1.for the data in the table, does y vary directly with x? if it does, write an equation for the direct variation.
x y
52 39
32 24
20 15
8 6

write the equation of the line in slope intercept form:
6. (4, 7); m = -
7. (-5, -5) & (-1, 3)

write the equation of the line in standard form:
8. (-3, 6); m =
9. (1, 0) & (-5, 3)

10. write the equation of the line perpendicular to 4x – 5y = -10 through the point (2, 3). write the equation in slope intercept form.

! for 6-10 could someone explain to me how to find the answer! i am completely lost!

1. y/x reduces to 3/4 in each case, so the variation is direct. The equation is
  y = (3/4)x

6–10: There are many useful forms of the equation for a line. It can be worthwhile to learn a few of them, or at least keep them where you have ready access.

6. The point-slope form is a good place to start. It can be simplified to slope-intercept form. For slope m and point (h, k), the point-slope equation can be written as
  y = m(x -h) +k
For point (4, 7) this is
  y = m(x -4) +7
When you substitute your value for m (not given here), you can find the slope-intercept equation to be
  y = mx +(7-4m)

7. The 2-point form of the equation for a line is a good place to start. For points (x1, y1) and (x2, y2), the equation of the line is
  y = (y2 -y1)/(x2 -x1)*(x -x1) +y1
  y = (3-(-5))/(-1-(-5))*(x -(-5)) +(-5)
  y = (8/4)(x +5) -5
  y = 2x + 5 . . . . . . . . slope-intercept form

8. You can start with the form
  mx -y = mh -k . . . . . . . . line with slope m through point (h, k)
To put this into standard form, you need to make sure the coefficient of x is a positive integer. You do that by multiplying the equation by the negative of the denominator of "m".
For your point, this is
  mx - y = 6 -3m
When you substitute your value of m (not given here), you will have to finishe the arithmetic on the right side of the equation.

9. There are several ways to write the standard form of equation for a line through two points. One place to start for points (x1, y1) and (x2, y2) could be ...
  (y2 -y1)x -(x2 -x1)y = (x1)(y2 -y1) -(y1)(x2 -x1)
For your points, this will be
  (3-0)x -(-5-1)y = 1*(3-0) -0(-5-1)
  3x +6y = 3
A factor of 3 must be removed to put this into standard form.
  x +2y = 1

10. To find the slope of the given line, solve for y.
  -5y = -4x -10
  y = (-4/-5)x +(-10/-5) = (4/5)x +2
The perpendicular line will have a slope that is the negative reciprocal of the slope of this line. That is,
  m = -1/(4/5) = -5/4
As above, it is convenient to start with the point-slope form, then simplify to the slope-intercept form.
  y = (-5/4)(x -2) +3
  y = (-5/4)x + 11/2

https://courses.lumenlearning.com/boundless-calculus/chapter/infinite-sequences-and-series/

Step-by-step explanation:

( i really need points srry)