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Vikki [24]
1 year ago
6

6x³ + 3x how do you factor out the GCF in this question??

Mathematics
2 answers:
shepuryov [24]1 year ago
7 0
Hope this helps, comment if you have any questions:)))
maw [93]1 year ago
4 0

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to factor out the GCF in the expression

\star~\star\mathrm{6x^3+3x}~\star~\star

\triangle~\fbox{\bf{KEY:}}

  • Divide every single term of the expression by the GCF (Greatest Common Factor).

So let's divide.

But first. we should work out the GCF.

Both terms have 3x in common, so we divide both terms by 3x:

\mathrm{6x^3+3x}

\mathrm{3x(2x^2+1)}

To ensure that we've factored correctly, we can always use the distributive property, distribute 3x and see whether or not we end up with \mathtt{6x^3+3x}.

So the answer is

\mathrm{3x(2x^2+1)}

Notice I put parentheses around <em>2x²+1</em>. This indicates that 3x is multiplied times both 2x² and 1.

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students :)

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3 years ago
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antoniya [11.8K]

Answer:

The answer is "experiment."

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4 0
3 years ago
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
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Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
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divide by w
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we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
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the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} &#10;
substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
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minimum value =- \frac{1}{3}


4 0
3 years ago
50 point available!!!!!!
g100num [7]

Answer:

2 miles is your answer hope this helps!!!!!!

Step-by-step explanation:


7 0
3 years ago
Read 2 more answers
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