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2 years ago

Please Help Me!

Mathematics

1 answer:

alexdok [17]2 years ago

3 0

Answer:

The answer to your question is below

Step-by-step explanation:

(2x⁴ + 7x³ - 3x + 9) - (6x⁴ - 3x³ - 6x² - 17)

Step 1. take out the parenthesis, in the second term, change the signs

2x⁴ + 7x³ - 3x + 9 - 6x⁴ + 3x³ + 6x² + 17

Step 2. Group like terms

(2x⁴ - 6x⁴) + (7x³ + 3x³) + (6x²) + (-3x) + (9 + 17)

Step 3. Simplify like terms

-2x⁴ + 10x³ + 6x² - 3x + 26

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$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ x &,& y~) 
% (c,d)
&&(~ -9 &,& -1~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-9+x}{2}~~,~~\cfrac{-1+y}{2} \right)=\stackrel{midpoint}{(8,14)}\implies 
\begin{cases}
\cfrac{-9+x}{2}=8\\\\
-9+x=16\\
\boxed{x=25}\\
-------\\
\cfrac{-1+y}{2} =14\\\\
-1+y=28\\
\boxed{y=29}
\end{cases}$

$\bf -------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ x &,& y~) 
% (c,d)
&&(~10 &,& 12~)
\end{array}\qquad
\\\\\\
\left( \cfrac{10+x}{2}~~,~~\cfrac{12+y}{2} \right)=\stackrel{midpoint}{(6,9)}\implies 
\begin{cases}
\cfrac{10+x}{2}=6\\\\
10+x=12\\
\boxed{x=2}\\
-------\\
\cfrac{12+y}{2} =9\\\\
12+y=18\\
\boxed{y=6}
\end{cases}$

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