the answer to this

Scilla [17]

420, 210, 140, 105, 84, 70, 60, 42, 35, 30, 28, 21, 20, 15, 14, 12, 10, 7, 6, 5, 4, 3, 2, 1

8 0

3 years ago

The line of reflection is blank and the coordinates of C' are blank ?

denpristay [2]

Answer:

I did this before it is C

Step-by-step explanation:

4 0

3 years ago