$$2y\le \frac{{y}^{2}}{{x}^{2}}+{x}^{2}$$

Hana Buck
2022-09-23
Answered

Prove that for any two real numbers $x,y$ with $x\ne 0$

$$2y\le \frac{{y}^{2}}{{x}^{2}}+{x}^{2}$$

$$2y\le \frac{{y}^{2}}{{x}^{2}}+{x}^{2}$$

You can still ask an expert for help

Kaiden Stevens

Answered 2022-09-24
Author has **12** answers

Hint:

Explore the following expression ${(x-\frac{y}{x})}^{2}$

Explore the following expression ${(x-\frac{y}{x})}^{2}$

mundocromadomg

Answered 2022-09-25
Author has **3** answers

By AM-GM we obtain:

$$\frac{{y}^{2}}{{x}^{2}}+{x}^{2}\ge 2\sqrt{\frac{{y}^{2}}{{x}^{2}}\cdot {x}^{2}}=2|y|\ge 2y.$$

Done!

$$\frac{{y}^{2}}{{x}^{2}}+{x}^{2}\ge 2\sqrt{\frac{{y}^{2}}{{x}^{2}}\cdot {x}^{2}}=2|y|\ge 2y.$$

Done!

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Solve the equation

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How can I prove that this function is bounded using the given hint?

$$f(x)=\frac{1-2\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x},$$

Hint:

1. $|A+B|\le |A|+|B|,$

2. $|\mathrm{sin}x|\le 1.$

$$f(x)=\frac{1-2\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x},$$

Hint:

1. $|A+B|\le |A|+|B|,$

2. $|\mathrm{sin}x|\le 1.$

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perform the indicated operations and express the answers in simplified form.

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I read in a set of memoranda that if $\frac{b-x}{x}=\frac{b}{a}$, then

$x=\frac{ab}{a+b}$

How is this true? I tried working it out but I could not understand. Please help.

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${2}^{\frac{2}{5}}$

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I apologise for asking something so fundamental, but what exactly does

${2}^{\frac{2}{5}}$

actually mean? I get raising a whole number to another whole number

${x}^{y}$

means you are multiplying x with itself y times, but what does it mean when y is a fraction?

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How to solve this proportion

Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.

$\frac{6.0\times {10}^{-6}}{{x}^{2}}=\frac{2.0\times {10}^{-6}}{(x-20{)}^{2}}$

What's confusing me is the solution manual to this problem lists writing the proportion as,

$\frac{(x-20{)}^{2}}{{x}^{2}}=\frac{2.0\times {10}^{-6}}{6.0\times {10}^{-6}}$

and then proceeds to solve the problem from there... but that doesn't seem right to me. Usually you would cross multiply a proportion and solve, but they seemed to do some illegal math or something. Could you guys work me through how to solve this? This answer is 47 by the way.

Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.

$\frac{6.0\times {10}^{-6}}{{x}^{2}}=\frac{2.0\times {10}^{-6}}{(x-20{)}^{2}}$

What's confusing me is the solution manual to this problem lists writing the proportion as,

$\frac{(x-20{)}^{2}}{{x}^{2}}=\frac{2.0\times {10}^{-6}}{6.0\times {10}^{-6}}$

and then proceeds to solve the problem from there... but that doesn't seem right to me. Usually you would cross multiply a proportion and solve, but they seemed to do some illegal math or something. Could you guys work me through how to solve this? This answer is 47 by the way.