You know a point (3,4) and can figure out a slope.
The slope intercept form is y = mx + b, where m is the slope and b is the y-intercept.
Thus the line y = 2x + 1 has a slope of 2.
Since all parallel lines have the same slope, the line that passes through (3,4) also has a slope of 2.
Knowing a point and the slope, use the point slope form: y - y1 = m(x - x1)
where m is the slope and (x1, y1) is the point.
For this problem, m = 2, x1 = 3, and y1 = 4.
y - 4 = 2(x - 3)
y - 4 = 2x - 6
y = 2x - 2
(There are other ways to solve this problem.)
Let L = length of the rectangle and W = width of the rectangle.
"length of a rectangle is 3 inches less than twice the width" translates to L = 2W-3.
Since the perimeter of a rectangle is: P = L + W + L + W or: P = 2L + 2W.
"perimeter of the rectangle is 18 inches: translates to: P = 18 or 2L + 2W = 18.
Using 2L + 2W = 18 and substituting 2W - 3 for L, we have:
2(2W-3) + 2W = 18
Multiplying out: 4W - 6 + 2W = 18
Simplifying: 6W - 6 = 18
Adding 6 to both sides: 6W = 24
Dividing both sides by 6: W = 4
Since L = 2W - 3, L = 2(4) - 3 ---> L = 5
When you graphed, you were told that the slope was the "change in y / change in x" or "rise over run".
For a slope to be 1/16, it has a vertical rise of 1 for every horizontal run of 16 .
If, on a piece of graph paper, you place a dot at the origin (0,0) (call this dot A); move over to the point (16,0) and put a dot there (call this dot C); and move up to the point (16,1) and put a dot there (call this dot B), you have drawn a right triangle (triangle ACB, with right angle at C) which represents a ramp with slope 1/16.
You need to find the size of the angle at A. Since you know the opposite side (1) and the adjacent side (16), use tangent (because tan = opp / adj). ---> tan ∠A = 1 /16 ---> ∠A = invtan (1/16) ---> ∠A = 3.576°.
Similarly for the slope of 1/20: tan ∠A = 1 /20 ---> ∠A = invtan (1/20) ---> ∠A = 2.862°.