MaxWong

Let h(x) be the function you need.

From the graph, you see h(4) = f(1). Knowing that f is translated horizontally to h, h(x) = f(x - 3).

\(\mathcal P(\text{same color}) = \dfrac29\cdot\dfrac18 + \dfrac39\cdot \dfrac28 + \dfrac49\cdot\dfrac38 = \dfrac5{18}\)

We can actually solve the equation to see how many real solutions there are.

\(\begin{cases}y = x^2 - 5\\x^2 + y^2 = 25\end{cases}\\ x^2 + (x^2 - 5)^2 = 25\\ x^4 - 9x^2 = 0\\ x^2 = 0 \text{ or } x^2 = 9\\ x = -3, 0, 3\)

Plugging in these values into the equation gives

\(y = 4, -5, 4\)

in this order.

Therefore there are 3 pairs of real solution, namely \((-3, 4), (0, -5), \text{ and }(3, 4)\)

You can use the definition of derivative on \(f'(x)\) to find \(f''(x)\) because \(f''(x) = \displaystyle\lim_{h\to0}\dfrac{f'(x + h) - f'(x)}{h}\).

If f is increasing on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f'(x) \geqslant 0\).

If f is decreasing on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f'(x) \leqslant 0\).

If f is concave downward on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f''(x) \leqslant 0\).

If f is concave upward on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f''(x) \geqslant 0\).

These are the definitions. I believe you can finish the rest with these.

The area is equal to \(\left|\mathbf{u} \times \mathbf{v}\right|\).

Notice that \(\mathbf{u}\times \mathbf{v} = \det\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&-2&0\\6&-2&0\end{bmatrix} = \det\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&-2&0\\1&0&0\end{bmatrix} = 2\mathbf{k}\)

Therefore \(\left|\mathbf{u} \times \mathbf{v}\right| = |2\mathbf{k}| = 2\)

Therefore the area of the required parallelogram is 2 sq. units.

Yes.

For the base case: n = 0.

\(\text{When }n = 0,\\ 5^{3^{0}} + 1 = 6\\ 3^{0 + 1} = 3\\ 3|6\)

Assume \(5^{3^k} + 1\text{ is divisible }3^{k + 1}\).

\(5^{3^{k + 1}} + 1 = 5^{3(3^k)} + 1 = (5^{3^k} + 1)\left(5^{2(3^k)} - 5^{3^k} + 1\right)\)

Now, \(5^{2(3^k)} \equiv 2^{2(3^k)} \equiv 4^{3^k} \equiv 1^{3^k} \equiv 1 \pmod 3\)

\(5^{3^k}\equiv 2^{3^k}\equiv 2^{2K + 1}\equiv 2 \pmod 3\) for some \(K\in \mathbb Z\)

Therefore \(3|\left(5^{2(3^k)} - 5^{3^k} + 1\right)\)

The proposition is proved.

P.S.: I cheated and used some modular arithmetic ^_^

The triangle is a right-angled triangle.

Radius of circumcircle = 26/2 = 13

Area of circumcircle = \(\pi \cdot 13^2 = 169\pi\)

Radius of incircle = \(\dfrac{2\times\text{Area}}{\text{Perimeter}} = 4\)

Area of incircle = \(\pi \cdot 4^2 = 16\pi\)

Therefore \(a - b = 169\pi - 16\pi = 153\pi\)

a. HCF = 2x. 

\(8x^3 - 30x = 2x(4x^3 - 15)\)

b. HCF = 2

\(-2a^2 + 4x + 10 = 2(-a^2 + 2x + 5)\)

c. HCF = wx

\(7wx^2 - wx = wx(7x - 1)\)

\(\quad f(0) + f(-1) + f(-2) + f(-3)\\ = 0 + 1 + 0 + (-3)^{-2}\\ = \dfrac{10}9\)