Y-2 = m (x-6), where m = (2-4)/(6-2) = -2/4 = -1/2
y-2 = 1/2 (x-6)
From the identity,
sin^2(x)+cos^2(x)=1
then
cos(x)=sqrt(1-sin^2(x))
where sin(x) is given as 0.42.
Note that the above answer applies to 0<x<90 degrees
4(r)^x=y
4(r)^(1)=2
r=1/2
y=4*(1/2)^x
Area of trapezoid = a + b/2 * h
a = 20, b = 9, h = 21/-16
20+9/2 * (21-16) = 29/2 * 5 = 72.5
The area of the trapezoid = 72.5 in^2
Area of the rectangle = l * w
l = 16, w = 20
16 * 20 = 320
The area of the rectangle: 320 in^2
Answer:
The distance covered is 113.75 m
Step-by-step explanation:
As per the question:
The initial velocity of the train, v = 20 m/s
The final velocity of the train, v' = 6 m/s
Uniform deceleration, a = 1.6
Or uniform acceleration, a = - 1.6
<em>Here, the body decelerates, i.e., slows down at a uniform rate thus we take acceleration with negative sign.</em>
Now, to find the distance covered, s:
Using the eqn of Kinemetics:



