Answer:
Step-by-step explanation:
<u>Given:</u>
Isolate the term of x from one side of the equation.
Use the distributive property.
<u>DISTRIBUTIVE PROPERTY:</u>

Each term within the parentheses can be multiplied by a factor outside the parentheses.
3(x+1)
3*x=3x
3*1=3
<u>Rewrite the expression down.</u>
3x+3
3x+3+2(3x+4)
<u>Multiply expand.</u>
2(3x+4)
2*3x=6x
2*4=8
6x+8
3x+3+6x+8
<u>Combine like terms.</u>
3x+6x+3+8
<u>Add the numbers from left to right.</u>
3+8=11
3x+6x+11
<u>Add.</u>
<u>Solutions:</u>
3x+6x=9x

- <u>Therefore, the final answer is 9x+11.</u>
I hope this helps. Let me know if you have any questions.
Well first if he buys 24 a month that equation is 24n n = months and if he already has 523 the explicit rule is an = 24 | 24 | 24 |
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523
Answer:
A polygon is regular when all angles are equal and all sides are equal (otherwise it is "irregular").
Answer:
the answer is 4 because 16×4 is 64
well, if that function f(x) were to be continuos on all subfunctions, that means that whatever value 7x + k has when x = 2, meets or matches the value that kx² - 6 has when x = 2 as well, so then 7x + k = kx² - 6 when f(2)
![f(x)= \begin{cases} 7x+k,&x\leqslant 2\\ kx^2-6&x > 2 \end{cases}\qquad \qquad f(2)= \begin{cases} 7(2)+k,&x\leqslant 2\\ k(2)^2-6&x > 2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 7(2)+k~~ = ~~k(2)^2-6\implies 14+k~~ = ~~4k-6 \\\\\\ 14~~ = ~~3k-6\implies 20~~ = ~~3k\implies \cfrac{20}{3}=k](https://tex.z-dn.net/?f=f%28x%29%3D%20%5Cbegin%7Bcases%7D%207x%2Bk%2C%26x%5Cleqslant%202%5C%5C%20kx%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20f%282%29%3D%20%5Cbegin%7Bcases%7D%207%282%29%2Bk%2C%26x%5Cleqslant%202%5C%5C%20k%282%29%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207%282%29%2Bk~~%20%3D%20~~k%282%29%5E2-6%5Cimplies%2014%2Bk~~%20%3D%20~~4k-6%20%5C%5C%5C%5C%5C%5C%2014~~%20%3D%20~~3k-6%5Cimplies%2020~~%20%3D%20~~3k%5Cimplies%20%5Ccfrac%7B20%7D%7B3%7D%3Dk)