Then known number in an algebraic expression is

1 answer:

7 0

Unsure of what you're asking.

If you have an algebraic expression in front of you and see a number there, you could label it as "known" or "a known quantity." Another word that comes to mind is "constant."

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The formula for evaluating the area of a sector with the given measure of the arc in degrees and radius is,

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Step-by-step explanation:

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Step-by-step explanation:

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To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)

Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}$

$\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill$