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juin [17]
2 years ago
13

PLEASE HELP 75 POINTS AND BRAINLIEST

Mathematics
1 answer:
sleet_krkn [62]2 years ago
6 0

I would say C ejejdkidjejdkdndkeiej

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What is the distance between the 2 points? Round to the nearest tenths place.
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Check the picture below.

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What is simplest form? Explain to me.
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2 years ago
Samantha rectangular gift is 10 inches. by 12 inches and is framed with a ribbon. She wants to use the same lengths of ribbon to
vitfil [10]

Answer:

The dimension of the rectangular gift is 10 by 12 inches so let us find the perimeter of this rectangle.

Perimeter of rectangular gift = 2 (L+ W) = 2 (10 +12) = 44 inches

Since we are to use the same length of ribbon to wrap a circular clock so the perimeter or circumference of the clock should be no more than 44 inches.

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6 0
2 years ago
Find the equation of the circle that has a diameter with endpoints located at (7, 3) and (7, –5).
posledela
So hmm check the picture below, that's about the circle and the endpoints, but notice, the endpoints make up a segment, namely the diameter of the circle, well.... let's see how long that is, because, the radius is half the diameter

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad 
%  (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{(7-7)^2+(-5-3)^2}\implies d=\sqrt{0+(-8)^2}\implies d=8
\\\\\\
\textit{the radius is half that, so is }\boxed{r=4}

now.. hmmm notice, the midpoint of the diameter, is the center of the circle, let's check that one out

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad 
%  (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 7+7}}}{2}\quad ,\quad \cfrac{{{ -5}} + {{ 3}}}{2} \right)\implies \left( \cfrac{14}{2}\ ,\ \cfrac{-2}{2} \right)\implies \boxed{(7,-1)}

now.. that we know what the center is, and what the radius is, well

\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k\quad }})\qquad 
radius=&{{ r}}\\
&7&-1&4
\end{array}

5 0
3 years ago
HELP, I can't figure out the first question
Margarita [4]
The answer is linearly
8 0
2 years ago
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