MaxWong

Let r(p) be the revenue.

r(p) = p(128-4p) = -4p^2 + 128p

Completing the square, r(p) = -4(p^2-32p) = -4(p-16)^2 + 1024.

Therefore, if the price is 16 dollars, the revenue will be maximized, which will be 1024 dollars.

Assuming you mean f(0), f(-2),f(3) instead of F being the antiderivative of f.

f(0) = 3(0) - 4 = -4

f(-2) = 3(-2) - 4 = -10

f(3) = 3(3) - 4 = 5

Now assume you mean the antiderivative.

F(0) = 3(0)^2 / 2 - 4(0) = 0

F(-2) = 3(-2)^2 / 2 - 4(-2) = 14

F(3) = 3(3)^2 / 2 - 4(3) = 3/2

Range of f(6x) is [-11,3]. So range of g(x)=f(6x)+1 is [-10,4]

\(r(x) = \dfrac{1}{(1-x)^2}\)

When x -> 1, r-> infinity. So the maximum value of r(x) is +infinity.

Now we just have to find the minimum value of r(x).

Horizontal asymptote of r(x) is y = 0, which means the r(x) is approaching 0 when x approachs +inf or -inf.

Therefore range of r(x) is (0, inf).

"In right triangle ABC with angle B = 90˚" implies that 0˚ < A < 90˚.

2 sin A = 3 cos A

2 tan A = 3

tan A = 3/2

A = arctan(3/2) = arccot(2/3)

sin A = sin(arccot(2/3)) = 1/sqrt(csc^2(arccot(2/3)) = 1/sqrt(1+cot^2(arccot(2/3))) = 1/sqrt(1+4/9) = \(\dfrac{3\sqrt{13}}{13}\)

We need to solve \(2(2^x-1)x^2+(2^{x^2}-2)x-2^{x+1}-2=0\).

Let \(f(x) = 2(2^x-1)x^2+(2^{x^2}-2)x-2^{x+1}-2\).

If you graph f(x) using any method, you find that it only touches x-axis at one point.

Graph link: https://www.desmos.com/calculator/wxie170yvi

We will use Newton-Rhapson to find solutions.

\(f'(x) = 4x(2^x-1)+2x^22^x\ln2+2^{x^2}-2+2x^2(2^{x^2}\ln2)-2^{x+1}\ln2\)

As this is too complicated, I solved it with a program.

Here is the source code in C++:
#include
using namespace std;
double f(double x){
    return 2*(pow(2,x)-1)*x*x+(pow(2,x*x)-2)*x-pow(2,x+1)-2;
}
double f_prime(double x){
    return 4*x*(pow(2,x)-1)+2*x*x*pow(2,x)*log(2)+pow(2,x*x)-2+2*x*x*(pow(2,x*x))*log(2)-pow(2,x+1)*log(2);
}
int main(){
    double x = 1.1;
    for (int i=0;i<1000;i++){ //iterate 1000 times
        x -= f(x)/f_prime(x);
    }
    printf("Solution: %100g", x);
}

Output: Solution: 1.32006

First we will need to find g(x).

\(g(3x-8) = 2x^2+5x-3\\ \quad g(x) \\= g\left(3\left(\dfrac{x+8}{3}\right)-8\right) \\= 2\left(\dfrac{x+8}{3}\right)^2+5\left(\dfrac{x+8}{3}\right)-3\\ =\dfrac{2}{9}\left(x^2+16x+64\right)+\dfrac{5}{3}(x+8)-3\\ =\dfrac{2}{9}x^2 +\dfrac{47}{9}x+\dfrac{221}{9}\)

Therefore g(-5) = \(\dfrac{2}{9}(-5)^2+\dfrac{47}{9}(-5)+\dfrac{221}{9}=\dfrac{50-235+221}{9} = \dfrac{36}{9} = 4\)

Ans = 4.

From \(g(3) = 9\), we know that \(f(9) = 3\).

From \(g(0) = 3\), we know that \(f(3) = 0\).

f(f(9)) = f(3) = 0.

Other informations are useless ones.

\(a\spadesuit 3 = 9\\\dfrac{a^2}{3} = 9\\a^2 = 27\\a = \pm\sqrt{27} = \pm3\sqrt{3} \)

Which means a = -3sqrt(3) or a = 3sqrt(3)

To make it less confusing, first know that: \(1-y = -(y-1)\).

Then, using the property of absolute value function, we have: \(|-x| = |x|\)

It is supposed to be \(-\ln|1-y| + C\), however, \(-\ln|1-y| = -\ln|-(y-1)| = -\ln|y-1|\).

Both \(-\ln|1-y| + C\) and \(-\ln|y-1| + C\) are correct answers.

Let \(\phi\) be the reference angle of \(\theta\).

\(\sin \theta = \dfrac{\sqrt3}{2}\\ \therefore \sin \phi = \dfrac{\sqrt3}{2}\\ \text{As }\phi\text{ is a reference angle, it always lies in Quadrant I.}\\ \therefore \phi = 60^{\circ}\\ \theta\text{ lies in Quadrant IV}\\ \therefore \theta = 360^{\circ}-60^{\circ} = 300^{\circ}\\\)

Triangle.

Imagine a clay(?) cone is cut into half from the top to the bottom. If you look at where you cut it into half, there should be a triangle.

Like this:

There you see the answer is triangle.