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Ira Lisetskai [31]
3 years ago
15

What is the 46th term of the arithmetic sequence shown below? -1,-5,-9,-13,-17,-21,...

Mathematics
1 answer:
Molodets [167]3 years ago
7 0

Answer:

-181

Step-by-step explanation:

According to the arithmetic calculator, it should be this.

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Nina’s total trip to and from work is 14 miles. If she works 243 days this year, how many miles will she drive to and from work?
beks73 [17]

Answer:

3,402

Step-by-step explanation:

Step 1:

243 × 14

Answer:

3,402

Hope This Helps :)

4 0
4 years ago
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.

8 0
2 years ago
Two sides of a triangle have lengths of 16 cm and 17 cm. Which could be the length of the third side?
Goryan [66]

The sum of the lengths of any two sides of a triangle must be greater than to the length of the third side.


x - third side


if x ≤ 16 < 17 or 16 ≤ x < 17 then


x + 16 > 17 |-16

x > 1


x ∈ (1; 16]


if 16 < 17 ≤ x then


16 + 17 > x

33 > x


x ∈ [17; 33)


Answer: x ∈ (1; 16] ∪ [17; 33) → 1 < x ≤ 16 or 17 ≤ x < 33

8 0
4 years ago
If two lines are perpendicular, then they form?
Delvig [45]

Answer:

a right angle?

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Anybody know the answer to number 2 !!!
MrRa [10]

Answer:

Well I can't put it on a number line but the order they should go in is: 2 \frac{1}{4}, 2.3 (repeating), 2.5,\sqrt{8}

Step-by-step explanation:

5 0
3 years ago
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