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USPshnik [31]
3 years ago
12

Can someone please explain how to find this answer? thanks

Mathematics
2 answers:
N76 [4]3 years ago
6 0
I would start with the standard form equation of the parallel line.
.. 5x -2y = 5(-2) -2(3) = -16
Then solve for y.
.. 2y = 5x +16 . . . add 2y+16; then divide by 2 for the next equation
.. y = (5/2)x +8 . . . . . . . corresponds to selection (B)
GrogVix [38]3 years ago
3 0
So, a line that is parallel to 5x-2y = -12, will have the same exact slope as that equation's, so what is it anyway?

\bf 5x-2y=-12\implies 5x+12=2y\implies \cfrac{5x+12}{2}=y
\\\\\\
\cfrac{5x}{2}+\cfrac{12}{2}=y\implies \stackrel{slope}{\cfrac{5}{2}}x+6=y

so, notice, from the slope-intercept form, it happens that the slope is 5/2, well, the line will have the same slope.

so we're looking for the equation of a line whose slope is 5/2 and runs through -2,3

\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
&&(~ -2 &,& 3~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=\cfrac{5}{2}[x-(-2)]
\\\\\\
y-3=\cfrac{5}{2}(x+2)\implies y-3=\cfrac{5}{2}x+5\implies y=\cfrac{5}{2}x+8
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According to the question, the line is through the points (0,5) and (-2,11)

To calculate the slope using standard formula that is :

m = \frac{(y(2)-y(1))}{(x(2)-x(1))} = \frac{(11-5)}{(-2-0)} =\frac{6}{-2} = -3

Now the slope equation will be: m(x(2)-x(1)) = (y(2)-y(1))

y-5 =(-3)(x-0) = y = -3(x)+5

The slope-intercept form is:y = -3x+5

What is the equation of the line?

To represent the set of points in an algebraic form is known as the equation of line. The set of points represent lines in the coordinate system. The parameters can be calculated and those parameters are slope and y-intercept.

To learn more about the equation of the line from the given link:

<u>brainly.com/question/14037984</u>

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Step-by-step explanation:

Domain is the set of x values that make the function defined. Allowed x values for the function (mapping).

The Range is the set of y values that make the function defined. Allowed y values for the function (mapping).

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Firstly, we need f(4), so we look for "4" in domain and see which number it corresponds to in range.

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