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3 years ago

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If the sum (3x ^ 2 - 2x + 12) + (x ^ 2 + 8x - 2) is multiplied by 2x ^ 2 what is the result written in standard form ?

1 answer:

alexandr402 [8]3 years ago

5 0

(3x^2 - 2x + 12) + (x^2 + 8x -2)
= (4x^2+6x+10)(2x^2)
= 4x^2(2x^2)+6x(2x^2)+10(2x^2)
= 8x^4 + 12x^3 + 20x^2

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The function f(x) = x^2 - 2x + 8 is transformed such that g(x) = f(x-2). Find the vertex of g(x).

Serjik [45]

Answer:

Step-by-step explanation:

Since g(x) is f(x-2), every time there is a "x" in the f(x) equation, substitute "x-2". So g(x) would be (x-2)^2-2(x-2)+8. Use the method of trial and error: substitute x as 1,-1, and 3, until you found the right vertex. The answer is B (3,7).

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3 years ago

Numbers that equal to 39

Natali5045456 [20]

Answer:

1 , 3 , 13 , and 39

Step-by-step explanation:

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2 years ago

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Five chips numbered 1 through 5 are placed in a bag. A chip is drawn and replaced. Then a second chip is drawn. Find the probabi

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The odds of drawing an even chip are $\frac{2}{5}$ .

The odds of drawing an odd chip are $\frac{3}{5}$ .

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Knowing you cannot simplify 6/25 any further, you now have your answer!

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3 years ago

The function h is defined by h (x) = 5x + 2

soldier1979 [14.2K]

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5 0

3 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago