MaxWong

So, this is the diagram of the question: 

The point I is the center of the incircle, i.e., the incenter of triangle ABC.

By tangent properties, we have \(\angle IYB = \angle IXB = 90^\circ\).

Since \(\angle IYB + \angle IXB = 180^\circ\), the quadrilateral IYBX is a cyclic quadrilateral, so \(\angle XIY = 180^\circ - \angle XBY = 180^\circ - 80^\circ = 100^\circ\).

Now, since \(IX\) and \(IY\) are both radii of a circle, \(\triangle IXY\) is isosceles. Using this information, we can find the size of \(\angle IYX\). Then just note that \(\angle AYI = 90^\circ\), and add that to \(\angle IYX\). You will get \(\angle AYX\) that way, which is the required angle.

Can you take it from here?

\(3x + 7 \equiv 2 \pmod{16}\\ 3x \equiv -5 \pmod{16}\\ 33x \equiv -55 \pmod{16}\\ x \equiv 9 \pmod{16}\)

Now you just take x = 9, then calculate the value of 2x + 11 + 4x. The answer is the remainder when 2x + 11 + 4x is divided by 16.

Consider the area of triangle ABC. It can be calculated two different ways.

\(\text{Area of }\triangle ABC = \dfrac{AD \times BC}2 = \dfrac{AC \times BE}{2}\)

So, by simplifying and substituting the values, we actually have \(5\times 18 = 10 \times BE\). That means BE = 9, not 15/2 which Guest suggested.

We need to list all the inequalities and solve the system of inequalities.

Let $x be the amount of the original pile of money.

Then \(\begin{cases}x \geq 200\\80 + \dfrac{4(x - 80)}5 > \dfrac{4x}5\end{cases}\).

Now it is your turn to solve each inequality and find the range of values of x so that both inequalities are satisfied.

This is asked so many times. Please see this thread for the answer: https://web2.0calc.com/questions/floor-function_20#r1

By similar triangles,

\(\dfrac{AB}{AC} = \dfrac{AS}{AT}\\ \dfrac{AT + BT}{AC} = \dfrac{AS}{AT}\\ AS = \dfrac{AT(AT + BT)}{AC} = \dfrac{11(11 + 9)}{35} = \boxed{\dfrac{44}7}\)

Actually, we can ignore the walking distances of the people living at A and E, because their walking distance, no matter where the point is, must add up to 14. Also, it is not optimal to place the point P between A and B or between D and E, because either the person living at B or the person living at D would have to walk a whole lot. We conclude that the optimal location of the point P must be somewhere between B and D. Using the same logic, the walking distances of the people living at B and D can actually be ignored because in that case, it would always add up to 6. We only need to care about how much the person who lives in C needs to walk. Therefore, the optimal strategy is to place the point P at C (i.e., all the friends meet at C). Then AP is just AC, which is 9.

y varies inversely as sqrt(x) means that y = k / sqrt(x) for some non-zero constant k.

Substitute x = 2 and y = 4 into y = k / sqrt(x),

\(4 = \dfrac k{\sqrt 2}\\ k = 4 \sqrt 2\)

So k is 4 * sqrt(2). When y = 16 and k = 4 * sqrt(2), we have 

\(16 = \dfrac{4\sqrt 2}x\\ \)

Can you take it from here?

Ok, so the function is not in the simplest form yet, let's simplify it.

Note that \(\arctan a + \arctan b = \arctan \left(\dfrac{a + b}{1 - ab}\right)\), so we have:

\(\arctan x + \arctan(1 - x) = \arctan\left(\dfrac1{1 - x(1 - x)}\right) = \arctan\left(\dfrac1{x^2 -x +1}\right)\)

What is the range of \(\dfrac1{x^2 - x + 1}\)? We can find that by completing the squares: 

\(x^2 - x + 1 = \left(x - \dfrac12\right)^2 + \dfrac34\\ \), so the range of x^2 - x + 1 is \(\left[\dfrac34, \infty\right)\).

Therefore, it means that the range of \(\dfrac1{x^2 - x + 1}\) is \(\left(0, \dfrac43\right]\).

(If you don't understand interval notations, that means \(\dfrac34 \leq x^2 - x + 1\) and \(0<\dfrac1{x^2 -x + 1} \leq \dfrac43\).)

Note that arctan is a strictly increasing function. As \(0<\dfrac1{x^2 -x + 1} \leq \dfrac43\), we have \(\arctan 0 < \arctan\left(\dfrac1{x^2 - x + 1}\right) \leq \arctan\left(\dfrac43\right)\), so \(0 < \arctan\left(\dfrac1{x^2 - x + 1}\right) \leq \arctan\left(\dfrac43\right)\)

The range of the function is \(\left(0, \arctan\left(\dfrac43\right)\right]\).

You can repeat my calculations 2 more times on your own, but here's how to do the first replacement of solutions:

After 25 gallons are drained, there are 75 gallons of solution with 50% alcohol. That means half of them is alcohol and half of them is other things. So we have 37.5 gallons of alcohol in the tank if you multiply 50% by 75 gallons. 

Now we pour the 25 gallons of 40% alcohol into the tank. That is actually 25(40%) = 10 gallons of extra alcohol and 25 gallons of extra total solution, so we have 100 gallons of solutions again, but the new concentration of alcohol is \(\dfrac{37.5 + 10}{100} \times 100\% = 47.5\%\) instead.

Can you repeat this process twice on your own? If you didn't understand the method, you can ask under this comment.