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vovikov84 [41]
2 years ago
3

Finding the Break-Even Point and the Profit Function Using Substitution Given the cost function C(x)=0.85x+35,000 and the revenu

e function R(x)=1.55x,find the break-even point and the profit function?
Mathematics
1 answer:
Alex73 [517]2 years ago
8 0

Answer:

x=50000

P(x)=0.7x-35000

Step-by-step explanation:

Given cost function is

C(x)=0.85x+35000

and revenue function is

R(x)=1.55x

At break even point, revenue is equal to cost

R(x)= C(x)

1.55x=0.85x+35000

Subtract 0.85 from both sides

0.7x=35000

divide by 0.7 on both sides

x=50000

Profit function

P(x)= R(x)- C(x)

P(x)=1.55x-(0.85x+35000)

P(x)=1.55x-0.85x-35000

P(x)=0.7x-35000

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The base of pyramid A is a rectangle with a length of 10 meters and a width of 20 meters. The base of pyramid B is a square with
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Given:
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Pyramid B: Base is square with 10 meter sides.
Heights are the same.

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Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
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ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
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Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
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   x² + y + 4i
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