Help pls i need this luv u guys <33

Mathematics

1 answer:

coldgirl [10]3 years ago

3 0

So this is so easy
the answer is33
pay attention what i’m saying:
if you have a circle like this and you have arc if the arc doesnt pass the center of the circle and fit to the environment of circle the measure of angle is half of the measure of arc

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A + b + c = 72

a = b - 1

c = b + 1

(b - 1) + b + (b + 1) = 72

Simplify

3b = 72

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a = 24 - 1

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Is sin theta=5/6, what are the values of cos theta and tan theta?

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let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$

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2 years ago

What is the peremeter of this polygon? (With picture)

lyudmila [28]

Check the picture below.

so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 2}})\quad 
% (c,d)
&({{ 2}}\quad ,&{{ 4}})\\
&({{ 2}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ 3}}\quad ,&{{ -2}})\\
&({{ 3}}\quad ,&{{ -2}})\quad 
% (c,d)
&({{ -2}}\quad ,&{{ -3}})\\
&({{ -2}}\quad ,&{{ -3}})\quad 
% (c,d)
&({{ -1}}\quad ,&{{ 2}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf -------------------------------\\\\
d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2}
\\\\\\
d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\
-------------------------------\\\\
d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\
-------------------------------\\\\
d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2}
\\\\\\
d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}$

$\\\\
-------------------------------\\\\
d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2}
\\\\\\
d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}$

so, those are their lengths, sum them all up, that's the polygon's perimeter.

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3 years ago