With a given parallel line and a given point on the line
we can use the point-line method: y-y0=m(x-x0)
where
y=mx+k is the given line, and
(x0,y0) is the given point.
Here
m=-10, k=-5, (x0,y0)=(-3,5)
=> the required line L is given by:
L: y-5=-10(x-(-3))
on simplification
L: y=-10x-30+5
L: y=-10x-25
It has a bunch of zeroes after the five so it is rounded unless you want to go to .18
Answer:
27
Step-by-step explanation:
First you need to find x.
It is shown that DE=EF, so we can use this.
5x-3=3x+7
5x=3x+10
2x=10
x=5
Then, you plug in the value of the variable and solve.
6(5)-3
30-3
27 = DF
Answer:
0
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(7-7)/(8-0)
m=0/8=0
Answer:
Death Valley California
C(t) = -0.30 (0)² + 40
C(t) = 40
C(t) = -0.30 (12)² + 40t – 12)2 + 40
65f
