How do you subtract $\frac{20}{33}-\frac{20}{33}$ ?

johnnyschile7da
2022-02-09
Answered

How do you subtract $\frac{20}{33}-\frac{20}{33}$ ?

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How do you write 0.4 as a fraction?

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How to solve for $x$ in $\sqrt{9+2x}-\sqrt{2x}=\frac{5}{\sqrt{9+2x}}$?

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Which step has the error? Describe what the error is

$\frac{5{r}^{2}}{2r+3}\xf7\frac{4r+6}{10}$

Step 1:$\frac{5{r}^{2}}{2r+3}\times \frac{2(2r+3)}{10}$

Step 2:$\frac{5{r}^{2}}{1}\times \frac{2}{10}=\frac{10{r}^{2}}{10}$

Step 3:$\frac{10{r}^{2}}{10}={r}^{2}$

Step 1:

Step 2:

Step 3:

asked 2021-09-26

1) Kareem lives $\frac{4}{10}$ of a mile from the mall. Write two equivalent fractions that show what fraction of a mile Kareem lives from the mall.

2) An electronics store sells a large flat screen television for 1,699. Last month, the store sold 8 of these television sets. About how much money did the televisions sell for?

2) An electronics store sells a large flat screen television for 1,699. Last month, the store sold 8 of these television sets. About how much money did the televisions sell for?

asked 2022-07-05

How did he get the fraction with fraction power?

So we have a simple equation that is from Kepler.

${\left(\frac{{\overline{r}}_{1}}{{\overline{r}}_{2}}\right)}^{3}={\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2}$

In an explanation of a physics book, you can resolve for ${r}_{2}$ like this:

${r}_{2}={r}_{1}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2/3}$

And I found

${r}_{1}=\sqrt[3]{\frac{{T}_{1}^{2}}{{T}_{2}^{2}}{r}_{2}^{3}}$

First question, is my approach correct? My second and main question is how did he get the ${r}_{2}$ equation that I stated first. The physics book doesn't explain how to get from the main equation to ${r}_{2}={r}_{1}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2/3}$. Can someone explain me, please? (By the way, of course the equation for ${r}_{1}$ and ${r}_{2}$ should be different).

Thank you!

So we have a simple equation that is from Kepler.

${\left(\frac{{\overline{r}}_{1}}{{\overline{r}}_{2}}\right)}^{3}={\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2}$

In an explanation of a physics book, you can resolve for ${r}_{2}$ like this:

${r}_{2}={r}_{1}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2/3}$

And I found

${r}_{1}=\sqrt[3]{\frac{{T}_{1}^{2}}{{T}_{2}^{2}}{r}_{2}^{3}}$

First question, is my approach correct? My second and main question is how did he get the ${r}_{2}$ equation that I stated first. The physics book doesn't explain how to get from the main equation to ${r}_{2}={r}_{1}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2/3}$. Can someone explain me, please? (By the way, of course the equation for ${r}_{1}$ and ${r}_{2}$ should be different).

Thank you!

asked 2021-09-22

When adding or subtracting fractions you ? add the denominators.

asked 2022-06-08

Partial fractions - alternative result

I have the following fraction:

$\frac{z}{(z-1)(z-2)}$

When I try to decompose it to partial fractions, I get:

$-\frac{1}{z-1}+\frac{2}{z-2}$

But the result in my book is:

$-\frac{z}{z-1}+\frac{z}{z-2}$

Both results are correct, but, how am I supposed to get the second one?

I have the following fraction:

$\frac{z}{(z-1)(z-2)}$

When I try to decompose it to partial fractions, I get:

$-\frac{1}{z-1}+\frac{2}{z-2}$

But the result in my book is:

$-\frac{z}{z-1}+\frac{z}{z-2}$

Both results are correct, but, how am I supposed to get the second one?