21x-33°=180°-135°
21x=12°
x=12/21
Answer:
25 remainder 11
Step-by-step explanation:
used a calc
Area of the parabolic region = Integral of [a^2 - x^2 ]dx | from - a to a =
(a^2)x - (x^3)/3 | from - a to a = (a^2)(a) - (a^3)/3 - (a^2)(-a) + (-a^3)/3 =
= 2a^3 - 2(a^3)/3 = [4/3](a^3)
Area of the triangle = [1/2]base*height = [1/2](2a)(a)^2 = <span>a^3
ratio area of the triangle / area of the parabolic region = a^3 / {[4/3](a^3)} =
Limit of </span><span><span>a^3 / {[4/3](a^3)} </span>as a -> 0 = 1 /(4/3) = 4/3
</span>
The solution is x<5
so
C,E,F, and G
We are already given with the function to solve for the area:
<span>A(θ) = 16 sin θ ⋅ (cos θ + 1)
</span>
We simply have to substitute the value of the angle into the function. So,
If <span>θ = 90°,
A(</span>90°) = 16 sin (90°) ( cos (<span>90°) + 1 )
Using the calculator or the definition of trigonometric functions at angle of </span><span>90°, we get the value of the area:
</span>A(<span>90°) = 16 square inches</span>
