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Vladimir79 [104]
3 years ago
5

I need to solve this problem in my exam:

Mathematics
1 answer:
mash [69]3 years ago
5 0
The solution to the problem is as follows:


Average cost is c(x)/x.
c(35) = 400 + 20(35) - 0.2(35)^2 = $855.
Then, the average cost is $855/35 = $24.43/radio


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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If f(x)=3x-15, then f^-1(x) =<br> solve for x please
amid [387]
1.  Replace f(x) with y:  y = 3x - 15
2.  Interchange x and y:  x = 3y - 15
3.  Solve this for y:  3y = x + 15, and y = (x + 15)/3
4.  Replace y with 

   -1                -1
 f   (x):           f   (x) = (x + 15) / 3            (answer)
5 0
3 years ago
A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into
algol [13]

Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

3 0
2 years ago
Which expression is equivalent to 350 · 22 ?
Semenov [28]

Answer:

7,700

Step-by-step explanation:

350 x 22= 7,700

4 0
3 years ago
Jean and Jericho who are playing in the school grounds decided to sit on a
Natalija [7]

We want to see underline the correct part in each statement.

  • 1) This situation illustrates (direct, <u>inverse</u>) variation.
  • 2) The two quantities that must vary in this situation are (<u>weight and </u>
  • <u>distance from the cente</u><u>r</u>, height and weight).
  • 3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the
  • center to balance the seesaw?
  • 4) When Jean moves farther from the center, Jericho tends to go (up,
  • <u>down</u>).
  • 5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

So, Jean and Jericho are playing on a seesaw.

Jean is heavier than Jericho.

Now, notice that a seesaw is a lever. So it "amplificates" the force that you apply in one end to lift the weight that is on the other end. Depending on the form of the lever and the weights, the force that you need to do changes.

If we define:

  • W₁ = Jean's weight.
  • d₁ = Jean's distance to the center.
  • W₂ = Jericho's weight.
  • d₂ = Jericho's distance to the center.

We must have:

W₁*d₁ = W₂*d₂

Then:

1) This situation illustrates (direct, <u>inverse</u>) variation.

d₁, the position of Jean, varies inversely with respect to Jean's weight.

2) The two quantities that must vary in this situation are (<u>weight and </u>

<u>distance from the center</u>, height and weight).

(by the equation above)

3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the

center to balance the seesaw?

By the given equation, we see that d₁ must be smaller than d₂.

And the last two are trivial:

4) When Jean moves farther from the center, Jericho tends to go (up,

<u>down</u>).

5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

If you want to learn more, you can read:

brainly.com/question/18320907

7 0
2 years ago
Can somebody please help me
a_sh-v [17]

Answer:

Step-by-step explanation:it’s number 4

8 0
2 years ago
Read 2 more answers
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