# Find the midpoint of the segrnent with the given encpoin's: (3/8,-(2/5) and (-(5/2), 3/2)

Question
Functions
Find the midpoint of the segrnent with the given encpoin's: $$\displaystyle{\left(\frac{{3}}{{8}},-{\left(\frac{{2}}{{5}}\right)}\right.}$$ and $$\displaystyle{\left(-{\left(\frac{{5}}{{2}}\right)},\frac{{3}}{{2}}\right)}$$

2021-01-09
We have to find the midpoint of the segment with the given endpoints
$$\displaystyle{\left(\frac{{3}}{{8}},-{\left(\frac{{2}}{{5}}\right)}\right.}$$ and $$\displaystyle{\left(-{\left(\frac{{5}}{{2}}\right)},\frac{{3}}{{2}}\right)}$$
The midpoint of two points, (x1,y1) and (x2,y2) is the point M found by the following formula: $$\displaystyle{M}={\left(\frac{{{x}{1}+{x}{2}}}{{2}},\frac{{{y}{1}+{y}{2}}}{{2}}\right)}$$
By applying midpoint formula, the midpoint of the segment with the given endpoints $$\displaystyle{\left({x}{1},{y}{1}\right)}={\left(\frac{{3}}{{8}},-{\left(\frac{{2}}{{5}}\right)}\right.}$$ and $$\displaystyle{\left({x}{2},{y}{2}\right)}={\left(-{\left(\frac{{5}}{{2}}\right)},\frac{{3}}{{2}}\right)}$$ is given by
PSKM=((x1+x2)/2,(y1+y2)/2) =(((3/8)-(5/2)/2),-((2/5)+(3/2)/2)) =(((3/8)-(20/8)/2),-((4/10)+(15/10)/2)) =(-17/16,19/20)ZSK

### Relevant Questions

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$$\displaystyle{a}.{f{{\left({x}\right)}}}={5}{x}−{x}{3}+{3}{x}{5}−{2}{f{{\left({x}\right)}}}={5}{x}-{x}^{{{3}}}+{3}{x}^{{{5}}}-{2}{f{{\left({x}\right)}}}={5}{x}−{x}{3}+{3}{x}{5}−{2}$$
$$\displaystyle{b}.{f{{\left({x}\right)}}}=−{22}{x}{3}−{8}{x}{4}−{2}{x}+{7}{f{{\left({x}\right)}}}=-{\frac{{{2}}}{{{2}}}}{x}^{{{3}}}-{8}{x}^{{{4}}}-{2}{x}+{7}{f{{\left({x}\right)}}}=−{22}​{x}{3}−{8}{x}{4}−{2}{x}+{7}$$
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Show transcribed image text A log 10 m long is cut at 1 meter intervals and itscross-sectional areas A (at a distance x from theend of the log) are listed in the table. Use the Midpoint Rule withn = 5 to estimate the volume of the log. (in $$\displaystyle{m}^{{{3}}}$$)
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$$x=\frac{-b}{2a*x}=\frac{-47.5}{2(-9.5)*x}=\frac{-47.5}{-19*x}=-(-2.5)x=2.5$$
$$y=-9.5(2.5)^{2}-47.5(2.5)+63y=59.375-118.75+63y=-115.125$$
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