Answer: y = v₀tgθx - gx²/2v₀²cos²θ
a = v₀tgθ
b = -g/2v₀²cos²θ
Step-by-step explanation:
x = v₀ₓt
y = v₀y.t - g.t²/2
x = v₀.cosθt → t = x/v₀.cosθ
y = v₀y.t - g.t²/2
v₀y = v₀.senθ
y = v₀senθ.x/v₀cosθ - g/2.(x/v₀cosθ)²
y = v₀.tgθ.x - gx²/2v₀²cos²θ
a = v₀tgθ → constant because v₀ and θ do not change
b = - g/2v₀²cos²θ → constant because v₀, g and θ do not change
I’m not sure if 1/6 is suppose to be 16 so if it is not let me know
A= 126
A= a+b/2 x h
12+4 / 3 x 16
Answer:
2b+9
Step-by-step explanation:
Twice the quantity b is 2*b = 2b
...and then you add 9 which is:
2b+9
8 post 4ft is the length 2ft ,yes if you dived 5 from 40 that is how many post , then you dived how many posts and that is the height then you dvide that and that is your width.
Let width = w
Let length = l
Let area = A
3w+2l=1200
2l=1200-3w
l=1200-3/2
A=w*l
A=w*(1200-3w)/2
A=600w-(3/2)*w^2
If I set A=0 to find the roots, the maximum will be at wmax=-b/2a which is exactly 1/2 way between the roots-(3/2)*w^2+600w=0
-b=-600
2a=-3
-b/2a=-600/-3
-600/-3=200
w=200
And, since 3w+2l=1200
3*200+2l=1200
2l = 600
l = 300
The dimensions of the largest enclosure willbe when width = 200 ft and length = 300 ft
check answer:
3w+2l=1200
3*200+2*300=1200
600+600=1200
1200=1200
and A=w*l
A=200*300
A=60000 ft2
To see if this is max area change w and l slightly but still make 3w+2l=1200 true, like
w=200.1
l=299.85
A=299.85*200.1
A=59999.985
