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3 years ago

I need help with this math problem

Mathematics

1 answer:

den301095 [7]3 years ago

8 0

Answer:

1). $\frac{2}{x^{2}-x-12 }=\frac{2}{(x+3)(x-4)}$

2). $\frac{1}{x^{2}-16 }=\frac{1}{(x-4)(x+4)}$

Step-by-step explanation:

In this question we have to write the fractions in the factored form.

Rational expressions are $\frac{2}{x^{2}-x-12 }$ and $\frac{1}{x^{2}-16 }$ .

1). $\frac{2}{x^{2}-x-12 }$

Factored form of the denominator (x² - x - 12) = x² - 4x + 3x - 12

= x(x - 4) + 3(x - 4)

= (x + 3)(x - 4)

Therefore. $\frac{2}{x^{2}-x-12 }=\frac{2}{(x+3)(x-4)}$

2). $\frac{1}{x^{2}-16 }$

Factored form of the denominator (x² - 16) = (x - 4)(x + 4)

[Since (a²- b²) = (a - b)(a + b)]

Therefore, $\frac{1}{x^{2}-16 }=\frac{1}{(x-4)(x+4)}$

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Please help ASAP!!!!!

HACTEHA [7]

It’s D, y = 4.
have a good day :)

6 0

3 years ago

Read 2 more answers

If this figure is a regular figure? How many degrees does each angle measure?

bearhunter [10]

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$\frac{(5 - 2) \times 180}{5} = \\$

$\frac{3 \times 18 \times 10}{5} = 3 \times 18 \times 2 = \\$

$54 \times 2 = 108$

Thus each angle has a measure of 108° degrees.

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5 0

2 years ago

Derivative of<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2

Anastaziya [24]

Answer:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

Step-by-step explanation:

we would like to figure out the derivative of the following:

$\displaystyle \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

to do so, let,

$\displaystyle y = \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

By simplifying we acquire:

$\displaystyle y = 3 - \frac{2}{x} - \frac{1}{ {x}^{2} }$

use law of exponent which yields:

$\displaystyle y = 3 - 2 {x}^{ - 1} - { {x}^{ - 2} }$

take derivative in both sides:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} (3 - 2 {x}^{ - 1} - { {x}^{ - 2} } )$

use sum derivation rule which yields:

$\rm\displaystyle \frac{dy}{dx} = \frac{d}{dx} 3 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

By constant derivation we acquire:

$\rm\displaystyle \frac{dy}{dx} = 0 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

use exponent rule of derivation which yields:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ - 1 -1} ) - ( - 2 {x}^{ - 2 - 1} )$

simplify exponent:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ -2} ) - ( - 2 {x}^{ - 3} )$

two negatives make positive so,

$\displaystyle \frac{dy}{dx} = 2 {x}^{ -2} + 2 {x}^{ - 3}$

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

$\displaystyle \frac{dy}{dx} = \frac{2 }{x^2}+ \frac{2}{x^3}$

simplify addition:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

and we are done!

5 0

3 years ago

Write the following expression in standered form.<br><br> -3•2•a+5(-7b)+(12•c•4)

Alex

Answer:

−6a − 35b + 48c

Step-by-step explanation:

You do pemdas and if you know Pemads you will get your anwser

6 0

2 years ago

A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formul

N76 [4]

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula $-5t^2+40t+45$ .

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

So, let's identify our coefficients. The standard quadratic form is:

$ax^2+bx+c$

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

$x=\frac{-(40)}{2(-5)}$

Multiply and reduce. So:

$x=-40/-10=4$

Now, substitute 4 for t to find the height. So:

$-5(4)^2+40(4)+45$

Evaluate:

$=-5(16)+40(4)+45$

Multiply and add:

$=-80+160+45=125\text{ meters}$

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

$-5t^2+40t+45=0$

First, let's divide everything by -5. This yields:

$t^2-8t-9=0$

We can factor:

$(t+1)(t-9)=0$

Zero Product Property:

$t+1=0\text{ or } t-9=0$

Solve for t:

$t=-1\text{ or } t=9$

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!

6 0

3 years ago