Answer:
(fog)(x) = 4x - 7
Step-by-step explanation:
f(x) = 2x-1 and g(x) = 2x-3;
(fog)(x) = f(g(x)
= f(2x - 3) = 2(2x - 3) - 1
= 4x - 6 - 1
= 4x - 7
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>
Time out = d/so = d/523
time back = d/sb = d/415
t = time out + time back
11 = d/523 + d/415 find common deneminator
11 = (415d + 523d)/(523*415)
11*523*415 = 415d + 523d
2387495 = 938d
d = 2545 miles
hope this help
Answer:
∠ B ≈ 66.42°
Step-by-step explanation:
Using the cosine ratio in the right triangle
cos B = 
= 
= 
, thus
B = 
( ) ≈ 66.42° ( to the nearest hundredth )
