f(3) = 3f(1) + 2f(2) = 3(-1) + 2(3) = 3
f(4) = 3f(2) + 2f(3) = 3(3) + 2(3) = 15
f(5) = 3f(3) + 2f(4) = 3(3) + 2(15) = 39
1) take the arcsin
arcsin (-0.61) = θ = -37.6°
However this is a 3rd quadrant angle
So.... θ = 180 + 37.6 = 217.6°
See heureka's excellent answer here :
https://web2.0calc.com/questions/geometry_38810
1
________________ =
(1 - t)^2 - ( 1 + t)^2
_________________________ =
(1 - 2t + t^2) - ( 1 + 2t + t^2)
_________________ ( the denominator cannot = 0...... so t cannot equal 0 )
- 4t
Domain = (-inf , 0) U ( 0 , inf)
p(1, -1) = (1) - 2(-1) = 1 + 2 = 3
p(-5, -2) = 3(-5) + (-2) = -15 - 2 = -17
So
p ( p (1,-1) , p(-5,-2) ) = p ( 3 , -17) =
(2) - 2(-17) =
36
Using the Law of Cosines
3.1^2 = 2^2 + 3.2^2 - 2 (2)(3.2) cos F
[ 3.1^2 - 2^2 - 3.2^2 ] / [- 2 * 2 * 3.2 ] = cos F
Using the cosine inverse
arccos ( [ 3.1^2 - 2^2 - 3.2^2 ] / [- 2 * 2 * 3.2 ] ) = F ≈ 68.8°
Simplifying the given expression we get
1/ b^2 - a^2
The first term will be greatest when b = 3 and the second term will be least when a = -2
So....the max value is
1/(3)^2 - (-2)^2 = 1/9 - 4 = -35/9
There is no value of r that makes this true
See the graph here (in x) of the function : floor (r) + r - 15.5 - floor (r/4)
https://www.desmos.com/calculator/sb9nm8plvj
If we set this function to 0, the graph never intercepts the x axis
So....no solution
OK....good deal !!!!
C
4 5
A 3 B
cos B = 6/10 = 3/5 → sin C = 3/5
sin B = 8/10 = 4/5 → cos C = 4/5
tan C = sin C / cos C = (3/5) / (4/5) = 3 / 4
