The axis of symmetry is at x = -3.
This can be found by looking at the basic form of vertex form:
y = (x - h)^2 + k
In this basic form the vertex is (h, k). By looking at what is plugged into the equation, it is clear that h = -3 and k = -4. This means the vertex is at (-3, -4).
It is a fact that the axis of symmetry is a vertical line of x = (vertex value of x). So we can determine that the axis of symmetry is at x = -3
Yes it is a function
Y=x(x-1)
X(x-1)
X to the 2nd power -1x
Hope that’s helps
Step-by-step explanation:
1. (2+4+1)/9 = 7/9
2. 2 1/3 + 2/3 = 2 + (1+2)/3 = 2 + 3/3 = 2+1 = 3
3. 1 1/5 + 2 3/5 = 1+2 + 1/5 + 3/5 = 3 + (1+3)/5 =
3 4/5
4. 5/6 + 2/10 + 1/5 = 5/6 + 1/5 + 1/5 = 5/6 + 2/5
= (5×5)/(6×5) + (2×6)/(5×6)
= 25/30 + 12/30
= (25+12)/30
= 37/30 = 1 7/30
5. 3 1/2 + 4 2/3
= 3+4 + 1/2 + 2/3
= 7 + (1×3)/(2×3) + (2×2)/(3×2)
= 7 + 3/6 + 4/6
= 7 + (3+4)/6 = 7 7/6 = 8 1/6
6. 9/13 - 5/13 = (9-5)/13 = 4/13
7. 7 6/8 - 5 2/8
= (7-5) + (6/8 - 2/8)
= 2 + 4/8
= 2 1/2
8. 2/3 - 3/7
= (2×7)/(3×7) - (3×3)/(7×3)
= 14/21 - 9/21 = (14-9)/21 = 5/21
9. 11 1/5 - 5 4/5
= 10 6/5 - 5 4/5
= (10-5) + (6/5 - 4/5)
= 5 + 2/5 = 5 2/5
10. 15 4/5 - 7 7/10
= (15-7) + (4/5 - 7/10)
= 8 + (4×2)/(5×2) - 7/10
= 8 + 8/10 - 7/10
= 8 + 1/10
= 8 1/10
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
