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## What is the greatest common factor of 60 and 72?

Parzival S.
Featured 2 months ago

12

#### Explanation:

I like to answer these questions by first doing prime factorizations:

$60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5$
$72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2 \times 2 \times 2 \times 3 \times 3$

The GCF will have all the numbers that are common to both 60 and 72.

#GCF=?#

Let's start with 2's first. 60 has two 2s, while 72 has three. Two 2s are common to both, so our GCF will have two 2s:

#GCF=2xx2xx?#

Now 3s. 60 has one 3 and 72 has two 3s. One 3 is common to both, so we'll put that in:

#GCF=2xx2xx3xx?#

And lastly to 5s - 60 has one but 72 doesn't, so there's nothing common there. And so we stop here:

$G C F = 2 \times 2 \times 3 = 12$

$60 = 12 \times 5$
$72 = 12 \times 6$

## How do you solve #(- 2) + ( - 8) + ( - 6)#?

Tony B
Featured 1 month ago

$- 16$

#### Explanation:

Sometimes the idea about subtraction is a little thought provoking.

$\textcolor{b r o w n}{\text{Very important fact}}$

The natural way to count on the number line is increasingly positive. Often drawn so that it is towards the right. The number line is the same sort of thing as the x-axis on a graph.

$\textcolor{b r o w n}{\text{Always think of starting to count to the right on the number line.}}$
$\textcolor{b r o w n}{\text{The minus sign means change direction of count}}$

$+ \text{ " -> "Count right}$
$- \text{ "-> "Reverse direction count so you count left}$

$- - \text{ "->"Reverse direction of count twice. You end up}$
$\text{ going in the positive direction}$

$\left(- 2\right) \to$ Count to the left

$+ \left(- 8\right) \to$ + is count right but the - instructs change direction. So you end up counting left. So this is the same as just $- 8$

$+ \left(- 6\right) \to$+ is count right but the - instructs change direction. So you end up counting left. So this is the same as just $- 6$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

$\left(- 2\right) + \left(- 8\right) + \left(- 6\right) \text{ "->" } - 2 - 8 - 6 = - 16$

## How do you evaluate #frac { 4^ { 10} \cdot 8^ { - 3} \cdot 16^ { - 2} } { 32}#?

EZ as pi
Featured 1 month ago

$= \frac{1}{2} ^ 2$

$= \frac{1}{4}$

#### Explanation:

As soon as there are indices, you cannot cancel any of the bases.

Working out each factor to its actual value is not practical - the values are far too big.

Write each base as the product of its prime factors.
You might have noticed that all the bases are powers of 2.

$\textcolor{red}{4 = {2}^{2}}$
$\textcolor{b l u e}{8 = {2}^{3}}$
$\textcolor{p u r p \le}{16 = {2}^{4}}$
$\textcolor{f \mathmr{and} e s t g r e e n}{32 = {2}^{5}}$

$\frac{{\textcolor{red}{4}}^{10} \cdot {\textcolor{b l u e}{8}}^{-} 3 \cdot {\textcolor{p u r p \le}{16}}^{-} 2}{\textcolor{f \mathmr{and} e s t g r e e n}{32}}$

$= \frac{{\textcolor{red}{\left({2}^{2}\right)}}^{10} \cdot {\textcolor{b l u e}{\left({2}^{3}\right)}}^{-} 3 \cdot {\textcolor{p u r p \le}{\left({2}^{4}\right)}}^{-} 2}{\textcolor{f \mathmr{and} e s t g r e e n}{\left({2}^{5}\right)}}$

Multiply the indices: ${\left({x}^{m}\right)}^{n} = {x}^{m \times n}$

$= \frac{\textcolor{red}{{2}^{20}} \cdot \textcolor{b l u e}{{2}^{-} 9} \cdot \textcolor{p u r p \le}{{2}^{-} 8}}{\textcolor{f \mathmr{and} e s t g r e e n}{\left({2}^{5}\right)}}$

Add the indices of like bases: ${x}^{m} \times {x}^{n} = {x}^{m + n}$

$= {2}^{3} / {2}^{5}$

$= \frac{1}{2} ^ 2$

$= \frac{1}{4}$

## How do you solve #y+ \frac { 17} { 23} = - \frac { 5} { 23}#?

Jim G.
Featured 1 month ago

$y = - \frac{22}{23}$

#### Explanation:

To solve for y, subtract $\frac{17}{23}$ from both sides of the equation.

Doing the same operation to both sides retains the ' balance' of the equation.

$y \cancel{+ \frac{17}{23}} \textcolor{red}{\cancel{- \frac{17}{23}}} = - \frac{5}{23} \textcolor{red}{- \frac{17}{23}}$

Since the fractions on the right side have a $\textcolor{b l u e}{\text{common denominator}}$ we can add the numerators.
That is - 5 - 17 = - 22

$\Rightarrow y + 0 = - \frac{22}{23}$

$\Rightarrow y = - \frac{22}{23}$

$\textcolor{b l u e}{\text{Checking this solution}}$

If we substitute this value into the left side and it is equal to the right side then it is the solution.

$\text{left side " =-22/23+17/23=-5/23=" right side}$

$\Rightarrow y = - \frac{22}{23} \text{ is the solution}$

## On Monday, it snowed a total of 15 inches. On Tuesday and Wednesday, it snowed an additional 4.5 inches and 6.75 inches, respectively. A weather forecaster says that over the last three days, it snowed over 21.5 feet. Is this a valid claim?

Gimpy C.
Featured 1 month ago

Get data into same units and compute

#### Explanation:

We can see that we have mixed units where some values are inches, and some are feet.

To properly reason about the problem, we must get everything in the same units.

We might be tempted to turn everything into feet, but since a foot is bigger than an inch, we will have fractional answers. That might make it more difficult, so it will be easier to convert feet to inches.

To do this, we must recall that there are 12 inches in 1 foot.

If it snowed 21.5 feet, and each foot has 12 inches, then it snowed $21.5 \cdot 12 = 258$ inches.

OK, well that might be reasonable. Let's look at the rest of our data.

Monday was 15 inches.
Tuesday was 4.5 inches.
Wednesday was 6.75 inches.

Just by looking at these values, we can see that we won't get anywhere near 258 inches, but let's be thorough.

$15 + 4.5 + 6.75 = 26.26$ inches.

We now know the weather reporter lied to us, and since we already converted the units, we can find out by how much.

$258 - 26.26 = 231.74$ inches

How many feet? Well, we just reverse the process. We can divide by 12 inches to get the total feet.

$\frac{231.74}{12} = 19.311$ feet.

But realistically, I think this problem might be one of those glance-and-solve types.

You quickly add up the inches to get 26.26 inches, and you already know that 21.5 feet is a great deal bigger than that, so you immediately answer no, it's not valid.

## What is 5/8 +7/8 ?

Jim G.
Featured 3 weeks ago

$\frac{12}{8} = \frac{3}{2}$

#### Explanation:

To add 2 fractions we require the $\textcolor{b l u e}{\text{denominators}}$ to be the same value.

In this case they are, both 8

We can therefore $\textcolor{b l u e}{\text{add the numerators}}$ while leaving the denominator as it is.

$\Rightarrow \frac{5}{8} + \frac{7}{8}$

$= \frac{5 + 7}{8}$

$= \frac{12}{8}$

We can $\textcolor{b l u e}{\text{simplify}}$ the fraction by dividing the numerator/denominator by the $\textcolor{b l u e}{\text{highest common factor}}$ of 12 and 8, which is 4

#rArr12/8=(12÷4)/(8÷4)=3/2larrcolor(red)" in simplest form"#

This process is normally done using $\textcolor{b l u e}{\text{cancelling}}$

$\Rightarrow \frac{12}{8} = {\cancel{12}}^{3} / {\cancel{8}}^{2} = \frac{3}{2} \leftarrow \textcolor{red}{\text{ in simplest form}}$

A fraction is in $\textcolor{b l u e}{\text{simplest form}}$ when no other factor but 1 divides into the numerator/denominator.

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