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3 years ago

PLEASE HELP PLEASE PLEASE PLEASE

Mathematics

1 answer:

fiasKO [112]3 years ago

7 0

Answer:

11,880π in^3

Step-by-step explanation:

The volume of a cylinder = π r^2 h

Large Cylinder V = π * 15^2*40 = 9,000π in^3

Small Cylinder V = π * 12^2 * (60-40) = 2880π in^3

Adding: the total volume of the model = 11,880π in^3

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An angle measures 148° more than the measure of its supplementary angle. What is the measure of each angle? AND DONT PLAGERISE!!

Usimov [2.4K]

Answer:

Step-by-step explanation:

Supplementary angles add up to 180°.

“θ is 148° more than its supplement”

Supplement of θ = 180°-θ

θ = (180°-θ) + 148°

2θ = 328°

θ = 164°

Supplement of θ = 180°-164° = 16°

6 0

2 years ago

PLEASE HELP ME! :(

snow_lady [41]

Here, I'll do the first one for you.
When they are talking about a number, use "x".
Since it says twice a number, you say 2x. Or 3x for three times the number
2x+12=3x-31
Then use algebra to find x. Get the numbers on one side and all the x's on the other.
2x+12+31=3x-31+31
2x+43-2x=3x-2x
x=43

Now do the rest on your own!

5 0

3 years ago

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

7 times as much as sum of 1/3 and 4/5

Salsk061 [2.6K]

So the formula is
7*(1/3+1/5) = x

First find the common denominator of the fractions. In this case the common denominator of 3 and 5 is 15 so we must convert them to 15ths. To do that we divide 15 by the denominator and take that answer and multiply the numerator by it.

So for 1/3 we take 15/3(denominator) = 5 and multiply the numerator (1) by it

5 * 1 = 5 so 1/3 converts to 5/15

Do the same for 1/5 we take 15/5 = 3 and multiply the numerator (4) by it

3*4 = 12 so 4/5 converts to 12/15

Now we add 5/15 and 12/15 = 17/15 so we have our formula now to

7*17/15

We make seven a fraction 7/1 and multiply across 7*17 = 119 and 15 * 1 = 15

We have 119/15 which when we divide and get 7 14/15 as your answer

8 0

3 years ago

Read 2 more answers

A random sample is made of 256 middle managers concerning their annual incomes. The sample mean is computed to be $35,420. If th

ziro4ka [17]

Answer: a. $128.125

Step-by-step explanation:

The standard error of the mean is given by :

$SE=\dfrac{\sigma}{\sqrt{n}}$ , where $\sigma$ = population standard deviation , n= sample size.