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3 years ago

I need help.........,

Mathematics

2 answers:

DerKrebs [107]3 years ago

8 0

Answer:

Dont worry your hero is here, the answer for p is 11

Step-by-step explanation:

15-4=11 or 4 plus 11 equals 15

Mrac [35]3 years ago

3 0

Answer:

Step-by-step explanation:

p=15-4

p= 11

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NEED HELP ASAP<br> Solve 4 – x = –8.

Anna71 [15]

Answer:

i think it is 12 positive

Step-by-step explanation:

i think

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3 years ago

Evaluate the function f(x) = x 2 + 1 for f(2

olasank [31]

F(x)= $x^{2} +1$
f(2)= $2^{2} +1$
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3 years ago

Divide the following polynomials using the long division model (4x ^ 4 - 5x ^ 3 + 2x ^ 2 - x + 5) + (x ^ 2 + x + 1)

densk [106]

Answer:

The ramainder is equal -86

Step-by-step explanation:

Since the polynomial degree Q(x) is 1, the remainder of the division must be a number. Therefore we only need to calculate the value of polynomial for x = -2

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3 years ago

Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2.

insens350 [35]

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

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

We have a negative value (-56.1); This will be called negative infinity

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

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

So, the behavior is:

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

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3 years ago

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Sever21 [200]

How old is sam? We need that first

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3 years ago

Read 2 more answers