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pogonyaev
4 years ago
5

What is the factorization of the trinomial below?x3 - 12x2 + 35x?

Mathematics
2 answers:
Fed [463]4 years ago
4 0
X³ - 12x² + 35x

x(x² - 12x + 35)

x(x-5)(x-7)
algol134 years ago
3 0

Answer: ×(x-7)(x-5)

Step-by-step explanation: they both of to be a negative because of the negative twelve

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Two cyclists, 80 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they
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The speed of the faster cyclist will be 30 miles/hour.

<em><u>Explanation</u></em>

One cyclist is 3 times as fast as the other.

Lets assume that the speed of the slower cyclist is  x miles/hour.

So, the <u>speed of the faster cyclist</u> will be:  3x miles/hour.

They start riding toward each other at the same time and meet 2 hours later.

We know that,  Distance= speed*time

So, the <u>distance traveled by slower cyclist</u> in 2 hours = (x*2)miles= 2x miles and the <u>distance traveled by faster cyclist</u> =(3x*2) miles = 6x miles.

Given that, they were 80 miles apart at the starting. So, the equation will be.....

2x+6x=80\\ \\ 8x=80\\ \\ x=10

So, the speed of the faster cyclist will be: (3×10)= 30 miles/hour.

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4 years ago
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B is the correct answer
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How much greater is 5 miles than 8,000 yards?
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10 POINTS!!! FULL ANSWER WITH FULL STEP BY STEP SOLUTION PLEASE
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A) Profit is the difference between revenue an cost. The profit per widget is
  m(x) = p(x) - c(x)
  m(x) = 60x -3x^2 -(1800 - 183x)
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Then the profit function for the company will be the excess of this per-widget profit multiplied by the number of widgets over the fixed costs.
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  P(x) = -3x^3 +243x^2 -1800x -50000

b) The marginal profit function is the derivative of the profit function.
  P'(x) = -9x^2 +486x -1800

c) P'(40) = -9(40 -4)(40 -50) = 3240
  Yes, more widgets should be built. The positive marginal profit indicates that building another widget will increase profit.

d) P'(50) = -9(50 -4)(50 -50) = 0
  No, more widgets should not be built. The zero marginal profit indicates there is no profit to be made by building more widgets.

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On the face of it, this problem seems fairly straightforward, and the above "step-by-step" seems to give fairly reasonable answers. However, if you look at the function p(x), you find the "best price per widget" is negatve for more than 20 widgets. Similarly, the "cost per widget" is negative for more than 9.8 widgets. Thus, the only reason there is any profit at all for any number of widgets is that the negative costs are more negative than the negative revenue. This does not begin to model any real application of these ideas. It is yet another instance of failed math curriculum material.

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