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3 years ago

Point C lies on segment AB . Find the length of line segment BC if AB = 1.5m and AC = 4.4 cm.

Mathematics

2 answers:

Alik [6]3 years ago

5 0

AB=1.5 m = 150 cm
BC=AB-AC
BC=150-4.4=145.6 cm = 1.456 m

lana [24]3 years ago

5 0

bearing in mind that AB is 1.5 Meters, and that there are 100 cm in 1 meter, therefore 150 cm in 1.5 m, so AB = 150 cm.

$\bf \underset{\leftarrow ~~\stackrel{1.5~m}{150~cm}~~\to }{\boxed{A}\stackrel{4.4~cm}{\rule[0.35em]{5em}{0.25pt}}C\stackrel{BC}{\rule[0.35em]{23em}{0.25pt}\boxed{B}}} \\\\\\ AB=AC+BC\implies \stackrel{AB}{150}=\stackrel{AC}{4.4}+BC\implies 150-4.4=BC\implies \stackrel{cm}{145.6}=BC$

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<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

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$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

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Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

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Read 2 more answers

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The slope within two sets of points is calculated using the slope equation $m = \frac{y_2 -y_1}{x_2 -x_1}$

<h3>What is slope?</h3>

The slope of a line or points is the rate of change of the line.

This in other words means that, the vertical change per unit horizontal change

Assume that the points are given as:

(x1, y1) and (x2, y2)

The slope (m) of the points is:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

Hence, the equation of two sets of points is $m = \frac{y_2 -y_1}{x_2 -x_1}$