MaxWong

The system is equivalent to

\(\begin{bmatrix}11&-1\\a-5&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}-14\\-11\end{bmatrix}\)

The system has no solutions when the determinant of \(\begin{bmatrix}11&-1\\a-5&0\end{bmatrix}\) is 0.

\(\det \begin{bmatrix}11&-1\\a-5&0\end{bmatrix} = 0\\ -(a - 5) = 0\\ \boxed{a = 5}\)

Therefore, the system has no solutions when a = 5.

\(\angle J\) and \(\angle B\) are complementary => \(\angle J + \angle B = 90^\circ\) => \(\angle B = 90^\circ - \angle J\)

\(\sin B = \sin(90^\circ - J) = \cos J = \dfrac 9 {12} =\dfrac34\)

Multiply through by \(x^2 + 5x - 24\).

For all real values of x except -8 and 3,

\(C(x + 8) + D(x - 3) = 4x - 23\\ (C + D)x + (8C - 3D) = 4x - 23\)

Comparing coefficients,

\(\begin{cases}C + D = 4\\8C - 3D = -23\end{cases}\)

Solving, (C, D) = (-1, 5).

Therefore CD = -5.

\((2x - 16) : 12 = 1 : 3\\ 2x - 16 = 4\\ x = 10\)

P(x) = -54 (x^2 - 3x + 2)

P(x) = -54 (x^2 - 3x + (3/2)^2 + 2 - 9/4)

P(x) = -54(x - 3/2)^2 + 27/2

Number of units = 1.5 hundred

Maximum profit = 13.5 thousand dollars

Try to do part (c) on your own.

\(x^2 + y^2 = 2x - 6y + 6 +4x - 8y\\ x^2 + y^2 = 6x - 14y + 6\\ x^2 - 6x + y^2 + 14y - 6 = 0\\ (x^2 -6x + 9) + (y^2 +14y + 49) = 64\\ (x - 3)^2 + (y + 7)^2 = 8^2 \)

The graph is a circle centred at (3, -7) and with radius 8 units.

The required area is \(\pi \cdot 8^2 = \boxed{64\pi\text{ sq. units}}\)

Just use a program to find it.

C++ source code: (G++ compiler, C++14)

#include
using namespace std;
int main(){
    int a = 0b1011110, b = 0b0100001;
    printf("Bitwise OR: %x\nBitwise AND: %x\nBitwise XOR: %x", a|b, a&b, a^b);
}

Output:

Bitwise OR: 7f
Bitwise AND: 0
Bitwise XOR: 7f

You may notice the output is in base 16. But converting base 16 to base 2 is easy.

Bitwise OR of the two numbers gives 1111111

Bitwise AND gives 0000000

Bitwise XOR gives 1111111

Definition of a probability density function: 

If f(x) is a probability density function, then \(\displaystyle \int^6_0 f(x)\,dx = 1\).

This means for the probability density function in the question, \( \displaystyle \int^6_0 Ax(6 - x)\,dx = 1\).

Now, notice that \( \displaystyle \int^6_0 x(6- x)\,dx = 36\).

Therefore A = 1/36.

For part B, sketching a quadratic function should be easy.

For part C, just find \(\displaystyle \int^6_4 f(x)\,dx\).

For part D, let W be the weight of a random jar. We can find \(\mathbb E(W) = \displaystyle \int^6_0 xf(x)\,dx\) and \(\mathbb E(W^2) = \displaystyle \int^6_0 x^2f(x)\,dx\) and​ then use the formula \(\operatorname{Var}(W) = \mathbb E(W^2) - \left(E(W)\right)^2\)

You should know how the standard deviation and the variance are related.

\(\dfrac{(3x - 2)^\circ - (x + 6)^\circ}{2} = 34^\circ\\ x - 4 = 34\\ \boxed{x = 38}\)

Let $x be the cost of an apple, $y be the cost of orange juice.

\(\begin{cases}3x + y = 1.35\\x + 2y = 1.2\end{cases}\\ \begin{cases}6x + 2y = 2.7\\x + 2y = 1.2\end{cases}\\ 5x = (6x + 2y) - (x + 2y) = 2.7 - 1.2 = 1.5\\ x = 0.3\\ 0.3 + 2y = 1.2\\ y = 0.45\)