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2 years ago

10

A pipe is at least 21 feet long and you want to cut it into 3 pieces. The second piece is to be twice as long as the first piece

and the third piece is to be 1 foot longer than the second piece. What is the minimum length of the first piece?

1 answer:

fgiga [73]2 years ago

6 0

Answer:

the minimum length of the first piece is 4 feet

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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

1 year ago

A number between 10 and 20 is both a foctor of 60 and<br>a multiple of 6.<br>What is the number?

Kisachek [45]

Answer:

12

Step-by-step explanation:

List out factors of 60 =

1 x 60 = 60

2 x 30 = 60

3 x 20 = 60

4 x 15 = 60

5 x 12 = 60

6 x 10 = 60

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 are factors of 60.

List out multiples of 6 from 10 to 20.

12 and 18.

Since 12 is seen on one of the factors of 60 and a multiple of 6 between 10 and 20, the answer is 12.

6 0

2 years ago

What is the slope of a line that is perpendicular to the line y=8x+5

statuscvo [17]

Answer:

-1/8

Step-by-step explanation:

Perpendicular slopes are negative reciprocals

The slope of y=8x+5 is 8, because it is in slope intercept form: y=mx+b form, where m is the slope.

Negative:

8/1--> -8/1

Reciprocal:

Flip the numerator and denominator

-8/1--> -1/8

So, the perpendicular slope of 8 is -1/8

4 0

2 years ago

Read 2 more answers

For what values of h and k does the linear system have infinitely many solutions?

Sophie [7]

Answer:

h=-32 and k=-24

Step-by-step explanation:

$4x_1 + 3x_2 = -1:a_1x_1 + b_1x_2=c_1\\hx_1 +kx_2=8:a_2x_1+b_2x_2=c_2$

For infinitely many solutions, lines must be same; $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\\=\frac{4}{h}=\frac{3}{k}=\frac{-1}{8}\\\\h=-32, k=-24$

7 0

2 years ago

The function f (x) has an average rate of change on the interval -2 < x < 3 of 3.4. If f (-2) = 9 then which of the follow

ad-work [718]

Answer:

The value of f(3)=12.4

Step-by-step explanation:

Average rate of change = 3.4

Interval: -2 < x < 3

f(-2) = 9

We need to find f(3)

The formula used to calculate Average rate of change is: $Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}$

here we have a=-2 and b=3

Now finding f(3)

$Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}\\3.4=\frac{f(3)-f(-2)}{3-2} \\3.4=\frac{f(3)-9}{1}\\3.4=f(3)-9\\f(3)=3.4+9\\f(3)=12.4$

So, The value of f(3)=12.4

3 0

2 years ago