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3 years ago

Can 12(x^3+y^2)+6(y^2+1) be the factored form of 12x^3+18y^2+6?

Mathematics

2 answers:

kirill [66]3 years ago

8 0

The correct answer is yes. just distribute 12 to all within first set then 6 to all in 2nd set then add like terms

Misha Larkins [42]3 years ago

4 0

Answer:

YES

Step-by-step explanation:

So how you can figure this out is you do the Distributive Property.

So it will be 12x^3+12y^2+6y^2+6.

The next thing you have to do is add together the ones that are similar which is 12y^2+6y^2 to get 18y^2.

Your answer is 12x^3+18y^2+6. So YES it can be factored into that form.

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Answer:

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2 years ago

What is the value of a1 of the geometric series? Sigma-Summation Underscript n = 1 Overscript infinity EndScripts 12 (negative o

worty [1.4K]

Answer:

Step-by-step explanation:

The given geometric series is

$\sum_{n = 1}^{ \infty} 12( { - \frac{1}{9} })^{n - 1}$

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Recall that the explicit formula is

$a_{n} = 12( { - \frac{1}{9} })^{n - 1}$

To find the first term, we put n=1 to get:

$a_{1} = 12( { - \frac{1}{9} })^{1 - 1}$

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bixtya [17]

Answer:

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Step-by-step explanation:

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