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How do you write the prime factorization of #20st#?

EZ as pi
Featured 3 months ago

$20 s t = 2 \times 2 \times 5 \times s \times t$

Explanation:

You need to write the given term or expression as the product of its prime factors. This is sometimes also called expanded form.

$20 s t = 2 \times 2 \times 5 \times s \times t \text{ } \leftarrow$ these are all prime factors.

The variables are really easy to do because even if there are indices, you can see exactly how many of each there are.

So $36 {s}^{3} {t}^{2}$ would be expanded as:

$2 \times 2 \times 3 \times 3 \times s \times s \times s \times t \times t$

How do i solve equations with negative numbers? including adding, subtracting, dividing, and multiplying?

Gimpy C.
Featured 3 months ago

For a while, explicitly put the -1

Explanation:

Let us agree on this $- 5 = \left(5\right) \left(- 1\right)$

Anything multiplied by 1 is itself, and anything multiplied by (-1) is its opposite.

Let us agree on this $5 = \left(- 1\right) \left(- 1\right) \left(5\right)$

A positive number multiplied by (-1) results in that negative number, and a negative number multiplied by (-1) results in that positive number.

A positive number multiplied by (-1) an even number of times results in that positive number (no change).

A negative number multiplied by (-1) an even number of times results in that negative number (no change).

Multiplying by (-1) twice undoes the multiplication.

So, with this in mind, let us consider addition.

$- 2 + 3$

We can write this as

$\left(- 1\right) \left(2\right) + 3$

Addition has a property that allows us to do it in any order, and still get the same result. It's called the communitive property.

$a + b = b + a$

Well, we just turned our problem into an addition problem. That means we can rearrange the terms. So, let's do that:

$3 + \left(- 1\right) \left(2\right)$

According to order of operations, we must multiply before adding, so let's multiply that (-1):

$3 - 2$

So, it appears that an addition problem with a negative in front is really a subtraction problem in disguise.

Let's try another one:

$- 2 - 3$

We will again replace the negatives with (-1), but it is important to remember that we are adding. We always add, but sometimes we add negative numbers.

$\left(- 1\right) \left(2\right) + \left(- 1\right) \left(3\right)$

Both terms are being multiplied by (-1), which brings us to another property. The distributive property says:

$a b + a c = a \left(b + c\right)$

Let us pull out the (-1) in like fashion.

$\left(- 1\right) \left(2 + 3\right) = \left(- 1\right) \left(5\right) = - 5$

Finally, we have

$2 - 3$

Again, this can be thought of as:

$2 + \left(- 1\right) \left(3\right)$

Move the bigger number to the front

$\left(- 1\right) \left(3\right) + 2$

Multiply first

$- 3 + 2$

We can pull out a (-1) here too because anything multiplied by 1 is itself and anything multiplied by (-1) is its opposite, so positive becomes negative in that case.

$\left(- 1\right) \left(3 - 2\right)$

$= \left(- 1\right) \left(1\right)$

$= \left(- 1\right)$

Now, it would be silly to do all of this every time. You will very rapidly internalize these ideas, but hopefully this will help in thinking about it.

We have covered addition, multiplication, and subtraction. Division might be a little tricky, but I know you can get it.

$- \frac{2}{3} = \frac{- 2}{3} = \frac{2}{-} 3$

Let us see them more clearly:

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

Remember that an even number of (-1) produces no change.

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

So, with this knowledge, let's solve an equation.

$\frac{- 5 + 5 - 2 + 7 \cdot - 2}{-} 2$

$= \frac{\left(- 1\right) 5 + 5 + \left(- 1\right) 2 + 7 \cdot \left(- 1\right) 2}{\left(- 1\right) 2}$

$= \frac{\left(- 1\right) \left(5 - 5\right) + \left(- 1\right) 2 + \left(- 1\right) 14}{\left(- 1\right) 2}$

$= \frac{\left(\cancel{- 1}\right) \left(0\right) + \left(\cancel{- 1}\right) 2 + \left(\cancel{- 1}\right) 14}{\left(\cancel{- 1}\right) 2}$
Note:All terms contain (-1), so it is a common factor.

$= \frac{\left(0\right) + 2 + 14}{2}$

$= \frac{2 + 14}{2}$

$= \frac{16}{2}$

$= 8$

Get the hang of it, and then abandon it. It will just be automatic.

Pam spent #1/2# of her money in a department store. She spent #1/4# of her remaining money in a stationary store. After she spent 60 cents in the snack shop, she had no money left. How much money did she have originally?

Gimpy C.
Featured 3 months ago

We can work backwards and use reciprocal of the fraction

Explanation:

The reciprocal of the fraction is just the flipped version of that fraction. So, $\frac{1}{2}$ becomes 2. And $\frac{1}{4}$ becomes 4.

A fraction multiplied by the reciprocal is always 1.

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

The question gives a specific value of 60 cents as a result of spending $\frac{1}{4}$ of her money after spending $\frac{1}{2}$ of her money.

So, if we take the reciprocal of the fractions, we can put the money back. For example, we go from some amount to 60 cents, by spending $\frac{1}{4}$ of the money. Let's give the money back.

$60 \cdot 4 = 240$

Now, before this she spent half the money, so again we can take the reciprocal, which is 2.

$240 \cdot 2 = 480$

She started with 480 cents. However, we might want to express this as dollars. So, we can divide by 100 because there are 100 pennies in a dollar.

$\frac{480}{100} = 4.80$

Pam started with $4.80. We want to double check though since we might be wrong. Let's work through the problem forward and see if we get 60 cents again. First, use the value that is in cents because the answer is in cents: 480. Next, she spent half the money: $480 \cdot \left(\frac{1}{2}\right) = 240$Now, she spent a fourth of the money. $240 \cdot \left(\frac{1}{4}\right) = 60$And this checks out. So, we can be sure that Pam started with$4.80.

How do you evaluate #2+6(9-3^2)-2#?

John D.
Featured 1 month ago

The answer is $0$.

Explanation:

Use the order of operations $\left(\text{PEMDAS}\right)$.

$\textcolor{red}{\text{P}}$ = parentheses
$\textcolor{b l u e}{\text{E}}$ = exponents
$\textcolor{\mathmr{and} a n \ge}{\text{MD}}$ = multiplying and dividing
$\textcolor{v i o \le t}{\text{AS}}$ = adding and subtracting

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

$2 + 6 \textcolor{red}{\left(9 - {3}^{2}\right)} - 2 \textcolor{w h i t e}{\text{XXX}}$ Evaluate the $\textcolor{red}{\text{P}}$arentheses

within parentheses, we start over with $\text{PEMDAS}$

$\left(9 - {3}^{2}\right) \textcolor{w h i t e}{\text{XXXXXXX}}$ No $\textcolor{red}{\text{P}}$arentheses
$\left(9 - \textcolor{b l u e}{{3}^{2}}\right) \textcolor{w h i t e}{\text{XXxxxxxxx}}$ Evaluate the $\textcolor{b l u e}{\text{E}}$xponents
$\left(9 - 9\right) \textcolor{w h i t e}{\text{xxXXxxxxxx}}$ No $\textcolor{\mathmr{and} a n \ge}{\text{M}}$ultiplication or $\textcolor{\mathmr{and} a n \ge}{\text{D}}$ivision
$\left(\textcolor{v i o \le t}{9 - 9}\right) \textcolor{w h i t e}{\text{xxXXxxxxxx}}$ Evaluate the $\textcolor{v i o \le t}{\text{A}}$ddition and $\textcolor{v i o \le t}{\text{S}}$ubtraction
$= \left(0\right)$

$2 + 6 \left(0\right) - 2 \textcolor{w h i t e}{\text{XXXXXX}}$ No $\textcolor{b l u e}{\text{E}}$xponents

$2 + \textcolor{\mathmr{and} a n \ge}{6 \left(0\right)} - 2 \textcolor{w h i t e}{\text{XXXXXX}}$ Evaluate the $\textcolor{\mathmr{and} a n \ge}{\text{M}}$ultiplication and $\textcolor{\mathmr{and} a n \ge}{\text{D}}$ivision

$\textcolor{v i o \le t}{2 + 0 - 2} \textcolor{w h i t e}{\text{XXXXXXX/}}$ Evaluate the $\textcolor{v i o \le t}{\text{A}}$ddition and $\textcolor{v i o \le t}{\text{S}}$ubtraction

$= 0$

What is the lowest common multiple for 8, 12, and 15?

The Lonely Donut
Featured 1 month ago

$120$

Explanation:

The easiest way to do this is to find the prime factorization of each number first.

$8 = 4 \cdot 2$
$8 = 2 \cdot 2 \cdot 2$
$8 = {2}^{3}$

$12 = 6 \cdot 2$
$12 = 3 \cdot 2 \cdot 2$
$12 = 3 \cdot {2}^{2}$

$15 = 5 \cdot 3$

Now that we have reduced all numbers to their prime factorizations, we will look for the factors with the highest exponent and multiply them all together to get our LCM.

There is ${2}^{3}$ from $8$, ${3}^{1}$ from $12$ and $15$, and ${5}^{1}$ from $15$.

So now we must do the following:

${2}^{3} \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3 \cdot 5$
$120$

Therefore, the LCM of $8$, $12$, and $15$ is $120$.

Explain why you can multiply both the numerator and denominator by the same number to make an equivalent fraction? Draw a model to show an example.

Shwetank Mauria
Featured 6 hours ago

Explanation:

A fraction, say $\frac{a}{b}$, where $a$ is called as numerator and $b$ is called denominator, assuming $a < b$, represents a part of a whole object, wherein the object is divided in $b$ equal parts, of whom $a$ are chosen.

For example, in the figure below, shows a full bar divided into $12$ parts of which $9$ parts (coloured blue) have been chosen and they represent $\frac{9}{12}$ of the whole bar.

Let us consider a simple example, say $\frac{1}{2}$, which is say object divided in two equal parts of which one is chosen, like in the figure below and this blue portion represents $\frac{1}{2}$ of whole.

What if we had divided the object in $4$ parts and then chosen $2$ parts. It would have appeared as shown below.

Although it is $\frac{2}{4}$, it is quite obvious that whether one gets $\frac{1}{2}$ of a whole or gets $\frac{2}{4}$ of the whole, there is no difference i.e. $\frac{1}{2} = \frac{2}{4}$.

What if we had divided the same in $6$ parts and for getting equivalent amount, one would have to choose $3$ parts (i.e. $\frac{3}{6}$) as is obvious from following figure.

Hence one can say that $\frac{3}{6} = \frac{2}{4} = \frac{1}{2}$.

Similarly dividing by $8$ and choosing $4$ out of it i.e. $\frac{4}{8}$ appears as below.

It is quite obvious that $\frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \frac{4}{8}$

Also observe that in this case we are just multiplying numerator and denominator same number (in above case $2$, $3$ and $4$) and

$\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{1 \times 3}{2 \times 3} = \frac{1 \times 4}{2 \times 4}$.

Now without actually drawing these figures consider $\frac{9}{12}$ shown in the first figure. Observe that had we divided the figure in $4$ parts, for equivalent portion, we would have selected $\frac{3}{4}$. Why?

Because $\frac{9}{12} = \frac{9 \div 3}{12 \div 3} = \frac{3}{4}$

Hence multiplying or dividing both numerator and denominator, make an equivalent fraction.

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