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Answer:</h2>
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Step-by-step explanation:</h2>
a. 2x^-3 • 4x^2
To solve this using only positive exponents, follow these steps:
i. Rewrite the expression in a clearer form
2x⁻³ . 4x²
ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.
In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to positive as follows;
iii. We then solve the result as follows;
=
Therefore, 2x⁻³ . 4x² =
b. 2x^4 • 4x^-3
i. Rewrite as follows;
2x⁴ . 4x⁻³
ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.
iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;
Therefore, 2x⁴ . 4x⁻³ =
c. 2x^3y^-3 • 2x
i. Rewrite as follows;
2x³y⁻³ . 2x
ii. Change position of terms with negative exponents;
iii. Now solve;
Therefore, 2x³y⁻³ . 2x =
Answer: no he is not correct
648
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162 4
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81 2 2 2
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3 3 3 3
Answer:
I don't think so.
Step-by-step explanation:
But what do I know? XD
Answer:
x=−6
Step-by-step explanationn
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)
m(x) = -3x^2 +243x -1800
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.
P(x) = x×m(x) -50,000
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.
