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Rudik [331]
4 years ago
8

Please help me. I can't figure this out. I've been doing math for 5 hours and that's not an exaggeration.

Mathematics
2 answers:
svetlana [45]4 years ago
6 0

Well you have to do PEMDAS

So we have an exponent so that will turn into a positive X

X+4x=10

So now we combine our X's together to get 5X=10

Now we divide both sides by X which gives us 2

Sorry read the problem wrong

Ask questions if you still need help :)


vazorg [7]4 years ago
3 0
-1 (X^2 - 4x = 10)

First divide by negative 1

-( x^2 - 4x + 4 = 10 + 4)

Second create a perfect square

(x - 2)^2 = 14

Third factor and add like terms

sqrt/ (x - 2)^2 = sqrt/ 14

Fourth find the square root of both sides

x - 2 = (+\-) sqrt/14

Fifth remember to factor in that the answer could be positive or negative!

x = 2 (+/-) sqrt/14

Sixth add two to both sides and conclude your answer is two plus or minus the square root of fourteen.


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Arte-miy333 [17]

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all work is shown and pictured

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Nesterboy [21]
<h3>Answer: </h3>

The GCF is 4

The polynomial factors to 4(3x^2+x+5)

==========================================================

Further explanation:

Ignore the x terms

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Factor each to their prime factorization. It might help to do a factor tree, but this is optional.

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  • 4 = 2*2
  • 20 = 2*2*5

Each factorization involves "2*2", which means 2*2 = 4 is the GCF here.

We can then factor like so

12x^2 + 4x + 20\\\\4*3x^2 + 4*x + 4*5\\\\4(3x^2 + x + 5)

The distributive property pulls out that common 4. We can verify this by distributing the 4 back in, so we get the original expression back again.

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4 0
2 years ago
Suppose that f(x+h)−f(x)= −6hx2−7hx−6h2x−7h2+2h3. <br> Find f′(x). <br><br> f′(x)=
Nataliya [291]
By definition,
f'(x)=Lim h->0  (f(x+h)-f(x))/h
We are already given
f(x+h)-f(x)=<span>−6hx2−7hx−6h2x−7h2+2h3=h(-6x^2-7x-6hx-7h+2h^2)
divide by h
</span>(f(x+h)-f(x))/h =h(-6x^2-7x-6hx-7h+2h^2)/h=(-6x^2-7x-6hx-7h+2h^2)
Finally, take lim h->0
f'(x)=Lim h->0  (f(x+h)-f(x))/h=(-6x^2-7x-0-0+0)=-6x^2-7x
=>
f'(x)=-6x^2-7x
8 0
3 years ago
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