Answer:
a=16
Step-by-step explanation:
16/4+2 = 6
Thus the answer is 16
No I do not remember you but it’s nice to meet you I suppose.
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
The median is 6, and the mean is 8.5
Answer:
(a) y = 9/4x -1/4
(b) (-2, -2), (-2, 2)
(c) see below
Step-by-step explanation:
(a) The derivative can be found from ...
2y·y' = 3x² +6x
At (x, y) = (1, 2), this is ...
4y' = 9
y' = 9/4
so the equation of the tangent is ...
y = (9/4)(x -1) +2
y = (9/4)x -1/4
__
(b) At y' = 0, we have ...
0 = 3x² +6x = 3x(x +2)
This says y' = 0 at x=0 or x=-2. (x = 0 is an extraneous solution.)
At x = -2, we have ...
y² = (-2)³ +3(-2)² = -8+12 = 4
y = ±2
So, the horizontal tangent points are (x, y) = (-2, ±2).