Answer:
Height of the fighter plane =1.5km=1500 m
Speed of the fighter plane, v=720km/h=200 m/s
Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, u=600 m/s
Time taken by the shell to hit the plane =t
Horizontal distance travelled by the shell =u
x
t
Distance travelled by the plane =vt
The shell hits the plane. Hence, these two distances must be equal.
u
x
t=vt
u Sin θ=v
Sin θ=v/u
=200/600=1/3=0.33
θ=Sin
−1
(0.33)=19.50
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.
H
max
=u
2
sin
2
(90−θ)/2g=600
2
/(2×10)=16km
Answer:

Step-by-step explanation:
1. Use the distributive property
5 ( x - 7 ) = 6 ( x + 2 ) → 5x - 35 = 6x + 12
2. Subtract 5x from both sides of the equation
5x - 5x -35 = 6x - 5x + 12 → -35 = x + 12
3. Subtract 12 from both sides
-35 - 12 = x + 12 - 12 → -47 = x
4. So, the answer is

Answer:

<h3>p=-2 is the right answer.</h3>
Answer:
as cheeks bxnxndnndnsnsndvbsncnf
Answer:
An exponential function in the form y=9(4ˣ)
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that exponential function
y = a bˣ ...(i)
The equation (i) passes through the point (0,9)
9 = a (b )⁰ = a
a =9
<u><em>Step(ii):-</em></u>
The equation (i) passes through the point (3,576)
576 = 9(b)³
⇒ 
⇒ 64 = b³
⇒ 4³ = b³

b = 4
<u><em>Final answer:-</em></u>
An exponential function in the form y=abˣ
An exponential function in the form y=9(4ˣ)
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