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2 years ago

Does anybody know the answer to this??

Mathematics

2 answers:

Angelina_Jolie [31]2 years ago

5 0

Answer:

a. 14 Cubic Units

Step-by-step explanation:

This is a complex shape made by a 2x3 and a 2x4 shape

The volume of each of them are 6 cubic units and 8 cubic units respectively.

Together they make a complex shape with the volume of 14 cubic units

lbvjy [14]2 years ago

4 0

Answer:

Step-by-step explanation:

i counted all of the cubes and came up with 14 hence 14 is the answer

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xenn [34]

Answer:

Yes by SAS

Step-by-step explanation:

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2 years ago

A farmer needs to construct a square fence with an area of 121 square feet to hold farm animals. How long does each side of the

erastovalidia [21]

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2 years ago

What is the are of a triangle if the B=12 <br> and the H=4?

Anuta_ua [19.1K]

24 units because 12*4=48 and 48/2=24

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3 years ago

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I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

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1 year ago

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natka813 [3]

E two sides added together must be larger then the last side

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2 years ago

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