Question
Mathematics
1. Given the linear equation $y=\frac{2}{3}x+2$y=(2)/(3)x+2, which table of values represents this equation?
A.
 $x$x $y$y -3 0 0 2 3 4
x y -3 0 0 2 3 4
B.
 $x$x $y$y 0 2 2 5 3 4
x y 0 2 2 5 3 4
C.
 $x$x $y$y -3 0 2 0 6 6
x y -3 0 2 0 6 6
D.
 $x$x $y$y -9 -6 -6 -2 0 2
x y -9 -6 -6 -2 0 2
1. What is the $y$y-intercept of the line that contains the points $\left(3,3\right)$(3,3) and $\left(6,-1\right)$(6,-1) ?
A. 5
B. 6
C. 7
D. 8
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1. To find the table of values that satisfies the equation $y=\frac{2}{3}x+2$y=(2)/(3)x+2, we can substitute different values of $x$x and solve for $y$y. Let’s check each table of values:
A)
When $x=-3$x=-3,
$y=\frac{2}{3}\left(-3\right)+2=-2$y=(2)/(3)(-3)+2=-2
When $x=0$x=0,
$y=\frac{2}{3}\left(0\right)+2=2$y=(2)/(3)(0)+2=2
When $x=3$x=3,
$y=\frac{2}{3}\left(3\right)+2=4$y=(2)/(3)(3)+2=4
B)
When $x=0$x=0,
$y=\frac{2}{3}\left(0\right)+2=2$y=(2)/(3)(0)+2=2
When $x=2$x=2,
$y=\frac{2}{3}\left(2\right)+2=\frac{8}{3}$y=(2)/(3)(2)+2=(8)/(3)
When $x=3$x=3,
$y=\frac{2}{3}\left(3\right)+2=\frac{10}{3}$y=(2)/(3)(3)+2=(10)/(3)
C)
When $x=-3$x=-3,
$y=\frac{2}{3}\left(-3\right)+2=0$y=(2)/(3)(-3)+2=0
When $x=2$x=2,
$y=\frac{2}{3}\left(2\right)+2=\frac{8}{3}$y=(2)/(3)(2)+2=(8)/(3)
When $x=6$x=6,
$y=\frac{2}{3}\left(6\right)+2=6$y=(2)/(3)(6)+2=6
D)
When $x=-9$x=-9,
$y=\frac{2}{3}\left(-9\right)+2=-4$y=(2)/(3)(-9)+2=-4
When $x=-6$x=-6,
$y=\frac{2}{3}\left(-6\right)+2=-2$y=(2)/(3)(-6)+2=-2
When $x=0$x=0,
$y=\frac{2}{3}\left(0\right)+2=2$y=(2)/(3)(0)+2=2
Therefore, the table of values that represents the equation $y=\frac{2}{3}x+2$y=(2)/(3)x+2 is table $\overline{)\mathbf{\text{(A)}}}$(A).
2. The $y$y-intercept of a line is the value of $y$y at which the line intersects the $y$y-axis (where $x=0$x=0). To find the $y$y-intercept of the line containing the points $\left(3,3\right)$(3,3) and $\left(6,-1\right)$(6,-1), we can use the point-slope formula:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$y-y_(1)=m(x-x_(1))
where $m$m is the slope of the line and $\left({x}_{1},{y}_{1}\right)$(x_(1),y_(1)) is one of the points on the line.
We can start by finding the slope:
$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{-1-3}{6-3}=-\frac{4}{3}$m=(y_(2)-y_(1))/(x_(2)-x_(1))=(-1-3)/(6-3)=-(4)/(3)
Using the point-slope formula with $\left({x}_{1},{y}_{1}\right)=\left(3,3\right)$(x_(1),y_(1))=(3,3):
$y-3=-\frac{4}{3}\left(x-3\right)$y-3=-(4)/(3)(x-3)
Simplifying:
$y-3=-\frac{4}{3}x+4$y-3=-(4)/(3)x+4
$y=-\frac{4}{3}x+7$y=-(4)/(3)x+7
The $y$y-intercept occurs when $x=0$x=0, so we substitute $x=0$x=0 into the equation:
$y=-\frac{4}{3}\left(0\right)+7=7$y=-(4)/(3)(0)+7=7
Therefore, the $y$y-intercept is $\overline{)\mathbf{\text{(C)}}7}$(C) 7.
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