Answer:
y = 14 
Step-by-step explanation:
For an exponential function of the form y = a
Use the given points to find a and b
Using (0, 14 ), then
14 = a
( = 1 ) , thus
a = 14
y = 14
Using (3, 3024 ) , then
3024 = 14 b³ ( divide both sides by 14 )
216 = b³ ( take the cube root of both sides )
b = ![\sqrt[3]{216}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216%7D)
= 6 , thus
y = 14 × 
← exponential function
<h3>Given:</h3><h3>Large cone:</h3>
<h3>Small cone:</h3>
<h3>Note that:</h3>
<h3>To find:</h3>
- The volume of the frustum of the given cone.
<h3>Solution:</h3>
- Frustum is a part of a cone formed by cutting off the top by a parallel plane.
Let's solve!
First, let's find the volume of the smaller cone.
Substitute the values according to the
formula.
Now, we can round off to the nearest hundredth.
The value in the thousandths place is smaller than 5 so we won't have to round up.
Next, let's find the volume of the bigger cone.
Substitute the values according to the formula.
Now, we can round off to the nearest hundredth.
The value in thousandths place is smaller than 5 so we won't have to round up.
Now, we can find the volume of the frustum.
We'll have to minus the volume of the smaller cone from the bigger cone.
<u>Hence, the volume of the frustum is 1172.86 cubic centimeters.</u>
The cow will produce 520.25 in one day.
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
