Given:

a)AA

b)SAS

c)SSS

d)ASA

sjeikdom0
2021-07-31
Answered

What similarity theorem is indicated by the figures at the right:

Given:

a)AA

b)SAS

c)SSS

d)ASA

Given:

a)AA

b)SAS

c)SSS

d)ASA

You can still ask an expert for help

irwchh

Answered 2021-08-01
Author has **102** answers

From the figure, in $\mathrm{\u25b3}PQR\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\mathrm{\u25b3}STU$ we have

$PQ\sim ST$

$QR\sim TU$

$RP\sim US$

$\Rightarrow \mathrm{\u25b3}PQR\sim \mathrm{\u25b3}STU$ (SSS Theorem)

Hence option (c) is the correct answer.

Hence option (c) is the correct answer.

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Find the volume of cylinder with base as the disk of unit radius in the xy plane centered at (1, 1, 0) and the top being the surface $z=((x-1{)}^{2}+(y-1{)}^{2}{)}^{3/2}.$.

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Find the volume of cylinder with base as the disk of unit radius in the xy plane centered at (1, 1, 0) and the top being the surface $z=((x-1{)}^{2}+(y-1{)}^{2}{)}^{3/2}.$.

I just knew that this problem uses triple integral concept but dont know how to start. I just need someone to suggest an idea to start. I will proceed then.

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I am trying to derive the entropy in normal distribution. Let $p(x)$ to be the probability density function of uniform normal distribution

$p(x)=\frac{1}{\sqrt{2\pi}\sigma}{e}^{-\frac{{x}^{2}}{2{\sigma}^{2}}}$

hence, by using integration by parts, we have

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}p(x)dx={x}^{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}p(x)dx-{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}2x({\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}p(x)dx)dx$

Because

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}p(x)dx=100\mathrm{\%}$

we have

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}p(x)dx={x}^{2}-{x}^{2}+C=C$

However, lots of relevant proofs online says that

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}p(x)dx={\sigma}^{2}$

Does anyone know the reason?

$p(x)=\frac{1}{\sqrt{2\pi}\sigma}{e}^{-\frac{{x}^{2}}{2{\sigma}^{2}}}$

hence, by using integration by parts, we have

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}p(x)dx={x}^{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}p(x)dx-{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}2x({\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}p(x)dx)dx$

Because

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}p(x)dx=100\mathrm{\%}$

we have

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}p(x)dx={x}^{2}-{x}^{2}+C=C$

However, lots of relevant proofs online says that

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}p(x)dx={\sigma}^{2}$

Does anyone know the reason?

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