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Tpy6a [65]
3 years ago
13

For the equation 2(4+x)+(13+x)=3x+k which value of k will create an equation with no solutions. .A. x B. 3x .C 15 D. 21

Mathematics
1 answer:
Artist 52 [7]3 years ago
6 0
<h3>Answer: choice C) 15</h3>

Simplify the left side to get

2(4+x)+(13+x)

2(4)+2(x) +13+x

8+2x+13+x

3x+21

------------

So the original equation

2(4+x)+(13+x) = 3x+k

turns into

3x+21 = 3x+k

------------

Subtract 3x from both sides

3x+21 = 3x+k

3x+21-3x = 3x+k-3x

21 = k

k = 21

-----------

If k = 21, then the original equation will have infinitely many solutions. This is because we will end up with 3x+21 on both sides, leading to 0 = 0 after getting everything to one side. This is a true equation no matter what x happens to be.

If k is some fixed number other than 21, then there will be no solutions. This equation is inconsistent (one side says one thing, the other side says something different). If k = 15, then

3x+21 = 3x+k

3x+21 = 3x+15

21 = 15 .... subtract 3x from both sides

The last equation is false, so there are no solutions here.

note: if you replace k with a variable term, then there will be exactly one solution.

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Answer:

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profit: $12209

Step-by-step explanation:

Given  the function:

p(x) = -x³ + 4x² + x

we want to maximize it.

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dp/dx = 0

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Using quadratic formula:

x = \frac{-b \pm \sqrt{b^2 -4(a)(c)}}{2(a)}

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x = \frac{-8 \pm 8.72}{-6}

x_1 = \frac{-8 + 8.72}{-6}

x_1 = -0.12

x_2 = \frac{-8 - 8.72}{-6}

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3 years ago
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