Answer:
6.19
Step-by-step explanation:
I hope this helps you
Slope of a line passing though point (x1,y1) nd (x2,y2) is
(y2-y1)/(x2-x1)
we have
(1,-3) and (3,-5)
x1=1
y1=-3
x2=3
y2=-5
slope=(-5-(-3))/(3-1)=(-5+3)/(2)=-2/2=-1
slope=-1
Answer:
10 m/s
Step-by-step explanation:
Velocity= Position/Time
You can take any point on the line to calculate velocity.
Lets take t=1s
At t=1, position=10m
velocity= 10m/1s
=10 m/s
You can take any other point too, velocity will remain same.
For eg.
At t=5s
position=50m
velocity=50/5
=10m/s
Hope it helps :-)
Assuming the equation is x^2 + y^2 = 1, then that's a circle, with radius 1, centered on the origin [0,0].
So there are two tangents at x = 0. They are y = 1, and y = -1 (horizontal lines).
There is one tangent at x = 1. It is x = 1 (a vertical line).
There is no tangent at x = 35, because the original equation has no solution at x = 35.
