What is the value of 14(38−14)+33÷9?
1 answer:
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Answer:
Altitude of the plane is 0.5 miles.
Step-by-step explanation:
From the figure attached,
An airplane A is at height h miles observes a small airstrip at D and a factory at F, 4.8 miles apart from D.
Angle of depressions for the airstrip is 13.1° and the factory is 4.1°.
We have to calculate the airplane's altitude h.
From ΔABF,
tan4.1 =
h = 0.07168(x + 4.8) -----(1)
From ΔABD,
tan13.1 =
h = 0.2327x -----(2)
From equation (1) and (2),
0.07168(x + 4.8) = 0.2327x
0.2327x - 0.07168x = 4.8×0.07168
0.161x = 0.344
x = miles
From equation (2),
h = 0.2327×2.137
h = 0.4972 miles
h ≈ 0.5 miles
Therefore, 0.5 miles is the altitude of the plane.
Answer:
V=25088π vu
Step-by-step explanation:
Because the curves are a function of "y" it is decided to take the axis of rotation as y
, according to the graph 1 the cutoff points of f(y)₁ and f(y)₂ are ±2
f(y)₁ = 7y²-28; f(y)₂=28-7y²
y=0; x=28-0 ⇒ x=28
x=0; 0 = 7y²-28 ⇒ 7y²=28 ⇒ y²= 28/7 =4 ⇒ y=√4 =±2
Knowing that the volume of a solid of revolution V=πR²h, where R²=(r₁-r₂) and h=dy then:
dV=π(7y²-28-(28-7y²))²dy ⇒dV=π(7y²-28-28+7y²)²dy = 4π(7y²-28)²dy
dV=4π(49y⁴-392y²+784)dy integrating on both sides
∫dV=4π∫(49y⁴-392y²+784)dy ⇒ solving ∫(49y⁴-392y²+784)dy
49∫y⁴dy-392∫y²dy+784∫dy =
V=4π( ) evaluated -2≤y≤2, or 2(0≤y≤2), also
⇒ V=25088π vu
