Volume of a cone, V = πr^2h/3
r = Sqrt 3V/πh = Sqrt [(58.9*3)/(3.14*9)] = 2.5 cm, minimum
Surface area, A = πr [r+ Sqrt (h^2+r^2)] = 3.14*2.5*[2.5+ Sqrt (9^2+2.5^2)] = 92.95 cm^2
Therefore, minimum amount of paper required is 92.95 cm^2
Answer:
The equation is:
f (t) = 4 + 5 (1 - cos (2pi t / 2))
Step-by-step explanation:
with the previous exercise we look for the equation for h = f (t)
So the data we have are
Wheel diameter = 10m (wheel radius = 5m)
1 wheel gets 1 revolution in 2 minutes.
the beginning of a entry will be related to that f (0) = 4
our wish is that f (z) get at least 4 with an amplitude of 5 (this value determines the radius of the wheel) for 2 minutes
with this the particle f (t) is transformed into
f (t) = 4 + 5 (1 - cos (2pi t / 2))
We know that the maximum value of cos in t will be 0, 1 -cos has minutes, the result will be as follows:
f (t) = 4 + 5 (1 - cos (2pi t / 2))
Answer:
B and D
7/12 cups
Step-by-step explanation:
2 1/3 + 2 1/4 = cups of flour needed
4 = cups of flour had
2 1/3 + 2 1/4
To have the same denominator, we take the lcm
2 + 2 + (4/12 + 3/12) =
2 4/12 + 2 3/12 - 4
Or
2 + 2 + (3 + 4) / 12 - 4
2 + 2 + (3 + 4) / 12 - 4 = 4 7/12 - 4
4 7/12 - 4 = 7/12 cups of flour needed
1st term = -10
2nd term = -6
3rd term = -2
4th term = 2
5th term = 6
6th term = 10
7th term = 14
The reason how I got 14 for the 7th term is because, i added 4 to each term.
Hope this helps!
Answer:
A reflection over the line x=3
Step-by-step explanation: