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2 years ago

1÷1+cosa+1÷1_cos a please solve question

Mathematics

1 answer:

zloy xaker [14]2 years ago

4 0

ok so THIS IS UR ANSWER I THINK PLEASE MAKE ME BRAINLIEST

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Andrew and Sam start counting from 150 to 200 including. Andrew counts by fives, and Sam counts by threes. What is the ratio of

Agata [3.3K]

Answer:

11/17

Step-by-step explanation:

The question says to find the ratio by which Andrew counts to the ratio Sam counts.

Andrew counts by 5s:

150,155,160,165,170,175,180,185,190,195,200. His total count is 11

Sam counts by 3s:

150,153,156,159,162,165,168,171,174,177,180,183,186,189,192,195,198. His total number of count is 17

The ratio of their count is 11:17

= 11/17

3 0

3 years ago

Read 2 more answers

70 is 25% of what number

Rina8888 [55]

I hope this helps you

70=?.25%

70=?.25/100

70=?.1/4

?=280

3 0

3 years ago

What is the density of a material that had a mass of 2500 grams and a volume of 500 cubic centimeters,in grams per cubic centime

lorasvet [3.4K]

2500/500=5

2500 grams/500 cm^3 = 5 g/cm^3

6 0

3 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

1 year ago

How is 35 to 10 proportional to 7 to 5

KiRa [710]

If you simplify 35/10 both side divide by 7 is equal to 7/5

6 0

3 years ago

1÷1+cosa+1÷1_cos a please solve question​

1÷1+cosa+1÷1_cos a please solve question