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harina [27]
2 years ago
12

Help me with this please.

Mathematics
1 answer:
andriy [413]2 years ago
4 0
1. bc/3 + a/xy + 3/ab
= b^2cxya + 3a^2b + 9xy/3xyab

2. (5/3y) + 2y - 12
=11/3y - 2

3. ?

Hope this helps and right! :)) Good luck
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The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum
wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information:P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000, where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2

Equating it to zero we get,

x^2 + 36x - 6400 = 0

We use the quadratic formula to find the values of x:

x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}, where a, b and c are coefficients of x^2, x^1 , x^0 respectively.

Putting these value we get x = -100, 64

Now, again differentiating

\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x

At x = 64,  \displaystyle\frac{d^2(P(x))}{dx^2} < 0

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0
3 years ago
an arc welder provides 250 amps of current when operating with a voltage of 40 volts. what is the power required by this welder
FinnZ [79.3K]
The power(P) of the circuits is the product of the voltage (V) in volts and current(I) in amperes. 
                                        P = IV

Substituting the known values to the equation,
                                     P = (250 amps) x (40 volts) = 10,000 Watts

Thus, the answer is 10,000 Watts. 

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2 years ago
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melisa1 [442]

Answer:

25

Step-by-step explanation:

square has 4 sides and 100 divided by 4 is 25

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Answer: the distance would be 75 mi

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