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bogdanovich [222]
3 years ago
6

Is y=x^2 a proportional relationship?

Mathematics
1 answer:
TiliK225 [7]3 years ago
6 0

Answer:

is y=x^2 a proportional relationship?

{ \sf{yes. \: constant \: of \: proportionality = 1}}

is y=2+x a proportional relationship?

{ \sf{no. \: unless \: y \: is \: proportinal \: to \: (2 + x)}}

is y=2/x a proportional relationship?

{ \sf{yes. \: where \: proportianality \: constant \: is \: 2}}

is y=2x a proportional relationship?

{ \sf{yeah. \: constant \: is \: 2}}

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If (m, 2) is on a circle with center (−3, 2) and radius 8, what is a value of m?
Talja [164]
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -3}}\quad ,&{{ 2}})\quad 
%  (c,d)
&({{ m}}\quad ,&{{ 2}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf \textit{we know the distance from m,2 and -3,2(center) is the radius}
\\\\\\
d=8\implies 8=\sqrt{[m-(-3)]^2+[2-2]^2}
\\\\\\
8=\sqrt{(m+3)^2+(0)^2}\implies 8=\sqrt{(m+3)^2}\implies 8^2=(m+3)^2
\\\\\\
64=(m+3)^2\implies \sqrt{64}=\sqrt{(m+3)^2}\implies 8=m+3
\\\\\\
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