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2 years ago

If f(x) = -3x - 2 find f(-2) + 9

Mathematics

2 answers:

ad-work [718]2 years ago

8 0

It shluld be 19 i think

wolverine [178]2 years ago

3 0

Answer:

f(-2) + 9 is 19

Step-by-step explanation:

$f(x) = - 3x - 2$

when x is -9:

$f( - 2) = - 3( - 2) - 2 \\ f( - 2) = 10$

sum up with 9, f(-2)+9:

$f( - 2) + 9 = 10 + 9 \\ = 19$

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Tell whether the square root of .25 is rational or irrational. Explain.

artcher [175]

The square root of .25 is rational because it is a terminating decimal.
√(.25)=05

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3 years ago

Choose the correct equation for the parabola based on the given information. Given: Focus:(2,8) Directrix: y = 4 a. 2(y-2)= (x-

Harman [31]

<h3>Answer: c. 8(y-6) = (x-2)^2</h3>

Explanation:

The directrix is horizontal, so the axis of symmetry is vertical. We'll have an x^2 term. The vertical distance from y = 4 to y = 8 is 4 units. Cut this in half to get 2, which is the focal distance p = 2.

The point (2,4) is directly below (2,8), and the point is on the directrix. The midpoint between (2,4) and (2,8) is (2,6). This is the vertex.

(h,k) = (2,6)

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8 0

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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ValentinkaMS [17]

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3 years ago

Read 2 more answers

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Sliva [168]

Lennon used the property o addition

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3 years ago