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Kitty [74]
3 years ago
13

PLEASE HELP!! Find the area of the sector.

Mathematics
1 answer:
Lunna [17]3 years ago
7 0

Answer:

B.

\frac{275\pi}{6}  {m}^{2}

Step-by-step explanation:

area \: of \: sector \: is \\  \frac{the \: angle}{360}  \times \pi {r}^{2}  \\ hence \: we \: have \\  \frac{165}{360}  \times  {(10m)}^{2}  =  \frac{275\pi}{6}  {m}^{2}

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Hi there! The answer is g(x) = 4^x.

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Which point is located on the line represented by the equation y+4=-5(x-3)?
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The answer would be (3,-4) This is because if you replace the variables with the numbers and solve the equation it will come out 0=0

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3 years ago
What is the center and radius of -10x = -121 - y2 - x2 -22y
OlgaM077 [116]

We need to find the center and the radius of

-10x\: =\: -121\: -\: y^2\: -\: x^2\: -22y

The general circle equation is the following

(x-h)^2+(y-k)^2=r^2

where

(h,k) is the center and

r is the radius

1. rearrange the equation

(x^2-10x)+(y^2+22y+121)=0

2. Add 25 on both sides

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What is the equation for a circle with a center at (-2,-4) that passes through the point (3,8)?
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Check the picture below.

so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
&&(~ 3 &,& 8~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
\\\\\\
\left( \cfrac{3-2}{2}~~,~~\cfrac{8-4}{2} \right)\implies \left( \frac{1}{2}~,~2 \right)\impliedby center

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
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%  (c,d)
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\end{array}~~~ 
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\\\\\\
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\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
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\\\\\\
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