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>
First solve the quadratic as you would an equation, so you will get two real zeroes p and q so that (x-p)(x-q)=0 is another way of expressing the quadratic. All quadratics can be represented graphically by a parabola, which could be inverted. When the x² coefficient is negative it’s inverted. If the coefficient of x² isn’t 1 or -1 divide the whole quadratic by the coefficient so that it takes the form x²+ax+b, where a and b are real fractions. The curve between the zeroes will be totally below the x axis for an upright parabola, and totally above for an inverted parabola. This fact is used for inequalities. An inequality will be <, ≤, > or ≥. This makes it easy to solve the inequality. If the position of the curve between the zeroes is below the axis then outside this interval it will be above, and vice versa. So we’ve defined three zones. x
q, and p
The answer is that c3x was the answer
15/3x5+4x8=40
1)15/3x9x8=40
2)15/27x8=40
3)15/216=40
4)14.4=40
Suppose a triangle has sides a, b, and c let theta be the angle opposite the side of length a. If cos theta < 0 The statement that is true is <span>b^2+c^2>a^2. The answer is letter A.</span>
