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3 years ago

How do u add monomials explain

1 answer:

7 0

To add or subtract monomials<span> that are like terms, you leave the variables as they are and you add or subtract the coefficients. If you have </span>monomials<span> that are not like terms, you add or subtract as many like terms as you can and you leave the terms that are not like terms in your answer.</span>

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What is the solution to the system of equations?

Sever21 [200]

Answer:

$(-2,\frac{5}{3})$

Step-by-step explanation:

Given:

The system of equations given are:

$y=\frac{2}{3}x+3\\x=-2$

Plug in $x=-2$ in the first equation and solve for $y$ .

$y=\frac{2}{3}(-2)+3\\y=\frac{-4}{3}+3\\y=\frac{-4}{3}+\frac{9}{3}\\y=\frac{-4+9}{3}=\frac{5}{3}$

Therefore, the solution is $(x,y)=(-2,\frac{5}{3})$

7 0

3 years ago

Read 2 more answers

Linear Systems Word Problem:

a_sh-v [17]

Answer:

x = 35 hotdogs

y = 52 sodas

There were 35 hotdogs and 52 sodas sold.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Alenkasestr [34]

We start with the expression at the left of the equation.

We can combine the terms as:

$\begin{gathered} \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}} \\ \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})} \end{gathered}$

We can now apply the distributive property for the both the numerator and denominator. We can see also that the denominator is the expansion of the difference of squares:

$\begin{gathered} \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2})^2-(\sqrt[]{2-\sqrt[]{3}}))^2} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})+(\sqrt[]{3}-2)\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{2^{}-(2-\sqrt[]{3})^{}} \\ \frac{\sqrt[]{2}\cdot(2+\sqrt[]{3})-\sqrt[]{2-\sqrt[]{3}}\cdot(2+\sqrt[]{3})+\sqrt[]{2}\cdot(\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}\cdot(\sqrt[]{3}-2)}{2-2+\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2+\sqrt[]{3}+\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}(-2-\sqrt[]{3}+\sqrt[]{3}-2)}{\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2\sqrt[]{3})+\sqrt[]{2-\sqrt[]{3}}(-4)}{\sqrt[]{3}} \\ 2\sqrt[]{2}-4\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{3}} \end{gathered}$

We then can continue rearranging this as:

7 0

1 year ago

What is the following product?

alexgriva [62]

Answer:

12.87

Step-by-step explanation:

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3 years ago

Find the value of angle L. HELP ASAP!!

AURORKA [14]

Answer:

B) 67

Step-by-step explanation:

4 0

4 years ago