MaxWong

Ignoring the toxic "discussion" above and actually solving the problem:

\(\dfrac{3 - z}{z + 1} \geq 1\\ \dfrac{3 - z}{z + 1} (z + 1)^2 \geq (z + 1)^2\text{ and }z +1 \neq 0\\ (3 - z)(z + 1) \geq (z + 1)^2\text{ and }z +1 \neq 0\\ -z^2 + 2z + 3\geq z^2 + 2z + 1 \text{ and }z +1 \neq 0\\ 2z^2 - 2 \leq 0 \text{ and }z +1 \neq 0\\ z^2 - 1 \leq 0 \text{ and }z +1 \neq 0\\ -1 \leq z \leq 1 \text{ and }z \neq -1\\ -1 < z \leq 1\)

Alternative solution:

The sum of coefficients of a polynomial \(p(x)\) is actually just \(p(1)\).

Proof: (You can omit this part if you just want the answer instead of the explanation.)

Let \(p(x) = a_n x^n + a_{n - 1}x^{n - 1} + a_{n - 2}x^{n - 2} + \cdots + a_2 x^2 + a_1 x + a_0\).

Then \(p(1) = a_n 1^n + a_{n - 1} 1^{n- 1} + \cdots + a_1 \cdot 1 + a_0 = a_n + a_{n -1} + a_{n-2} + \cdots + a_1 + a_0\), which is exactly the sum of coefficients of p(x).

Therefore, we just find \(g(1) = f(1 + 7) = f(8)\). The sum of coefficients of g(x) is \(f(8) = 8^7 - 3\cdot 8^3 + 2\).

7 revolutions in 14 minutes means an average of 1 revolution per 2 minute.

That means the angular velocity of the Ferris wheel is \(\omega = \dfrac{2\pi}{120}\text{ rad s}^{-1} =\dfrac\pi{60} \text{ rad s}^{-1}\).

Using the formula \(v = r \omega\), we have the linear velocity = \(v = (30)\left(\dfrac\pi{60}\right) = \dfrac\pi2 \text{ ft s}^{-1}\).

Edit: That's 1.6 ft/s if you round it off to the nearest tenth.

Let P be the point on the tangent at the bottom of the image.

Note that \(\angle PXZ = \angle XZW\) since they are alternate angles of the pair of parallel lines.

Also, \(\angle PXZ = \angle XWY\) since they are angles in alternate segments.

Therefore \(\angle XZW = \angle XWY\). Also, \(\angle ZXW = \angle WXY\) since they are the same angle.

By AA postulate, \(\triangle XZW \sim \triangle XWY\).

Let YZ = t. Then by similar triangles, \(\dfrac{XZ}{XW} = \dfrac{XW}{XY}\). i.e., \(\dfrac{14 - t}{6} = \dfrac{6}{14}\).

Can you take it from here?

Since A + 2 = C + 2 = D + 2, A = C = D.

Since A + 2 = B - 2, B = A + 4.

Then A + (A + 4) + A + A = 36. i.e., A = 8.

Then A = C = D = 8 and B = 12. ABCD = 8^3 * 12 = 6144.

Note that the right hand side is just a constant. 

Take log on both sides:

\(\log_9 (9^y) = \log_9 \left(\dfrac{3^5 \cdot 9 \cdot 6}{27 \cdot 3}\right)\\ y = \log_9 \left(\dfrac{3^5 \cdot 9 \cdot 6}{27 \cdot 3}\right)\)

I will leave the simplification to you.

Multiply by (x^2 - 1) on both sides gives:

\(A(x + 1) + B(x - 1) = x + 8\\ (A + B)x + (A - B) = x + 8\)

Note that this holds true for any real number x.

Therefore, we can compare the coefficients of x on both sides.

\(\begin{cases}A + B &=& 1\\A - B &=& 8\end{cases}\)

Can you take it from here?

Consider the coefficient of x^2 on both sides.

\(5(10) + (-6)(t) + 7(4) = 60\\ -6t + 78 = 60\\ t = 3\)

Yeah, Vinculum's answer is correct.

That means \(2\lfloor r \rfloor + 2r = 31 + 2\lfloor 4r \rfloor\).

Now, the right-hand side of the equation is an integer and \(2\lfloor r \rfloor\) is an integer. That means \(2r\) is also an integer.

Case 1: Let \(r = x + 0.5\), where x is an integer.

Then 

\(2x + \dfrac12 = 15.5 + 4x + 2\\ 2x = 17 + 4x\\ 2x = -17\\ x = -\dfrac{17}2\)

which is not an integer. Therefore, there are no roots of such form.

Case 2: \(r\) is an integer.

\(2r = 15.5 + 4r\)

\(r = -\dfrac{31}4\)

which is not an integer.

Therefore, there are no such \(r\) satisfying the equation.

By Menelaus' theorem,

\(\dfrac{QO}{ON} \cdot \dfrac{NR}{RP}\cdot \dfrac{PM}{MQ} = 1\\ \dfrac{QO}{ON} \cdot \dfrac12 \cdot \dfrac11 = 1\\ \dfrac{QO}{ON} = 2\\ QO = 2 ON\)

Since QN = 12, QO = 8 and ON = 4.

Since PN = NR and PR = 18, NR = 9.

Then by Pythagorean theorem, 

\(OR = \sqrt{NR^2 + ON^2} = \sqrt{4^2 + 9^2} = \sqrt{97}\)