Answer:
49m/s
59.07 m/s
Step-by-step explanation:
Given that :
Distance (s) = 178 m
Acceleration due to gravity (a) = g(downward) = 9.8m/s²
Velocity (V) after 5 seconds ;
The initial velocity (u) = 0
Using the relation :
v = u + at
Where ; t = Time = 5 seconds ; a = 9.8m/s²
v = 0 + 9.8(5)
v = 0 + 49
V = 49 m/s
Hence, velocity after 5 seconds = 49m/s
b) How fast is the ball traveling when it hits the ground?
V² = u² + 2as
Where s = height = 178m
V² = 0 + 2(9.8)(178)
V² = 0 + 3488.8
V² = 3488.8
V = √3488.8
V = 59.07 m/s
1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
Simplify 1/3(6x - 15) to 6x - 15/3
6x - 15/3 = 1/2(10x - 4)
Factor out the common term; 3
3(2x - 5)/3 = 1/2(10x - 4)
Cancel out 3
2x - 5 = 1/2(10x - 4)
Simplify 1/2(10x - 4) to 10x - 4/2
2x - 5 = 10x - 4/2
Factor out the common term; 2
2x - 5 = 2(5x - 2)/2
Cancel out 2
2x - 5 = 5x - 2
Subtract 2x from both sides
-5 = 5x - 2 - 2x
Simplify 5x - 2 - 2x to 3x - 2
-5 = 3x - 2
Add 2 to both sides
-5 + 2 = 3x
Simplify -5 + 2 to -3
-3 = 3x
Divide both sides by 3
- 1 = x
Switch sides
<u>x = -1</u>
have you figured out the answer i need help on the problem to
