Factory overhead is a _____ account.

Mathematics

1 answer:

ladessa [460]3 years ago

6 0

Answer:

Factory overhead is a liability account

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What is the solution to a two-variable equation? pls answer

zloy xaker [14]

In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true. In other words, it is where the two graphs intersect, what they have in common. So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system

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3 years ago

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Find 3 equivalent ratios for<br> 4/12.

Neko [114]

Answer:

8:24, 24: 72, 16:48

Step-by-step explanation:

There's a lot just multiply them by the same number

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A rectangle's length is seven times its width. The perimeter of the rectangle is one hundred sixty feet.

Arada [10]

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Area=700 square ft

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Classify each pair of numbered angles.

kifflom [539]

Adjacent angles are angles with a common side and vertex, linear pairs are adjacent angles that are supplementary, and vertical angles are angles made by the same two lines but on opposite sides.

For the first one, 5 and 6 clearly do not share a side but they are made up by the same 2 lines and are opposite of each other, making them vertical.

For the next one, since the angles only share 1 line (and not a side) they can't be any of the above.

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Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.

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3 years ago