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:
The volume of the cylinder = π r² h
where r is the radius of the cylinder and h is the height of the cylinder.
This is formula is applied for the right cylinder as figure 1 and oblique cylinder as figure2.
<u>The volume of the cylinder of figure 1:</u>
r = 6 and h = 7
volume = π r² h = π * 6² * 7 = 252π units³
<u>The volume of the cylinder of figure 2:</u>
r = 11 and h = 15
volume = π r² h = π * 11² * 15 = 1,815π units³
If you are 12 years old, just measure your own height, then multiply your height by 1.19 if you’re a male, or 1.07 if you’re a female.
Answer:
FD≈25.94.. rounded = 26
Step-by-step explanation:
FD²=12²+(4x+11)²
FD²=144+16x²+88x+121
FD²=265+16x²+88x
also
FD²=12²+(13x-16)²
FD²=144+169x²-416x+256
FD²=400+169x²-416x
thus
265+16x²+88x = 400+169x²-416x
16x²-169x²+88x+416x+265-400 = 0
-153x²+504x-135 = 0
we will solve this quadratic equation by suing the quadratic formula to find x
x=(-504±sqrt(504²-4(-153)(-135)))/2(-153)
x=(-504±
)/2(-153)
x=(-504±
)/-306
x=(-504±
)/-306
x=(-504±414)/-306
x=(-504+414)/-306 and x=(-504-414)/-306
x=-90/-306 and x=-918/-306
x= 5/17 , 3
substituting x by the roots we found
check for 5/17:
4x+11 = 4×(5/17)+11 = (20/17)+11 = (20+187)/17 = 207/17 ≈ 12.17..
13x-16 = 13×(5/17)-16 = (65/17)-16 = (65-272)/17 = -207/17 ≈ -12.17..
check for 3:
4x+11 = 4×3+11 = 12+11 = 23
13x-16 = 13×3-16 = 23
thus 3 is the right root
therfore
ED=23 and CD=23
FD²=FE²+ED² or FD²=FC²+CD²
FD²=12²+23²
FD²=144+529
FD²=673
FD=√673
FD≈25.94.. rounded = 26
Answer:
94% so C
Step-by-step explanation:
I did the assignment
Hope it helps