(x^2 + 8)(2x + 4)(3x + 6) I kinda need help with this...

Mathematics

2 answers:

STatiana [176]2 years ago

7 0

Answer:

6x^4+24x^3+72x^2+192x+192

Step-by-step explanation:

Licemer1 [7]2 years ago

6 0

I hope this helps,Just remember the rules.

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Very easy! PLEASE HELP I will mark brainliest

Sladkaya [172]

Answer:

B i think let me solve

Step-by-step explanation:

8 0

3 years ago

Solve for x. What is the solution to the system of equations?<br> ( <br> , <br> )

labwork [276]

(3,-2) hope this helps Yall

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2 years ago

a recipe calls for 6 cups of brown sugar for every 2 cups of white sugar. How many cups of brown sugar is required for every cup

sammy [17]

Answer:

You need 3 cups of brown sugar for every 1 cup of white sugar.

Step-by-step explanation:

To find the smallest amount possible, divide both numbers by the smaller number. So, 6 (the amount of brown sugar) Divided by 2 equals 3, and 2 (the amount of white sugar) Divided by 2 equals 1.

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3 years ago

Find d for the arithmetic series with S17=-170 and a1=2

Irina18 [472]

So, we know the sum of the first 17 terms is -170, thus S₁₇ = -170, and we also know the first term is 2, well

$\bf \textit{ sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
n=17\\
S_{17}=-170\\
a_1=2
\end{cases}
\\\\\\
-170=\cfrac{17(2+a_{17})}{2}\implies \cfrac{-170}{17}=\cfrac{(2+a_{17})}{2}
\\\\\\
-10=\cfrac{(2+a_{17})}{2}\implies -20=2+a_{17}\implies -22=a_{17}$

well, since the 17th term is that much, let's check what "d" is then anyway,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=17\\
a_{17}=-22\\
a_1=2
\end{cases}
\\\\\\
-22=2+(17-1)d\implies -22=2+16d\implies -24=16d
\\\\\\
\cfrac{-24}{16}=d\implies -\cfrac{3}{2}=d$

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3 years ago

35 out of 100 simplified

marshall27 [118]

Answer:

7/20

Step-by-step explanation:

5 0

3 years ago

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