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3 years ago

Point b is the midpoint of ac-. if ab= 4x+2 and ac=10x-6, find the length of ac

Mathematics

1 answer:

Natasha2012 [34]3 years ago

3 0

let's bear in mind that B is the midpoint and thus it cuts a segment into two equal halves.

$\bf \underset{\leftarrow \qquad \textit{\large 10x-6}\qquad \to }{\boxed{A}\stackrel{4x+2}{\rule[0.35em]{10em}{0.25pt}} B\stackrel{\underline{4x+2}}{\rule[0.35em]{10em}{0.25pt}\boxed{C}}} \\\\\\ AC=AB+BC\implies 10x-6=(4x+2)+(4x+2)\implies 10x-6=8x+4 \\\\\\ 2x-6=4\implies 2x=10\implies x=\cfrac{10}{2}\implies x= 5 \\\\[-0.35em] ~\dotfill\\\\ AC=(4x+2)+(4x+2)\implies AC=[4(5)+2]+[4(5)+2] \\\\\\ AC=22+22\implies AC=44$

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$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

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$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

Before performing any calculation it's good to recall a few properties of integrals:

$\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}$

$\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }$

So we apply the first property in the first expression given by the question:

$\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}$

And we solve the second integral:

$\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }$

Then we take the last equation and we subtract 10 from both sides:

$\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}$

And we divide both sides by 2:

$\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}$

Then we apply the second property to this integral:

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}$

Then we use the other equality in the question and we get:

$\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}$

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}$

We substract 8 from both sides:

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}$

• $\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}$

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