algebra help?

the line with x-intercept (12, 0) and y-intercept (0, 4).
m = (4-0)/(0-12)=-1/3
y=-1/3x+c
0=-1/3(12)+c
c = 4
so, the line y= -1/3x + 4

Given f(x) = –x – 4, find f(a – 3).
f(a-3) = -(a-3)-4 = -a-1

perpendicular to the line through points (6, 3) and (7, 7).
m = (7-3)/(7-6) = 4
the gradient of pependicular line = -1/4

If the line pass (6,3),
y=-1/4x+c
3=-1/4(6)+c
c = 9/2
the line y = -1/4x + 9/2

If the line pass (7,7),
y=-1/4x+c
7=-1/4(7)+c
c = 35/4
the line y = -1/4x + 35/4

x intercept and y intercept
slopem=(4-0)/(0-12) ( m=(y2-y1)/(x2-x1)


m=4/-12=-1/3
in general line formula
y=mx+b
m=slope
b=y intercept
answer is y=(-1/3)x+4


f(a-3)
you just replace every x by a-3
so f(a-3)= -(a-3)-4
answer is f(a-3)= -a-1

last question i dont get what do you want
i justhave a question is f(x)=-x-4 is related to lastquestion.
if yes. so the answer is
from f(x)=-x-4
so slope of this line m=-1
if they are perpendicular
so slope is m=1 ( slope perpendicular is -1/slope of line which is -1/-1=1)

Note!!!!!!!!!
(if they are parallel slope of the line and slope parallel will be the same)
line y=x+b
plugging the point(6,3) they didnt matter which point
3=6+b
b=-3
answer is y=x-3


if the middle question is not relative to the last question
so we find slope
m=(7-3)/(7-6)=4
slope of perpendicular m=-1/4
y=-1/4x+b
plugging point(6,3)
3=-(6/4)+b
3+3/2=b
b=9/2
so answer is
y= -(1/4)x+9/2
i dont really know yor question in middle have relative to last question or not..
if they have relative so answer is y=x-3
if they dont have relative( last question by themself)
the answer is y= -(1/4)x+9/2
good luck

1) gradient m = (4-0)/(0-12)
= 1/3
y incept c= 3
the line y = 1/3x + 3 ===> 3y - x -9 = 0

2) f(a-3) = -a+3-4 =-a-1

3)i assume that the perpendicular line goes through (6,3)
gradient of line goes through the point = (7-3)/(7-6) =4
therefore gradient of the perpendicular line is -1/4

eqn of the line Y = 1/4x +c, determine c by substituting (6,3)

a) Assume that it is equation of line that is required.
Gradient , m = - 4/12 = - 1/3
Line passes thro` (12,0)
Equation of line is given by y - 0 = (-1/3)(x - 12)
y = (-1/3)x + 4

b) f (x) = - x - 4
f(a - 3) = - (a - 3) - 4
f(a - 3) = - a + 3 - 4
f(a - 3) = - a - 1

c) Gradient of line thro` (6,3) and (7,7) is
m = (7 - 3) / (7 - 6) = 4 / 1 = 4
Gradient of perpendicular = -1/4

the line with x-intercept (12, 0) and y-intercept (0, 4).
slope: m= 4-0/0-12=-2/3
y-4= -2/3(x-0)
3y-12= -2x
3y=-2x+12
y=-2/3x+4



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Given f(x) = –x – 4, find f(a – 3).
First, you have to replace x with a-3
f(a-3)= -(a-3)-4
f(a-3)= -a+3-4
f(a-3)= -a-1

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perpendicular to the line through points (6, 3) and (7, 7).
first, you have to find the slop.
m=7-3/7-6=4
perpendicular 4= -1/4
point-slope form: y-3= -1/4 (x-6) times the whole equation by 4
4y-12= -(x-6)
4y-12= -x+6
4y= -x+18
y=-1/4x+9/2