Answer:
Make a coordinate plane and you have to solve around the origin say 4,4 and 4,1.
Step-by-step explanation:
I think it’s between 100-200 like 150 cause it’s near a 100 and more then a 100 too.
<span>et us assume that the origin is the floor right below the 30 ft. fence
To work this one out, we'll start with acceleration and integrate our way up to position.
At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.
To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60
To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4
If the ball clears the fence, it means x = 0, y > 30
x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds
for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft
which is less than 30ft, so it means that the ball will NOT clear the fence.
Just for fun, let's check what the speed should have been :)
x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4
x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s
v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s
So he needed to hit the ball at at least 123.34 ft/s to clear the fence.
You're welcome, Thanks please :)
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*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*
(21)
Area of a Regular Hexagon: 
square units
(22)
Similar to (21)
Area = 
square units
(23)
For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:
Hence, area of the hexagon will be: 
square units
(24)
Given is the inradius of an equilateral triangle.
Substituting the value of inradius and calculating the length of the side of the equilateral triangle:
Side = 16 units
Area of equilateral triangle = 
square units
