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nikitadnepr [17]
2 years ago
11

(14 + 11x) + (x ^ 2 - 15x + 17)​

Mathematics
1 answer:
Sidana [21]2 years ago
4 0

Answer:

VERY EASY

SIMPLIFIED

Let's simplify step-by-step.

14+11x+x²−15x+17

=14+11x+x²+−15x+17

Combine Like Terms:

=14+11x+x²+−15x+17

=(x²)+(11x+−15x)+(14+17)

=x²+−4x+31

<h2>Answer:</h2><h2>=x2−4x+31</h2>

BRAINLIEST PLEASE

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36-6/3
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2 years ago
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7th grade math help me please :(.
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3 years ago
What is y=1/2x+4 in standard form?
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<h3>Answer:  x-2y = -8</h3>

===========================================

Explanation:

Multiply both sides by 2 to clear out the fraction

y = (1/2)x+4

2y = 2[ (1/2)x + 4 ]

2y = 2*(1/2)x + 2*4

2y = x + 8

Then move the x term over to the left side

2y = x+8

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Optionally we can multiply both sides by -1

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This is in standard form Ax+By = C with A = 1, B = -2, C = -8

The reason why I multiplied both sides by -1 was to make A > 0 which is what some textbooks use as convention. Of course -x+2y = 8 is equally valid too.

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Given

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We have to set the restraint

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because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

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(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

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