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
Answer:
112
Step-by-step explanation:
Answer:
-6
Step-by-step explanation:
2×-y=6
subtract the 2x
-y=-2×+6
divided by negative 1 on both sides
y=2×-6
<span>(2*h)/15 = 20 // - 20
(2*h)/15-20 = 0
2/15*h-20 = 0 // + 20
2/15*h = 20 // : 2/15
h = 20/2/15
h = 150
h = 150</span>
Answer:The answer is 12
Step-by-step explanation:
I cheated
