# Make the internet a better place to learn

2

## How do you simplify -16+9-5?

CountryGirl
Featured 2 months ago

-12

#### Explanation:

Adding and subtracting negative numbers is a pretty tricky thing to learn (trust me, I know!), but if you go through it slowly and use a couple tricks, you can figure it out.
Here's our equation:
-16 + 9 - 5 = ?

The first two numbers we are going to deal with are $16$ and $9$. Now these aren't just normal $16$ and $9$, one is positive and one is negative. I just put some parentheses around them, so that we remember that:
$\left(- 16\right) + \left(+ 9\right)$

So the easiest way to learn how to do this type of problem is make a number line like this:
Obviously, this number line doesn't go far enough, but you could draw it to go all the way to $- 16$. Now put your finger at $- 16$ and count to the right 9 lines. Where is your finger at now? It should be at $- 7$.

( If you need extra explanation: You moved 9 spaces to the right because that is that direction positive numbers are. Since 9 is positive, you will get closer to 0, not further away.)

Okay! So now we have:
-7 - 5 = ?
(-7) + (-5) = ? (I added parentheses again)
Put your finger back on the $- 7$ and move to the left 5 spaces. Because $5$ is negative, we move to the left, or further away from 0.
You should now be at $- 12$, which is the answer of this equation.

2

## How you test 69,902 for divisibility by 2, 3, 5, 9, or 10?

Shwetank Mauria
Featured 2 months ago

$69902$ is divisible only by $2$ and not by $3 , 5 , 9$ or $10$. Please see below for details.

#### Explanation:

Divisibility by $2$ can be tested by checking the digit at unit's place. If we have $0 , 2 , 4 , 6$ or $8$ at unit's place, then the number is divisible by $2$. Here we have $2$ at unit's place in $69902$, and hence it is divisible by $2$.

Divisibility by $3$ can be tested by checking that sum of all the digits is divisible by $3$ or not. Here sum of digit's in $69902$ is $6 + 9 + 9 + 0 + 2 = 26$, which is not divisible by $3$, therefore $69902$ is not divisible by $3$.

Divisibility by $5$ can be tested by checking the digit at unit's place. If we have $0$ or $5$ at unit's place, then the number is divisible by $5$. Here we have $2$ at unit's place in $69902$, and hence it is not divisible by $5$.

Divisibility by $9$ can be tested by checking that sum of all the digits is divisible by $9$ or not. Here sum of digit's in $69902$ is $6 + 9 + 9 + 0 + 2 = 26$, which is not divisible by $9$, therefore $69902$ is not divisible by $9$. Note that if a number is not divisible by $3$, it is also not divisible by $9$.

Divisibility by $10$ can be tested by checking the digit at unit's place. If we have $0$ at unit's place, then the number is divisible by $10$. Here we have $2$ at unit's place in $69902$, and hence it is not divisible by $10$.

2

## How do you find the prime factorization of 196?

Shwetank Mauria
Featured 2 months ago

$196 = 2 \times 2 \times 7 \times 7$

#### Explanation:

As the last two digits in $196$ are divisible by $4$, $196$ too is divisible by $4$.

Dividing by $4$ we get $49$ and hence

$196 = 4 \times 49$

but factors of $4$ are $2 \times 2$ and that of $49$ are $7 \times 7$. Further $2 ' s$ and $7 ' s$ are prime numbers and cannot be factorized.

Hence prime factors of $196$ are

$196 = 2 \times 2 \times 7 \times 7$

Note : This method of factorization, in which we first find identifiable factors and then proceed until all prime factors are known is called tree method. This is graphically described below.

2

## How do you simplify 739div 41?

EZ as pi
Featured 2 months ago

$18.024$

#### Explanation:

Do long division ..

$\textcolor{w h i t e}{\ldots \ldots . l} 18.024$
$41 | \overline{739.000}$
$\textcolor{w h i t e}{l l l l m} \underline{41} \text{ } \leftarrow 73 \div 41 = 1 \mathmr{and} 1 \times 41 = 41$
$\textcolor{w h i t e}{l l l l m} 329 \text{ } \leftarrow 43 - 41 = 32$ and bring down the $9$
$\textcolor{w h i t e}{l l l l m} \underline{328} \text{ } \leftarrow 329 \div 41 = 8 \mathmr{and} 8 \times 41 = 328$
$\textcolor{w h i t e}{l l l l m l l l} 100 \text{ } \leftarrow 329 - 328 = 1$ and bring down the $00$
$\textcolor{w h i t e}{l l l l l l l l l m} \underline{82} \text{ } \leftarrow 100 \div 41 = 2 \mathmr{and} 2 \times 41 = 82$
$\textcolor{w h i t e}{l l l l l l l l l m} 180 \text{ } \leftarrow 100 - 82 = 18$ and bring down the $0$
$\textcolor{w h i t e}{l l l l l l l l l m} \underline{164} \text{ } \leftarrow 100 \div 41 = 4 \mathmr{and} 4 \times 41 = 164$

You can continue for as many decimal places as you wish.

Or simply use a calculator if you need the answer immediately.

2

## Which measurement is the greatest between 6 km, 60 m, 600 cm, and 6,000 mm?

NJ
Featured 1 month ago

$6 \text{ km}$

#### Explanation:

This is a metric prefix problem. Some of the metric prefixes are:

In this case, you have:

$\implies 6 \text{km" = 6 xx10^3 "m" = 6000 "m}$
$\implies 60 \text{m}$
$\implies 600 \text{cm" = 600 xx 10^(-2) "m" = 6 "m}$
$\implies 6000 \text{mm" = 6000 xx 10^(-3) "m " = 6 "m}$

We see that the largest is $6000 \text{m}$, which corresponds to our $6 \text{km}$.

2

## How do you use a number line to divide a fraction by a whole number?

Tony B
Featured 1 month ago

See the construction tips in the explanation.

#### Explanation:

$\textcolor{red}{\text{Measuring is counting}}$

$\textcolor{b l u e}{\text{Step 1}}$

Draw a number line ( A to B) of some easily divisible length. Perhaps 15 lots of $\frac{1}{2}$ cm. We will end up counting in ${15}^{\text{ths}}$

Draw the line BG of some length that is easily divided into 5 equal parts. The angle does not matter as long as it is sensible.

Draw the line GA. Then the parallel lines from F,E,D and C

This has provided a full set of ${5}^{\text{ths}}$ from A to B

Count $\frac{4}{5}$ from A towards B and mark that point (H).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Step 2}}$

In the same way as in Step 1 divide AH into 3 parts
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$

In the same way divide AB into ${15}^{\text{th}}$. You only need to subdivide AJ to make your point. You will have $\frac{4}{15}$ matching:

$\frac{4}{5} \div 3 \to \frac{4}{5} \times \frac{1}{3} = \frac{4}{15}$

2

## Which speed is the fastest ? A)18 feet in 20 minutes B)90 feet in 2.5 hours C)20 yards in 1.5 hours D)3 2/3 yards in 15 minutes

Sean
Featured 1 month ago

color(red)A) is the fastest.

#### Explanation:

.

In order to compare, we need to convert all four speeds to a common unit. Let's convert them all to feet per hour:

color(red)A) $18$ feet in $20$ minutes

There are $60$ minutes in each hour. We divide $18$ by $20$ to find the speed as feet per minute. Then we multiply the result by $60$ to get feet per hour:

$\frac{18}{20} \cdot 60 = \frac{9}{10} \cdot 60 = 9 \cdot 6 =$ $\textcolor{red}{\text{54 feet per hour}}$

color(red)B) We divide $90$ by $2.5$ to get feet per hour:

$\frac{90}{2.5} =$ $\textcolor{red}{\text{36 feet per hour}}$

color(red)C) There are $3$ feet in each yard.

$20$ yards is equal to $20 \cdot 3 = 60$ feet. We divide $60$ by $1.5$ to get feet per hour:

$\frac{60}{1.5} =$ $\textcolor{red}{\text{40 feet per hour}}$

color(red)D) $3 \frac{2}{3} = \frac{11}{3}$. We multiply $\frac{11}{3}$ yards by $3$ to get feet:

$\frac{11}{3} \cdot 3 = 11$ feet

We divide $11$ feet by $15$ to get feet per minute and multiply the result by $60$ to get feet per hour:

$\frac{11}{15} \cdot 60 =$ $\textcolor{red}{\text{44 feet per hour}}$

Therefore, $A$ is the fastest.

2

## What is the least common multiple of 7 and 24?

Tony B
Featured 1 month ago

A teacher will expect the prime number method. Just for the hell of it this is a different approach!

168

#### Explanation:

We have two numbers ; 24 and 7

I am going to count the 24's. However lets look at this value.

24 can be 'split' into a sum of 7's with a remainder. So each 24 consists of:

$24 = \left(7 + 7 + 7 + 3\right)$

If we sum columns of these we will get the 3 summing to a value into which 7 will divide exactly. When this happens we have found our least common multiple.

REMEMBER WE ARE COUNTING THE 24's

$\text{ count "color(white)("dd") "The 24's}$
$\textcolor{w h i t e}{\text{ddd") 1color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 2color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 3color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 4color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 5color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
$\textcolor{w h i t e}{\text{ddd") 6color(white)("ddd}} \left(\textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 7 + \textcolor{w h i t e}{.} 3\right)$
color(white)("ddd") 7color(white)("ddd")ul( (color(white)(.)7+color(white)(.)7+color(white)(.)7+color(white)(.)3)larr" Add"
$\textcolor{w h i t e}{\text{ddddddd.}} 49 + 49 + 49 + \underbrace{21}$
$\textcolor{w h i t e}{\text{dddddddddddddddddddd}} \downarrow$
color(white)("dddddddddddddd")" exactly divisible by 7"

We have a count of 7 so the value is $7 \times 24 = 168$

4

## If the temperature at 7am was -7 Celicas degrees and the temperature rose 8 Celicas degrees during the morning, what was the temperature by noon. Use a number line to illustrate your answer?

CountryGirl
Featured 1 month ago

$1$ Celsius.

#### Explanation:

Let's use a number line and start at $- 7$ Celsius.

Now since the temperature rose by 8, we know that it is going to move to the right towards the positive numbers. It is always the same on a number line:

$\text{moving right = increase}$
$\text{moving left = decrease}$

Put your finger on the $- 7$ and move $8$ spaces to the $\text{right}$.

You should be on the $1$.

So the temperature is now $1$ Celsius.

2

## What is the GCF and LCM of 150,175 and 200?

EZ as pi
Featured 2 weeks ago

$H C F = 25 \mathmr{and} L C M = 4 , 200$

#### Explanation:

Write each number as the product of its prime factors.

Then you see what are made up of and what they have in common:

$\text{ } 150 = 2 \textcolor{w h i t e}{\times \times \times} \times 3 \times \textcolor{b l u e}{5 \times 5}$
$\text{ } 175 = \textcolor{w h i t e}{\times \times x . \times \times} \times \textcolor{b l u e}{5 \times 5} \times 7$
" "200 = ul(2xx2xx2color(white)(x..x)xxcolor(blue)(5xx5)" ")

$\text{ } H C F = \textcolor{w h i t e}{\times \times x . \times \times x} \textcolor{b l u e}{5 \times 5} = 25$

$\text{ } L C M = 2 \times 2 \times 2 \times 3 \times 5 \times 5 \times 7 = 4 , 200$