Answer:
if by distance around the hula hoop you mean circumference, then -
circumference = 2 pi r
= 2 X 22/7 X 14
= 4 X 22
= 88in
Answer:
If the stand-alone method would be used the startup would pay $60,000.00.
Step-by-step explanation:
The current cost for Spring Harbor Corporation is $180,000.00, but they use only 70% of the corporate suite.
If the start-up divide the corporate suite, they will use only 30%, but the total cost will be 180,000.00 + 20,000.00 = 200,000.00
We add the values because 180,000.00 counts the cost of maintenance paid before.
Using the percentages to find the cost for the start-up:
So, $60,000.00 will be allocated to the start-up business.
Answer:
Step-by-step explanation:
for the first one is
Domain:
(−∞,∞),{x|x∈R}
Range:
(−7,∞),{y|y>−7}
Horizontal Asymptote:
y=−7
y-intercept(s):
(0,−6)
the second one is
y-intercept(s):
(0,8)
Horizontal Asymptote:
y=2
Domain:
(−∞,∞),{x|x∈R}
Range:
(2,∞),{y|y>2}
Answer:
1. 15
2. 8
Step-by-step explanation:
The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)
The nth term of a GP is;
Tn = ar^(n-1)
Where;
a = first term
r = common ratio
For the first sequence;
The common ratio r is
r = T3/T2 = 540/90 = 6
r = 6
T2 = ar^(2-1) = ar
T2 = 90 = ar
Substituting the values of r;
90 = a × 6
a = 90/6
a = 15
First term = 15
2. The sam method applies here.
Common ratio r = T3/T2 = 128/32 = 4
r = 4
T2 = ar^(2-1) = ar
T2 = 32 = ar
Substituting the values of r;
32 = a × 4
a = 32/4
a = 8
First term = 8
<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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