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Orlov [11]
3 years ago
14

2 pyramids have a volume of 8 and the larger volume is 343 what is the scale factor

Mathematics
1 answer:
Mice21 [21]3 years ago
7 0

Answer

i think its 42.875

step-by-step explanation:

divide 343 by 8 i think

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Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]​
Schach [20]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

\rm :\longmapsto\:n +  {n}^{2} +  {n}^{3}  +  -  -  -  +  {n}^{n}

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

\rm \:  =  \: \dfrac{n( {n}^{n}  - 1)}{n - 1}

\rm \:  =  \: \dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }

Now, Consider Denominator, we have

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {n}^{n}

can be rewritten as

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {(n - 1)}^{n} +   {n}^{n}

\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

Now, Consider

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

So, on substituting the values evaluated above, we get

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{ {n}^{n}  - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

Now, we know that,

\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x}  =  {e}^{k}}}}

So, using this, we get

\rm \:  =  \: \dfrac{1}{1 +  {e}^{ - 1}  + {e}^{ - 2} +  -  -  -  -  \infty }

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

\rm \:  =  \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }

\rm \:  =  \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }

\rm \:  =  \: \dfrac{1}{\dfrac{e}{e - 1} }

\rm \:  =  \: \dfrac{e - 1}{e}

\rm \:  =  \: 1 - \dfrac{1}{e}

Hence,

\boxed{\tt{ \displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} } =  \frac{e - 1}{e} = 1 -  \frac{1}{e}}}

3 0
2 years ago
Please help me!! Will get brainliest!
makvit [3.9K]
Completing the square follows the principle of taking a x^{2} +bx+c and converting it into (x+ \frac{b}{2})^2+d where d is the 'correctional number' as I like to call it - i.e. the number that converts the expanded bracket into the +c, since the expanded bracket will give us a x^{2} +b.

In this case, 2/2=1 so we have the first part: (w+1)^2.
Expanding this gives us w^2+2w+1. We need c to be 9, so we can just add 8.
Putting this together:

w^2+2w+9=(w+1)^2+8

Now we can solve it more easily.
Rearranging:
(w+1)^2=-8 \\ w+1= +-\sqrt{-8} = +-\sqrt{8}i=+-2 \sqrt{2}i   \\ \boxed{w=-1+ 2\sqrt{2}i\ or\ -1-2 \sqrt{2}i  }
5 0
3 years ago
Answer choices<br> A. 23<br> B. 25<br> C. 35<br> D. 50
Tasya [4]
35? I’m not sure what the question is asking
3 0
3 years ago
Read 2 more answers
ILL GIVE EVERYTHING IF U GET IT RIGHTTT
Veronika [31]

Answer:

A

Step-by-step explanation:

Because if you did the math it's A

I'm 100% Sure

6 0
3 years ago
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Sandy typed 5 1/2 pages in 15 minutes. at this rate how many pages would sandy type per hour
Brums [2.3K]

Answer:

22 pages per hour.

Step-by-step explanation:

5.5/15 = 0.36666

0.36666 * 60 = 22

6 0
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