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Fudgin [204]
3 years ago
9

Heyy!! Can someone help me please!!

Mathematics
2 answers:
abruzzese [7]3 years ago
5 0

Answer:

7x + 14

Step-by-step explanation:

the first thing to do is expand the parentheses/brackets.

3(5x + 2) -2(4x - 4) will be

3(5x) + 3(2) -2(4x) -2(-4)

= 15x + 6 - 8x + 8

collect like terms

15x - 8x + 6 + 8 = 7x + 14

the answer is 7x + 14

nadezda [96]3 years ago
5 0

Answer:

3(5x+2)-2(4x-4)

15x+6-8x+8

15x-8x+6+8

7x+14

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Answer:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

Step-by-step explanation:

Given

p(x) = \frac{x^2-2x-3}{x+2} -- Missing from the question

Required

The behavior of the function around its vertical asymptote at x = -2

p(x) = \frac{x^2-2x-3}{x+2}

Expand the numerator

p(x) = \frac{x^2 + x -3x - 3}{x+2}

Factorize

p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}

Factor out x + 1

p(x) = \frac{(x -3)(x + 1)}{x+2}

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

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

As x approaches -2 implies that:

x -> -2^{-} Say x = -3

p(x) = \frac{(x -3)(x + 1)}{x+2}

p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

x->-2^{-}, p(x)->-\infty

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: x>-2

Say x = -2.1

p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1

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x->-2^{+}, p(x)->-\infty

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x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

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3 years ago
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Step-by-step explanation:

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your welcome

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