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## What is \frac { \frac { 1} { 3} } { \frac { 3} { 9} }?

Tony B
Featured 2 months ago

A different approach. Tends not to be used as the shortcut method is very much faster. Explained why the shortcut method works.

$\frac{1}{3} \div \frac{3}{9} = 1$

#### Explanation:

A fractions structure is such that you have:

$\left(\text{count")/("size indicator")" "->" "("numerator")/("denominator}\right)$

You can not $\underline{\text{directly divide the counts}}$ unless the size indicators are the same.

.......................................................................................................................
The shortcut method adopts a sort of 'indirect division' in that first it divides the counts then multiplies by a conversion factor.
.......................................................................................................................
$\textcolor{b l u e}{\text{Using first principles}}$

Initial condition: $\to \frac{1}{3} \div \frac{3}{9}$

Multiply by 1 and you do not change the value of the fraction. However, 1 comes in many forms.

$\textcolor{g r e e n}{\left[\frac{1}{3} \textcolor{red}{\times 1}\right] \div \frac{3}{9} \text{ "->" } \left[\frac{1}{3} \textcolor{red}{\times \frac{3}{3}}\right] \div \frac{3}{9}}$

$\text{ } \left[\frac{3}{9}\right] \div \frac{3}{9}$

Now the 'size indicators' (denominators) are the same you can directly divide the counts;

$\text{ } 3 \div 3 = 1$

$\textcolor{g r e e n}{\text{ Try this with different values. It really does work!}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Short cut method}}$

$\text{ } \frac{\textcolor{g r e e n}{1}}{\textcolor{b r o w n}{3}} \div \frac{\textcolor{g r e e n}{3}}{\textcolor{b r o w n}{9}}$

Turn the divisor upside down and multiply:

$\text{ } \frac{\textcolor{g r e e n}{1}}{\textcolor{b r o w n}{3}} \times \frac{\textcolor{b r o w n}{9}}{\textcolor{g r e e n}{3}}$

This is the same as:

$\text{ } \textcolor{g r e e n}{\frac{1}{3}} \times \textcolor{b r o w n}{\frac{9}{3}}$

$\text{ "color(green)(1/3)" "xx" } \textcolor{b r o w n}{\frac{9}{3}}$

$\text{ "uarr" } \uparrow$

$\text{Division of counts conversion factor}$

$\text{ } = 1$

## How do you solve x/6=8?

Tony B
Featured 1 month ago

$x = 48$

#### Explanation:

$\textcolor{b l u e}{\text{Shortcut method that jumps steps from first principle method}}$

$\textcolor{b r o w n}{\text{Shortcut methods are much faster than first principles.}}$$\textcolor{b r o w n}{\text{Which is why people use them. However the shortcut outcome is}}$$\textcolor{b r o w n}{\text{based on the outcome of the first principle method.}}$

$\frac{x}{6} = 8$

The 6 has the process of divide applied to it in that
$\frac{x}{6} = x \times \frac{1}{6} = x \div 6$

As the 6 has divide applied to it on the left of = it becomes the opposite of multiply when we move it to the other side of the =. So we have:

$x \div 6 = 8 \text{ "->" } x = 8 \times 6$

$x = 48$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{3 very important facts used in first principle method}}$

1. An equation by definition is stating that what is on one side of = has the same value as on the other side. It is just that they may look different. So if you change one side in some way you have to change the other side in exactly the same way

2. For multiply and divide: to move something to the other side of the = change it to 1 as $\textcolor{b r o w n}{1 \times \text{something = something}}$.

3. For add and subtract: to move something to the other side of = change it to 0 as $\textcolor{b r o w n}{0 + \text{ something = something}}$.
...................................................................................................................
$\textcolor{b l u e}{\text{Answering your question using first principles}}$

$\frac{x}{6} = 8$

$\frac{x}{6}$ is the same as $\frac{1}{6} \times x$

So to get $x$ on its own we need to change $\frac{1}{6} \text{ into } 1$

Multiply both sides by $\textcolor{red}{6}$

$\textcolor{g r e e n}{\frac{x}{6} = 8 \text{ "->" "color(red)(6xx)x/6" "=" } \textcolor{red}{6 \times} 8}$

$\text{ "color(green)((color(red)(6))/6 xx x" "=" } 48$

" "color(green)(1xx x" "=" "48)

But $1 \times x \text{ gives just } x$

$\text{ "color(green)(x" "=" } 48$

## Answer is 4 which order to solve? 3^0 (5^0 - 6^-1 * 3)/ 2^-1 I don't know if I should bring the 6^-1 down first or multiply by 3 first.

sente
Featured 3 weeks ago

${3}^{0} \left(\frac{\left({5}^{0} - {6}^{- 1} \cdot 3\right)}{2} ^ \left(- 1\right)\right) = 1$

#### Explanation:

Following the order of operations, with the parentheses made explicit:

First, perform any operations within parentheses.

${3}^{0} \textcolor{red}{\left(\frac{\left({5}^{0} - {6}^{- 1} \cdot 3\right)}{2} ^ \left(- 1\right)\right)}$

Within those parentheses, we treat numerators and denominators as having parentheses around them, and so perform operations within those first.

$= {3}^{0} \left(\frac{\textcolor{red}{\left({5}^{0} - {6}^{- 1} \cdot 3\right)}}{2} ^ \left(- 1\right)\right)$

Evaluate any exponents. Recall that if $x \ne 0$, then ${x}^{0} = 1$ and that ${x}^{-} a = \frac{1}{x} ^ a$.

$= {3}^{0} \left(\frac{\left(\textcolor{red}{{5}^{0}} - {6}^{- 1} \cdot 3\right)}{2} ^ \left(- 1\right)\right)$

$= {3}^{0} \left(\frac{\left(1 - \textcolor{red}{{6}^{- 1}} \cdot 3\right)}{2} ^ \left(- 1\right)\right)$

$= {3}^{0} \left(\frac{\left(1 - \textcolor{red}{\frac{1}{6} ^ 1} \cdot 3\right)}{2} ^ \left(- 1\right)\right)$

$= {3}^{0} \left(\frac{\left(1 - \frac{1}{6} \cdot 3\right)}{2} ^ \left(- 1\right)\right)$

Perform any multiplication or division, going left to right.

$= {3}^{0} \left(\frac{\left(1 - \textcolor{red}{\frac{1}{6} \cdot 3}\right)}{2} ^ \left(- 1\right)\right)$

$= {3}^{0} \left(\frac{\left(1 - \textcolor{red}{\frac{3}{6}}\right)}{2} ^ \left(- 1\right)\right)$

$= {3}^{0} \left(\frac{\left(1 - \frac{1}{2}\right)}{2} ^ \left(- 1\right)\right)$

Perform any addition or subtraction, going left to right.

$= {3}^{0} \left(\frac{\left(\textcolor{red}{1 - \frac{1}{2}}\right)}{2} ^ \left(- 1\right)\right)$

$= {3}^{0} \left(\frac{\frac{1}{2}}{2} ^ \left(- 1\right)\right)$

All operations in the numerator have been completed. Moving to the denominator, we have an exponent to evaluate.

$= {3}^{0} \left(\frac{\frac{1}{2}}{\textcolor{red}{{2}^{- 1}}}\right)$

$= {3}^{0} \left(\frac{\frac{1}{2}}{\textcolor{red}{\frac{1}{2} ^ 1}}\right)$

$= {3}^{0} \left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)$

We now perform the remaining division. Recall that any nonzero number divided by itself is $1$.

$= {3}^{0} \left(\textcolor{red}{\frac{\frac{1}{2}}{\frac{1}{2}}}\right)$

$= {3}^{0} \left(1\right)$

All operations within parentheses have been evaluated. Going back, we now evaluate the remaining exponents.

$= \textcolor{red}{{3}^{0}} \cdot 1$

$= 1 \cdot 1$

And finally, we perform the remaining multiplication.

$= 1$

## How do you solve -\frac { 2} { 5} = - \frac { 2} { 3} + s?

Jim G.
Featured 2 weeks ago

$s = \frac{4}{15}$

#### Explanation:

Isolate s by adding $\frac{2}{3}$ to both sides of the equation.

Adding the same value to both sides retains the 'balance' of the equation.

$- \frac{2}{5} \textcolor{red}{+ \frac{2}{3}} = \cancel{- \frac{2}{3}} \textcolor{red}{\cancel{+ \frac{2}{3}}} + s$

$\Rightarrow s = - \frac{2}{5} + \frac{2}{3} \leftarrow \textcolor{m a \ge n t a}{\text{reversing the equation}}$

Before we can add these fractions we require them to have a
$\textcolor{b l u e}{\text{common denominator}}$

Multiplying the numerator/denominator of the first fraction by 3 and multiplying the numerator/denominator of the second fraction by 5, gives a common denominator.

$\Rightarrow s = \left(- \frac{2}{5} \times \frac{3}{3}\right) + \left(\frac{2}{3} \times \frac{5}{5}\right)$

$\textcolor{w h i t e}{\times \times} = - \frac{6}{15} + \frac{10}{15} \leftarrow \textcolor{red}{\text{ common denominator}}$

Now the denominators are the same we can add the numerators.

$\Rightarrow s = \frac{- 6 + 10}{15} = \frac{4}{15}$

## If the ratio of boys to girls is 3:2 and there are 25 students in a class, how do you make equal ratios to show how many students in the class are boys and how many are girls?

Tony B
Featured 2 weeks ago

The explanation is much longer than doing the mathematics.

The ratio of 15 boys to 10 girls is equivalent to the ratio of 3:2

#### Explanation:

Consider the starting point:
We have 3 boys and 2 girls. This gives a total count of 5

So we need to see how many lots of 5 will fit into 25.

5 lots of 5 gives 25

So we have 5 lots of the ratio 3:2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Method 1}}$

5 lots of 3:2 $\to 5 \times \left(3 : 2\right) = \left(5 \times 3\right) : \left(5 \times 2\right) = 15 : 10$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Method 2}}$

Write the ratio in fractional form

$\textcolor{m a \ge n t a}{\text{We can do this as long as we do not view and treat it like a fraction}}$

It does not matter in this case which we put on the top. I chose:

$\left(\text{boys")/("girls}\right) \to \frac{3}{2}$

Multiply a value (or system) by 1 and you do not change the value. However, 1 comes in many forms.

$\textcolor{g r e e n}{\frac{3}{2} \textcolor{red}{\times 1} \text{ "->" } \frac{3}{2} \textcolor{red}{\times \frac{5}{5}}}$

" "=color(green)((3color(red)(xx5))/(2color(red)(xx5))

" "=15/10 =("boys")/("girls")

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The ratio of 15 boys to 10 girls is equivalent to the ratio of 3:2

$15 : 10 \equiv \left(15 \div 5\right) : \left(10 \div 5\right) = 3 : 2$

where $\equiv$ means equivalent to

## How do you evaluate -( \frac { 1} { 2} ) ^ { 2} + 3+ 2?

Gimpy C.
Featured 2 hours ago

$\frac{19}{4}$

#### Explanation:

$- {\left(\frac{1}{2}\right)}^{2} + 3 + 2$

We want to follow order of operations. The first thing to check is parentheses. There's a (1/2) in there, but that's as far as that can go. So, we check the next thing, which is exponents.

We must square both the numerator and denominator.

$- {1}^{2} / {2}^{2} + 3 + 2$

$- \frac{1}{4} + 3 + 2$

We are adding everything now, and that operation is communitive (we can do it in any order), but we have a fraction. Let's take care of the simple addition first.

$- \frac{1}{4} + 5$

With that out of the way, we need to make the 5 compatible by multiplying it by a factor of 1. Since our fraction is in fourths, we will multiply by $\frac{4}{4}$

$- \frac{1}{4} + 5 \cdot \frac{4}{4}$

$- \frac{1}{4} + \frac{20}{4}$

This can be written as

$\frac{20}{4} - \frac{1}{4}$

$\frac{19}{4}$

19 is prime, so we cannot do anything else. If we are not allowed to leave it as an improper fraction (larger numerator than denominator), we must convert to a mixed fraction.

To do this, we divide the numerator by the denominator to get the whole number, and the remainder is under the denominator unchanged.

$\frac{19}{4} = 4 r 3$

$4 \frac{3}{4}$

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