The distance from the center $(-6, -7)$ to the inside point $(-6, 5)$:

$d_1 = \sqrt{(-6-(-6))^2 + (5-(-7))^2} = \sqrt{0 + 144} = 12$

Since $(-6, 5)$ is inside the circle: $m > 12$.

The distance from the center $(-6, -7)$ to the outside point $(8, -7)$:

$d_2 = \sqrt{(8-(-6))^2 + (-7-(-7))^2} = \sqrt{196 + 0} = 14$

Since $(8, -7)$ is outside the circle: $m < 14$.

So $12 < m < 14$, and $m$ is an integer, therefore $m = \mathbf{13}$.

• \(25 = 5^2\)
• \(30 = 2 \times 3 \times 5\)
• \(35 = 5 \times 7\)
• \(40 = 2^3 \times 5\)

Square Pyramid analysis:

• Top view: A square base with diagonals → appears as a square ✓
• Front view: Triangular elevation ✓
• Side view: Triangular elevation ✓

A cone would give a circular top view. A triangular prism would give a rectangular side view. A cylinder gives circular front/side views.

All three views match a square pyramid perfectly.

Let $I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x}{1+2^x}dx$. Use the property $\int_{-a}^a f(x)dx$: replace $x\to-x$:

$I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x}{1+2^{-x}}dx=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{2^x\sin^2x}{1+2^x}dx$

Adding: $2I=\displaystyle\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{1-\cos2x}{2}dx=\dfrac{\pi}{2}$

$I=\dfrac{\pi}{4}$

Cyclosis (cytoplasmic streaming) is the directed flow of cytosol that moves organelles like mitochondria. It is driven by actin-myosin interactions.