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3 years ago

Divide (x^2+9x+19)÷(x+5)

Mathematics

1 answer:

gayaneshka [121]3 years ago

6 0

Since this can’t be factored easily, you have to use synthetic division. This makes the answer x+4+(-1/x+5)

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The point A(–1, 4) is translated five units right and three units down. Which rule describes the translation?

lozanna [386]

Answer: SECOND OPTION.

Step-by-step explanation:

Given the following point identified as "A":

$A(-1,4)$

You can identify that the x-coordinate of the point is:

$x=-1$

And its y-coordinate is:

$y=4$

According to the exercise, this point is translated five units right and three units down. This means that, in order to find the new coordinates, you need to add 5 to the original x-coordinate and subtract 3 from the original y-coordinate.

Therefore, you can conclude that the rule that best describe the translation of the point $A(-1,4)$ is the following:

$(x,y)$ → $(x+5,y-3)$

Then, the point translated is:

$A'(-1+5,\ 4-3)$ → $A'(4,1)$

6 0

3 years ago

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:) got to love math lol

zzz [600]

Oof im bad at math XD

7 0

3 years ago

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Select the correct answer. Which expression is equivalent to 8x^2^3 sqrt 375x + 2^3 sqrt 3x^7, if x=0?

natima [27]

Answer:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}$

Step-by-step explanation:

The question is poorly formatted. The original question is:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7$

We have:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7$

Open bracket

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

Express 8 as 2^3

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^{3* \frac{2}{3}} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^2 *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

Express 2^3 as 8

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{75x^3} + 8\sqrt{ 3x^7}$

Expand each exponent

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2 *3x} + 8\sqrt{ 3x * x^6}$

Split

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 *5x *\sqrt{3x} + 8\sqrt{3x} * x^3$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =20x^3\sqrt{3x} + 8x^3\sqrt{3x}$

Factorize

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}$

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3 years ago

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PLEASE HELP!!<br> 4/3n - 8 = -6

iris [78.8K]

Answer:

n=3/2

Step-by-step explanation:

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3 years ago

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To measure the distance across a pond, Sandy and Garrett mark a spot at on

DerKrebs [107]

Answer:

distance across the pond: 13 yd

distance walked by Garret: 12 yd

Step-by-step explanation:

Data:

distance across the pond: x yd

distance walked by Sandy: 5 yd

distance walked by Garret: x-1 yd

A right triangle is formed where the distance across the pond is the hypotenuse. From Pythagorean theorem:

x² = 5² + (x-1)²

x² = 25 + x² - 2x + 1

0 = 26 - 2x

x = 13

4 0

3 years ago