Edmund Adams
2021-11-06
Answered

To simplify the expression $2\sqrt{3}+5\sqrt{3}$ .

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Alicia Washington

Answered 2021-11-07
Author has **23** answers

Calculation:

When two similar irrational numbers are added or subtracted, the rational parts of the two numbers are added or subtracted.

Given$2\sqrt{3}+5\sqrt{3}$

$2\sqrt{3}+5\sqrt{3}=(2+5)\sqrt{3}$

$2\sqrt{3}+5\sqrt{3}=7\sqrt{3}$ .

When two similar irrational numbers are added or subtracted, the rational parts of the two numbers are added or subtracted.

Given

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Multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form.

$(x+y)}^{2$

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"A rational function is defined as the quotient of polynomials in which the denominator has a degree of at least 1"

If we are talking merely about x, then I get the concept. A rational function f(x) could be written as "

The issue that I'm having is that of talking about rational functions of n variables. For instance, what would be the meaning of 'f(x,y) is a rational function of

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Logarithm question-I donot know but this question may be solved by any other way also.

Let $({x}_{0},{y}_{0})$ be the solution of the following equations.

$(2x{)}^{\mathrm{ln}2}=(3y{)}^{\mathrm{ln}3}$

${3}^{\mathrm{ln}x}={2}^{\mathrm{ln}y}$

Then ${x}_{0}$ is

A) $\frac{1}{6}$

B) $\frac{1}{3}$

C) $\frac{1}{2}$

D) $6$

I have tried this problem by taking log on both sides of the two equations. But, finally I could not make up to get the values of $x$ and $y$

Let $({x}_{0},{y}_{0})$ be the solution of the following equations.

$(2x{)}^{\mathrm{ln}2}=(3y{)}^{\mathrm{ln}3}$

${3}^{\mathrm{ln}x}={2}^{\mathrm{ln}y}$

Then ${x}_{0}$ is

A) $\frac{1}{6}$

B) $\frac{1}{3}$

C) $\frac{1}{2}$

D) $6$

I have tried this problem by taking log on both sides of the two equations. But, finally I could not make up to get the values of $x$ and $y$

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Compute 837+179.

asked 2022-05-23

Let

$f(z)={\displaystyle \frac{z-a}{z-b}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z\ne b\ne a$

be a complex valued rational function.

How can I show that, if $|a|,|b|<1,$, then there is a complex number ${z}_{0}$ satisfying $|{z}_{0}|=1$ and $f({z}_{0})\in \mathbb{R}$ ?

I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that

$\frac{a-b}{z-b}}={\displaystyle \frac{\overline{a}-\overline{b}}{\overline{z}-\overline{b}}}.$

I could make a quadratic equation by using the fact that $\overline{z}={\displaystyle \frac{1}{z}},\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}z\in \mathrm{\partial}\mathbb{D}.$ Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.

Also, I would like to know that what happen if (both or atleast one) $a,b\notin \mathbb{D}.$Any comment or hint will be welcome. Thank you.

$f(z)={\displaystyle \frac{z-a}{z-b}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z\ne b\ne a$

be a complex valued rational function.

How can I show that, if $|a|,|b|<1,$, then there is a complex number ${z}_{0}$ satisfying $|{z}_{0}|=1$ and $f({z}_{0})\in \mathbb{R}$ ?

I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that

$\frac{a-b}{z-b}}={\displaystyle \frac{\overline{a}-\overline{b}}{\overline{z}-\overline{b}}}.$

I could make a quadratic equation by using the fact that $\overline{z}={\displaystyle \frac{1}{z}},\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}z\in \mathrm{\partial}\mathbb{D}.$ Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.

Also, I would like to know that what happen if (both or atleast one) $a,b\notin \mathbb{D}.$Any comment or hint will be welcome. Thank you.