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bogdanovich [222]

3 years ago

6

Is y=x^2 a proportional relationship?

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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Answer:

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Read 2 more answers

If (m, 2) is on a circle with center (−3, 2) and radius 8, what is a value of m?

Talja [164]

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

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

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Prove that 1/3 root2 is irrational

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Let us assume that 1/2+root 3 is rational . So 1/2+root 3 = a/b where a and b are irrationals. since rhs is a rational number root 3 should be also rational .

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