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kow [346]
3 years ago
12

5m<6m+6 pls help me ​

Mathematics
1 answer:
torisob [31]3 years ago
4 0

Let's solve your inequality step-by-step.

5m<6m+6

Step 1: Subtract 6m from both sides.

5m−6m<6m+6−6m

−m<6

Step 2: Divide both sides by -1.

−m / −1 < 6 / −1

m>−6

Answer:

m>−6

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5 0
3 years ago
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
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Answer:

i like how to tell us to answer something with no file

Step-by-step explanation:

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Aleksandr [31]
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3 years ago
The sum of 6 consecutive even numbers is 126.<br> What is the fourth number in this sequence?
lilavasa [31]

Answer:

The 6 numbers are: 16, 18, 20, 22, 24, and 26 so the answer would be 22.

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3 years ago
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