7/8 = .875
So
[ 56 + 7/8 ] % =
[56 + .875 ] % =
56.875 %
2
x^2 + 10x + 16 [ 2x^2 + 15x - 8 ]
2x^2 + 20x + 32
_____________
-5x - 40
So.......the answer is 2 + [ - 5x - 40 ] / [ x^2 + 10x + 16 ]
We can use this identity :
tan (2t) = [2 tan ( t) ] / [ 1 - tan^2 (t) ] =
2 (12/5) / [ 1 - (12/5)^2 ] =
(24 / 5) / [ 1 - 144 / 25 ] =
(24 / 5) / ( [25 - 144] / 25 ) =
(24 / 5) / ( -119/25 ) =
(24 / 5) * ( -25 / 119) =
(-24 * 5) / 119 =
-120 / 119
e^x = -x
A graph might be the easiest way to solve this
Here is the solution : https://www.desmos.com/calculator/bjsnjcbch7
It occurs at x ≈ -.567
√2 cos(2 θ ) = 1 divide both sides by √2
cos (2 θ ) = 1 / √2
Let 2 θ = x
cos ( x) = 1 / √2 and this happens at x = pi/4 , 7pi/4 , 9pi/4 , 15pi/4
2 θ = pi/4, 7pi/4 , 9pi/4, 15pi/4 divide by 2
θ = pi / 8 , 7pi/ 8, 9pi / 8, 15pi / 8 answer "B"
sin A = 0.3154
Take the inverse sine
arcsin (0.3154) ≈ 18.38° = 18° or [180 - 18.38]° ≈ 161.62° = 162°
The area =
(1/2) (15) (44) sin (66) ≈ 301.47 square feet
cos^2(20°) + cos^2(52°) + cos^2(38°) + cos^2(70°)
Note cos (20°) = sin (70”)
And cos(52°) = sin (38°)
So.....we have
sin^2(70°) + sin^2(38°) + cos^2(38°) + cos^2(70°) rearrange as
[sin^2 (70°) + cos^2(70°)] + [ sin^2(38°) + cos^2(38°) ] =
1 + 1 = 2
To test whether we have a right triangle....this must be true.....
18^2 + 25^2 = 33^2
949 = 1089 ???? ....... no......not a right triangle
cos^2(x) + cos^2(pi - x)= 1
cos^2(x) + [ cos (pi)cos (x) + sin(pi) sin(x) ] ^2
cos^2(x) + [ - cos(x) ] ^2
cos^2(x) + cos^2(x)
2cos^2(x)
This is not an identity !!!
