1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
Answer:
a. V = (20-x)
b . 1185.185
Step-by-step explanation:
Given that:
- The height: 20 - x (in )
- Let x be the length of a side of the base of the box (x>0)
a. Write a polynomial function in factored form modeling the volume V of the box.
As we know that, this is a rectangular box has a square base so the Volume of it is:
V = h *
<=> V = (20-x)
b. What is the maximum possible volume of the box?
To maximum the volume of it, we need to use first derivative of the volume.
<=> dV / Dx = -3
+ 40x
Let dV / Dx = 0, we have:
-3
+ 40x = 0
<=> x = 40/3
=>the height h = 20/3
So the maximum possible volume of the box is:
V = 20/3 * 40/3 *40/3
= 1185.185
Answer:
75000 centigrams
Step-by-step explanation:
At the end of the year, Juan has 52.71 more than 4 times his balance at the beginning. Okay, let's set this up.
4x + 52.71
(4 times) (52.71 more)
His ending was 172.90, so
4x + 52.71=172.90
4x= 120.19
x= 30.05
He had $30.05 at the beginning of the year.