MaxWong

Area of that cross section = Area of triangle ABC.

Using Pythagoras theorem, BC = \(\sqrt{17^2-8^2} = 15\)

Area of ABC = 8*15/2 = 60 sq ft

Area of the cross section = 60 sq ft :)

The longest line segment that can be drawn inside an arc is the red line below:

Diameter = 12cm --> Radius = 6cm

Central angle of one arc = 360 deg / 3 = 120 deg

l^2 = 6^2 + 6^2 - 2(6)(6)cos120 deg (Law of Cosines)

I^2 = 72(1-cos(120)) = 108.

\(\boxed{\text{Degree to radian: } \theta_{\text{rad}}=\theta_{\text{deg}}\cdot\dfrac{\pi}{180^{\circ}}}\)

\(\quad 510^{\circ}\\ =\dfrac{510^{\circ}\cdot\pi}{180^{\circ}}\\ =\dfrac{17\pi}{6}\text{rad}\)

\(\quad\sqrt{75x^3y^6}\\ =\sqrt{75}\cdot\sqrt{x^3}\cdot\sqrt{y^6}\\ =\sqrt{25}\cdot\sqrt3\cdot(x\cdot\sqrt{x})\cdot(y^3)\\ =5xy^3\sqrt{3x}\)

Therefore the fourth option is the answer.

\(\begin{align} f(x)&=x^3+x^2-9x-9\\ f(x)&=x^2(x+1)-9(x+1)\\ f(x)&=(x^2-9)(x+1)\\ f(x)&=(x+3)(x+1)(x-3)\\ \end{align}\\ \therefore \text{-3, -1 and 3 are zeros of the polynomial.}\)

\(\begin{align}\dfrac{1}{4}\left(\dfrac{5}{12}x-\dfrac{3}{4}\right)& > x+\dfrac{1}{12}\\ \dfrac{5}{48}x-\dfrac{3}{16}& > x + \dfrac{1}{12}\\ 5x-9& >48x+4\\ 43x&<-13\\ x&<-\dfrac{13}{43} \end{align}\)

\(\quad(x+2)(-x^2+2x-6)\\ =x(-x^2+2x-6)+2(-x^2+2x-6)\\ =-x^3+2x^2-6x-2x^2+4x-12\\ =-x^3-2x-12\)

Therefore the answer is the third option.

3x(x+4)(x-6) = 0

x = 0 or x+4 = 0 or x-6 = 0

x = 0 or x = -4 or x = 6

I think the question is asking all roots(x) are real, not a.

Therefore, shouldn't we solve the general root first, then find the probability that all x are real?

I mean:

\(x^3+(2a+1)x^2+(4a-1)x+2 = 0\\ x = -2 , \dfrac{\sqrt{4a^2-4a-3}-2a+1}{2}, \dfrac{-\sqrt{4a^2-4a-3}-2a+1}{2}\)

So the question becomes probability that \(\sqrt{4a^2-4a-3}\) is real.

Minimum value of f(x) = |0| - 4 = -4

Maximum is +infinity.

Therefore the range of f(x) is [-4,+infinity).