__The following questions can be answered in the next four days!__

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

## Questions

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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## Achievements

The Badges will still be continued further.

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