The following questions can be answered in the next four days!

Hi, everyone! Here are some brand new questions to answer!


1. Which of the following are true? There are multiple answers. [30 points]

Try not to use the calculator, though. They can all be done with simple calculations without even knowing the exact values of them.

(A) sin 50° < cos 50°
(B) tan 50° < cot 50°
(C) tan 50° < sec 50°
(D) sin 230° < cos 230°
(E) tan 230° < cot 230°

2. On a two-dimensional plane, where do $ y = 4^x $ and $ y = 2^{(3x+2)} $ intersect? [45 points]No lucky prime hint today. Though, you can still guess it!

Answers for last time

For number 1: The answer is easily found by simplifying $ \frac{1}{n} + \frac{2}{n} + \cdots + \frac{10}{n} $ into $ \frac{1 + 2 + 3 + \cdots + 10}{n} = \frac{55}{n} $, and therefore n must be 1, 5, 11, or 55.

For number 2: Correct! The answer is 4. This can be found by simply multiplying 2 over and over again and find that it loops at each 20 numbers. The same happens for $ 3^n $, actually, looping every 20 numebrs.

For number 3: I should have used this one as a checkpoint question since there is some difficulty with this one.

$ \begin{cases} x + 2y + az = 1 & \text{(1)}\\ 3x + 4y + bz = -1 & \text{(2)}\\ 2x + 10y + 7z = c & \text{(3)} \end{cases} $

A) (If there is a solution) Exactly one set of answers is available.

B) (If there is a solution) 11a - 3b ≠ 7

C) (If there is a solution) c = 14

D) (If there are no solutions) 11a - 3b = 7

E) (If there are no solutions) c ≠ 14

First, a breakdown to the equations.

$ \begin{cases} (3) - (1) \times 2: 6y + (7-2a)z = c - 2 \\ (1) \times 3 - (2): 2y + (3a-b)z = 4 \end{cases} $

With the simplifications, it is now easier to go through the options:

A) Now we can see that there is a possibility for multiple sets of answers, or no sets of answers at all. For example, a = 1/2, b = -(1/2). Then if c = 6, then there are multiple solutions; otherwise, no solutions possible.

B) If there is one solution, $ \frac{6}{2} \ne \frac{7 - 2a}{3a - b} \Rightarrow 11a - 3b \ne 7 $

Still, keep in mind that if the result of a, b, and c causes there to be multiple solutions, $ \frac{6}{2} = \frac{7 - 2a}{3a - b} = \frac{c-2}{4} \Rightarrow 11a - 3b = 7 \text{and that } c = 14 $

This one is counted as correct because I worded the original option as "a solution", which, originally the question was worded "If solutions are possible". In this QOTD's case, the answer is correct.

C) As described "a solution", c does not always equates to 14 if there is exactly one solution. Only when there are multiple solutions can this be defined.

D) E) If there are no solutions, $ 11a - 3b = 7 \text{and that } c \ne 14 $

User:3primetime3 earned 16 points!


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