How do you convert 3/20 into a decimal and percent?

percibaa8
2021-12-17
Answered

How do you convert 3/20 into a decimal and percent?

You can still ask an expert for help

asked 2022-06-13

Expressing $\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ with rational Denominator

asked 2022-08-21

Prove inequality $\sum \frac{{a}^{3}-{b}^{3}}{{\left(a-b\right)}^{3}}\ge \frac{9}{4}$

Given $a,b,c$ are positive number. Prove that

$\frac{{a}^{3}-{b}^{3}}{{(a-b)}^{3}}+\frac{{b}^{3}-{c}^{3}}{{(b-c)}^{3}}+\frac{{c}^{3}-{a}^{3}}{{(c-a)}^{3}}\ge \frac{9}{4}$

$\iff \sum \frac{3(a+b{)}^{2}+(a-b{)}^{2}}{(a-b{)}^{2}}\ge 9$

$\iff \frac{(a+b{)}^{2}}{(a-b{)}^{2}}+\frac{(b+c{)}^{2}}{(b-c{)}^{2}}+\frac{(c+a{)}^{2}}{(c-a{)}^{2}}\ge 2$

Which

$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}=-1$

Use ${x}^{2}+{y}^{2}+{z}^{2}\ge -2(xy+yz+zx)$ then inequality right

I don't know why use ${x}^{2}+{y}^{2}+{z}^{2}\ge -2(xy+yz+zx)$ then inequality right?

P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$

We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1)=>xy+yz+zx=-1$

Given $a,b,c$ are positive number. Prove that

$\frac{{a}^{3}-{b}^{3}}{{(a-b)}^{3}}+\frac{{b}^{3}-{c}^{3}}{{(b-c)}^{3}}+\frac{{c}^{3}-{a}^{3}}{{(c-a)}^{3}}\ge \frac{9}{4}$

$\iff \sum \frac{3(a+b{)}^{2}+(a-b{)}^{2}}{(a-b{)}^{2}}\ge 9$

$\iff \frac{(a+b{)}^{2}}{(a-b{)}^{2}}+\frac{(b+c{)}^{2}}{(b-c{)}^{2}}+\frac{(c+a{)}^{2}}{(c-a{)}^{2}}\ge 2$

Which

$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}=-1$

Use ${x}^{2}+{y}^{2}+{z}^{2}\ge -2(xy+yz+zx)$ then inequality right

I don't know why use ${x}^{2}+{y}^{2}+{z}^{2}\ge -2(xy+yz+zx)$ then inequality right?

P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$

We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1)=>xy+yz+zx=-1$

asked 2021-12-10

How to change 1.25 into a fraction?

asked 2022-08-16

Factorial equality $\text{}\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{2}^{k}2!}$$\text{}=\frac{(2k)!}{{2}^{k}{2}^{k}k!k!}$

In a generating function identity proof in my textbook there is a step that I can't wrap my head around.

$\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{2}^{k}2!}$

$=\frac{(2k)!}{{2}^{k}{2}^{k}k!k!}$

How does one get from the left side of the equation to the right side? Is there an intuitive explanation as for why this makes sense?

In a generating function identity proof in my textbook there is a step that I can't wrap my head around.

$\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{2}^{k}2!}$

$=\frac{(2k)!}{{2}^{k}{2}^{k}k!k!}$

How does one get from the left side of the equation to the right side? Is there an intuitive explanation as for why this makes sense?

asked 2021-09-24

I am having trouble with these type of problems. $\frac{3}{5}(9y-4)+\frac{9}{16}(15y+12)$ I cant figure out how to simplify with fractions?

asked 2022-07-07

Among the following, which is closest to $\sqrt{0.016}$?

Among the following, which is closest in value to $\sqrt{0.016}$?

A. $0.4$

B. $0.04$

C. $0.2$

D. $0.02$

E. $0.13$

My Approach:

$(\frac{16}{1000}{)}^{\frac{1}{2}}=(\frac{4}{250}{)}^{\frac{1}{2}}=\frac{2}{5\cdot \sqrt{10}}=\frac{\sqrt{2}}{5\cdot \sqrt{5}}$

But I don't know how to reach to answer.

Thank You!

Among the following, which is closest in value to $\sqrt{0.016}$?

A. $0.4$

B. $0.04$

C. $0.2$

D. $0.02$

E. $0.13$

My Approach:

$(\frac{16}{1000}{)}^{\frac{1}{2}}=(\frac{4}{250}{)}^{\frac{1}{2}}=\frac{2}{5\cdot \sqrt{10}}=\frac{\sqrt{2}}{5\cdot \sqrt{5}}$

But I don't know how to reach to answer.

Thank You!

asked 2022-05-27

Simplify $\frac{5x}{{x}^{2}-x-6}+\frac{4}{{x}^{2}+4x+4}$

How come the answer is left as $\frac{5x}{(x+2)(x-3)}+\frac{4}{(x+2{)}^{2}}$. Why don't we go any further?

How come the answer is left as $\frac{5x}{(x+2)(x-3)}+\frac{4}{(x+2{)}^{2}}$. Why don't we go any further?