MaxWong

First, we see the g.c.d of the terms is 4x. We factor this out:

\(4x^3+8x^2-96x = 4x(x^2+2x-24)\).

Then we factor the quadratic. (If you don't know how, you can try expanding the expression in each option.)

\(4x^3+8x^2-96x = 4x(x^2+2x-24)=4x(x+6)(x-4)\).

So we see the answer is the second option.

You are saying that when you draw arbitrary radii of a circle, the angle sum at the center is 360 degrees. And when you draw all diagonals of a square, the angle sum at the center is also 360 degrees.

This is always true, because at any point, the angle sum at the same point is always 360 degrees. But as you say a triangle "is 180 degrees", I think you meant the interior angle sum of triangle is 180 degrees, which is half of 360 degrees. But this does not mean a circle is a square and a triangle is half a circle.

\(f(x) = x^2-6x+8, g(x) = x-2\)

You see that when x = 2, f(x)=g(x)=0 and when x = 5, f(x)=g(x)=3.

So f(x) = g(x) when x = 2 and x = 5.

From the table, P(1) = 0.04.

6.

By considering the discriminant of the equation,

\(\Delta = b^2 - 4(2)(18) = b^2 - 108\).

we know that b² - 108 must be 0 because the equation has a double root.

\(b^2 - 108 = 0\\ b = \pm \sqrt{108}\\ b= \pm6\sqrt 3\)

5. 

\(3x^2 - 4x + 15 = 0\\ x^2 - \dfrac{4}{3} x + 5 =0\\ \therefore (b,c) = \left(\dfrac{-4}{3},5\right)\)

\(\text{Let }A(-2,-2),B(4,-2),C(5,1),D(2,3)\).

Connect BD.

It is obvious that \(\text{Area of }\triangle ABD = \dfrac{6\cdot 5}{2} = 15 \text{ unit}^2\).

Now add point E(5, 3), F(2, -2), G(5, -2) on the same coordinate system.

Consider rectangle DEFG.

\(\text{Area of }\triangle BCD = \text{Area of rectangle } DEFG - \text{Area of }\triangle DEC - \text{Area of }\triangle CBG - \text{Area of }\triangle DFB\)

\(\text{Area of }\triangle BCD = 3\cdot 5 - \dfrac{2\cdot 5 + 3\cdot 2+1\cdot 3}{2} = 15 - \dfrac{19}{2}=\dfrac{11}{2} \text{ unit}^2\).

Therefore the total area of the figure is 15 + 11/2 = 20.5 unit².

\(f(x) = 3x^5 + 6x^2 - 5\\ g(x) = 2x^4+7x^2-x+16\\ (f-g)(x) = (3x^5+6x^2-5)-(2x^4+7x^2-x+16)\\ (f-g)(x) = (3x^5+6x^2-5)-2x^4-7x^2+x-16\\ (f-g)(x) = 3x^5-2x^4-x^2+x-21\)

max(g(x)) = 4 and max(f(x)) = 2, so the first statement is true.

One of the x-intercepts of g(x) is 5, while one of the x-intercetps of f(x) is also 5, so the second statement is true.

When x=0, g(x) = 3 and f(x) = -1. That means the y-intercept of f(x) is smaller than that of g(x). That means the third statement is false.

f(x) has 2 x-intercepts, while g(x) also has 2 x-intercepts. Therefore the fourth statement is false.

Conclusion: Only the first and the second statements are true.

1) Odd

2) Neither Even nor Odd

3) Neither Even nor Odd