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3 years ago

The surface area "S" of a cube with side length e is S= 6e^2.

Mathematics

1 answer:

LuckyWell [14K]3 years ago

8 0

Just plug in the numbers.
S=6e^2
150=6e^2 divide by 6 to both sides
25=e^2 take the square root of both sides
e=5

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Use the tax tables to determine the tax for the given filing status and taxable income amount:

NeX [460]

Answer:

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2 years ago

ASAP brainliest if right

kifflom [539]

Answer:

option A

Step-by-step explanation:

y = k . f( x )

• if k > 1, the graph of y = k•f (x) is the graph of f (x) vertically stretched by multiplying each of its y-coordinates by k.

• if 0 < k < 1 (a fraction), the graph is f (x) vertically shrunk (or compressed) by multiplying each of its y-coordinates by k.

• if k < 0, the vertical stretch or shrink is followed by a reflection across the x-axis.

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2 years ago

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marusya05 [52]

Answer:

See below

(B) and (C) are correct.

Step-by-step explanation:

We have the following limit

$\lim \limits_{n\rightarrow \infty} \left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} }, \forall x>0$

I am not sure about methods concerning the quotient, but in this type of question I would try to convert this limit into integration.

Considering the numerator, we have

$(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right) = \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)$

- I didn't forget about $n^n$

Considering the denominator, we have

$(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right) = \prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)$

- I didn't forget about $n!$

Therefore,

$\left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} } = \left(\dfrac{n^2 \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)}{ n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)} \right)$

$= \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

Now we have

$\lim \limits_{n\rightarrow \infty} \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}, \forall x>0$

This is just the notation change so far.

What I want to do here is apply definite integrals using Riemann Integrals (We will write the limit as an definite integral). A nice way to do it is using logarithms. Therefore, we can apply the natural logarithm in both sides.

Now, recall two properties of logarithms:

$\boxed{\log_a mn = \log_a m + \log_a n}$

$\boxed{\log_a m^p = p\log_a m}$

$\boxed{\log_a \left(\dfrac{m}{n} \right) = \log_a m- \log_a n}$

Thus,

$\ln f(x) = \lim \limits_{n\rightarrow \infty} \ln \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n}\ln\left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln \left(n^n \prod_{k=1}^n \left(x+\dfrac{n}{k} \right) \right)-\ln \left( n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right) \right) \right]$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln n^n + \prod_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)-\ln n! -\prod_{k=1}^n \ln\left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Considering

$\lim \limits_{n\rightarrow \infty} \frac{x}{n} (\ln n^n - \ln n! ) = \lim \limits_{n\rightarrow \infty} \frac{x}{n} (n\ln n - \ln n! )= \lim \limits_{n\rightarrow \infty} \frac{x \cdot\ln\frac{n^n}{n!} }{n}$

Using Stirling's formula

$\dfrac{n^n}{n!}\underset{\infty}{\sim} \dfrac{n^n}{\sqrt{2n \pi}\left(\dfrac{n}{e}\right)^n}=\dfrac{e^n}{\sqrt{2n \pi}}$

then

$\ln\left(\frac{n^n}{n!}\right)\underset{\infty}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right)$

$\implies \frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$

This shows our limit equals 1 as $\frac{\log(2\pi n)}{2n} \rightarrow 0$ and $\ln(e)=1$

Employing a Riemann sum in the main limit, we have

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[ \sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) - \sum_{k=1}^n\ln \left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Now dividing the terms inside the parenthesis by $\dfrac{n}{k}$ in $\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)$

we have

$\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) = \sum_{k=1}^n \ln \left(\frac{kx}{n} +1\right)$

Now dividing the terms inside the parenthesis by $\dfrac{n^2}{k^2}$ in $\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right)$

we have

$\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right) = \sum_{k=1}^n \ln \left(\frac{(kx)^2}{n^2} +1\right)$

Therefore

$= \frac xn\sum_{k=1}^n\ln\dfrac{z+1}{z^2+1}$

for $\dfrac{kx}{n} = z$

Using Riemann Integral,

$\lim \limits_{n\rightarrow \infty} \int_0^1\ln\frac{z+1}{z^2+1}dz$

From

$\frac{f'(x)}{f(x)}=\ln\frac{z+1}{z^2+1}$

We can see that the function is increasing for , but because of the denominator, it is negative for .

Therefore,

(A) is false because $\dfrac{1}{2} < 1$

(B) is true because

(D) is false, just consider $\ln\frac{z+1}{z^2+1}$ for $z=1$ and $z=2$

7 0

2 years ago

) Trong số 45 học sinh của lớp 10A có 15 bạn xếp loại học lực giỏi, 20 bạn xếp loại hạnh kiểm tốt, trong đó có 10 bạn vừa học lự

pychu [463]

Answer:

20 học sinh

Step-by-step explanation:

8 0

2 years ago

Which of the following values cannot be probabilities? 3/5, 5/3, 0, 1, 1.15, StartRoot 2 , -0.57, 0.09

Ugo [173]

Answer:

$\sqrt{2},-0.57$

Step-by-step explanation:

Probability of any incidence can only be a rational number.

Porbability can never be less than zero. It is basically defined as the ratio of

chance of occurence of an event to the total possible ways of occurence of the event. So an event cannot occur and the probability can be zero but never negative.

$\sqrt{2}$ being a irrational number cannot be a possibkle vlue for probability.

Hence values which cannot be possible are $\sqrt{2},-0.57$ .

7 0

3 years ago