Answer:
A = 27π cm²
Step-by-step explanation:
Area of the shaded region = area of full circle of O - Area of the unshaded sector
= (πr²) - (θ/360*πr²)
Where,
r = 6 cm
θ = 360
Area of shaded region = (π*6²) - (90/360*π*6²)
= (36π) - (¼*π*36)
= 36π - 9π
Area of shaded region = 27π cm²
The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
Answer:
A,C,D
Step-by-step explanation:
Answer:
1/2 X + X -15 + 1/2 X + 100 + X -25 = 540
1/2x + x + 1/2x + x -15 + 100 -25 = 540
3 x + 60 = 540
3x + 60 - 60 = 540 - 60
3/3x = 480/3
x = 160
Step-by-step explanation:
1/2 X + X -15 + 1/2 X + 100 + X -25 = 540
1/2x + x + 1/2x + x -15 + 100 -25 = 540
3 x + 60 = 540
3x + 60 - 60 = 540 - 60
3/3x = 480/3
x = 160
