1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-2);y=-x-2
D.y=-x
2.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-1);y=-3/2x-6
C.y=-3/2x+2
3.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (4,2);x=-3
D.y=4
4.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (-2,3);y=1/2x-1
B.y=-2x-1
5.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (5,0);y+1=2(x-3)
D.y=-1/2x+5/2
Answer:
2777.7777... subs
Step-by-step explanation:
11000+400+1300+300=13000
6-1.32=4.68
13,000/4.68
=2777.777777...
Youll need to provide more information
You have to get the x to one side so you'll subtract 3x from -6x and 3x from 3x, the 3x will cancel out and -6x - 3x is -9x-7=11, you have to add 7 from both sides the 7+7 cancels out and also add 11+7 which leaves -9x=18, you have to divide -9x/-9 and do 18/-9, which gets you x=-2. Here's the work shown if you didn't what i said...
-6x-7=3x+11
-3x -3x
-9x-7=11
+7 +7
-9x=18
-9 -9
x=-2
Answer:
8 and 12
Step-by-step explanation:
Sides on one side of the angle bisector are proportional to those on the other side. In the attached figure, that means
AC/AB = CD/BD = 2/3
The perimeter is the sum of the side lengths, so is ...
25 = AB + BC + AC
25 = AB + 5 + (2/3)AB . . . . . . substituting AC = 2/3·AB. BC = 2+3 = 5.
20 = 5/3·AB
12 = AB
AC = 2/3·12 = 8
_____
<em>Alternate solution</em>
The sum of ratio units is 2+3 = 5, so each one must stand for 25/5 = 5 units of length.
That is, the total of lengths on one side of the angle bisector (AC+CD) is 2·5 = 10 units, and the total of lengths on the other side (AB+BD) is 3·5 = 15 units. Since 2 of the 10 units are in the segment being divided (CD), the other 8 must be in that side of the triangle (AC).
Likewise, 3 of the 15 units are in the segment being divided (BD), so the other 12 units are in that side of the triangle (AB).
The remaining sides of the triangle are AB=12 and AC=8.
