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zhenek [66]
3 years ago
11

State the independent variable and the dependent variable in the linear relationship. Then find the rate of change for the situa

tion. The cost of admission admission is ​$4848 for forfour pets pets and ​$9696 for eighteight pets pets. Determine the independent variable.
Mathematics
1 answer:
nataly862011 [7]3 years ago
8 0

We have been given that the cost of admission is ​$48 for four pets and ​$96 for eight pets. We are asked to determine the independent variable and the dependent variable in the given linear relationship.

We can see that as the number of pets is increasing cost of admission is also increasing. This means that cost depends on number of pets.

Therefore, cost of admission is dependent variable and number of pets is independent variable.

To find rate of change, we will use slope formula.

We have been given two points on line that are (4,48) and (8,96).

m=\frac{y_2-y_1}{x_2-x_1}

Upon substituting the coordinates of our given points in slope formula, we will get:

m=\frac{96-48}{8-4}

m=\frac{48}{4}

m=12

Therefore, the rate of change is $12 per pet.

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Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo
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Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
%  (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2}  \right)}

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now, a perpendicular line to RS, will have a negative reciprocal slope to it.  Well, what is the slope of RS anyway?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
%   (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}

and let's check the reciprocal negative of that,

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies  +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
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