1. To solve this we are going to use the slope formula: [tex]m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} } [/tex]
where
[tex]m[/tex] is the slope
[tex](x_{1},y_{1})[/tex] are the coordinates of the first point
[tex](x_{2},y_{2})[/tex] are the coordinates of the second point
We know from our problem that the first point is (-1,3) and the second point is (0,5). So lest use our slope formula:
[tex]m= \frac{5-3}{0-(-1)} [/tex]
[tex]m= \frac{2}{1} [/tex]
[tex]m=2[/tex]
We can conclude that slope of the line that passes through the points (-1,3) and (0,5) is 2.
2. [tex]x=5[/tex] and [tex]y=15[/tex] give us the point (5,15). Lets use the origin (0,0) as the first point of our line. First we are going to find its slope:
[tex]m= \frac{15-0}{5-0} [/tex]
[tex]m= \frac{15}{5} [/tex]
[tex]m=3[/tex]
Next, we are going to use the point slope formula: [tex]y-y_{1}=m(x-x_{1})[/tex]:
[tex]y-0=3(x-0)[/tex]
[tex]y=3x[/tex]
When [tex]x=9[/tex]:
[tex]y=3(9)[/tex]
[tex]y=27[/tex]
We can conclude that the direct relation equation that relates [tex]x[/tex] and [tex]y[/tex] is [tex]y=3x[/tex]; also, when [tex]x=3[/tex] [tex]y=27[/tex].
3. Firs we are going to use the slope formula to find the slope between our points (-3,4) and (1,4)
[tex]m= \frac{4-4}{1-(-3)} [/tex]
[tex]m= \frac{0}{1+3} [/tex]
[tex]m=0[/tex]
Now, we are going to use the point slope formula to fin our line:
[tex]y-4=0[x-(-3)][/tex]
[tex]y-4=0(x+3)[/tex]
[tex]y-4=0[/tex]
[tex]y=4[/tex]
We can conclude that the equation of the line that passes through the points (-3,4) and (1,4) is [tex]y=4[/tex]
4. We know that the slope of our line is [tex] \frac{2}{3} [/tex], so [tex]m= \frac{2}{3} [/tex]. We also know that our line passes through the point (-3,-1), so lest use the point slope formula to find the equation of our line:
[tex]y-(-1)= \frac{2}{3} [x-(-3)][/tex]
[tex]y+1= \frac{2}{3} (x+3)[/tex]
[tex]y+1= \frac{2}{3} x+2[/tex]
[tex]y= \frac{2}{3} x+1[/tex]
We can conclude that the equation of the line that has slope [tex] \frac{2}{3} [/tex] and passes through the point (-3,-1) is [tex]y= \frac{2}{3} x+1[/tex]
5. To express [tex]y=- \frac{2}{3} x-1[/tex] in standard form [tex]Ax+By=C[/tex] using integers, we are going to follow this steps:
Step 1. Multiply both sides of the equation by 3:
[tex]3y=3(- \frac{2}{3} x-1)[/tex]
[tex]3y=-2x-3[/tex]
Step 2. Add [tex]2x[/tex] to both sides of the equation:
[tex]3y+2x=-2x+2x-3[/tex]
[tex]3y+2x=-3[/tex]
We can conclude that the line of the equation in standard form using integers is [tex]3y+2x=-3[/tex]
6. Remember that if two lines are parallel, they have the same slope; therefore, the slope of our line will be [tex]m=5[/tex]. We also know that our line passes through the point (2,-1), so lets use our point-slope formula to find its equation:
[tex]y-(-1)=5(x-2)[/tex]
[tex]y+1=5x-10[/tex]
[tex]y=5x-11[/tex]
We can conclude that the equation of the line that passes through the point (2,-1) and is parallel to the line [tex]y=5x-2[/tex] is [tex]y=5x-11[/tex]
7. Remember that two lines are perpendicular if the slope of the second line is the negative reciprocal of the first line; since the slope of the first line is [tex]-3[/tex], the slope of the second line will be: [tex]m=- \frac{1}{-3} = \frac{1}{3} [/tex].
Now that we have the slope of our line, lets use the point-slope formula to find its equation:
[tex]y-5= \frac{1}{3} (x-3)[/tex]
[tex]y-5= \frac{1}{3} x-1[/tex]
[tex]y= \frac{1}{3} x+4 [/tex]
We can conclude that the line that passes trough the point (3,5) and is perpendicular to the line [tex]y=-3x+7[/tex] is [tex]y= \frac{1}{3} x+4[/tex]
8. The graph of the absolute value equation [tex]y=|x-2|-3[/tex] is the graph of the parent absolute value equation [tex]y=|x|[/tex] shifted three units right and two units down.
Check the graph of both functions in the first picture.
9.
a. Check the scatter plot in the second picture.
b. We can conclude that we have a negative correlation in our scatter plot; in other words the, variables move in opposite directions. For our scatter plot we can infer that as time increases depth decreases. In addition, the best fit for our scatter pot is a line.
c. To find the depth at 15 seconds, we need to find the equation of our scatter plot first. Since our scatter plot is a line, we can use tow points, find the slope, and use the point-sole formula to find its equation.
Lets use the points (2,47) and (22,7):
[tex]m= \frac{7-47}{22-2} [/tex]
[tex]m= \frac{-40}{20} [/tex]
[tex]m=-2[/tex]
[tex]y-47=-2(x-2)[/tex]
[tex]y-47=-2x+4[/tex]
[tex]y=-2x+51[/tex]
Now to fin the depth at 15 seconds, we just need to replace [tex]x[/tex] with 15:
y=-2(15)+51
[tex]y=-30+51[/tex]
[tex]y=21[/tex]
We can conclude that the depth at 15 seconds is 21 centimeters.
