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2 years ago

PLEASE ANSWER ASAP :)

Mathematics

1 answer:

Gala2k [10]2 years ago

8 0

Answer:

(x+6)( x^2 -2)

Step-by-step explanation:

x^3 +6x^2 -2x-12

We can factor by grouping

x^3 +6x^2 -2x-12

Factor x^2 out of the first group and -2 out of the the second group

x^2 (x+6) -2 (x+6)

Now factor out (x+6)

(x+6)( x^2 -2)

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If Oliver actually worked a total of fourteen hours doing his 14 shifts at Slow Food to Go, how many hours was he scheduled to w

fiasKO [112]

Answer:

His shifts were 1 hour each

Step-by-step explanation:

If Oliver worked a total of 14 hours in a span of 14 shifts, that would mean he worked 1 hour per shift because 1(14) = 14.

14/14 = 1 hour

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2 years ago

Which expression is equivalent to the expression 6(-2n + 7) - 3(5n + 9)?

timurjin [86]

-27n+15 is your answer

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3 years ago

If X = 2 centimeters, Y = 4 centimeters, and Z = 6 centimeters, what is the area of the object?

Tatiana [17]

Answer:

Step-by-step explanation:

the shape is a trpizoid with height y

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5 0

3 years ago

Read 2 more answers

How do I differentiate this?

disa [49]

Answer:

$\frac{8}{\left(-x+2\right)^2}$ & $x = 1, x = 3$

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

$u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

Now remember the sum rule:

$\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right) = 0\\\frac{d}{dx}\left(3x+2\right) = 3$

For this second bit we apply the chain rule:

$\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}$

If we substitute these values back into the expression... $\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

...we get the following:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)$

The rest is just pure simplification:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}$

Now let's equate this to equal 8 for the second bit and solve for x:

$\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3$

4 0

3 years ago

Just want to double check our answer.

bearhunter [10]

0 divided by 2 is 0.

You can use the multiplicative identity property: the product of any number and zero is still zero.

Hope this helped :)

3 0

3 years ago