TheXSquaredFactor

Yes, Cphill, I must concede to you and your wolframalpha...

What I was trying to do was find a number in the expression \(\frac{a}{2}\) such that\(a\in \mathbb{Z}\hspace{1mm}\text{and}\hspace{1mm}-4<\frac{a}{2}\leq -3\). And of course, the only two numbers that work is -7 and -6. 

\(\frac{123344564788977}{12234455590000}\approx10.0817\)

Is this what you want to know?

The equation \(\left \lfloor{t}\right \rfloor=2t+3\) can be a tad difficult to parse. Sometimes, it is better to think of an easier problem before attempting one a harder one. Let's think about a basic equation that contains the floor function.

\(\left \lfloor{t}\right \rfloor=5\)

Before we can solve this, though, we have to understand the floor function. The floor function truncates any decimal. The solution set for this simplified equation, represented in a compound inequality, is \(5\leq t<6\). Any value for t that satisfies this restriction is a solution. We can use this logic to generalize the floor function, actually to \(\left \lfloor{x}\right \rfloor=a\rightarrow a\leq x < a+1\) . However, you must have caution when doing this, however. Not every solution is actually a solution. I'll touch on this later. Knowing this will be key to solving this equation. Knowing this, the floor function transforms from  \(\left \lfloor{t}\right \rfloor=2t+3\) to  \(2t+3\leq t<2t+3+1\). Let's solve this compound inequality. I'll solve each one separately.

Now, let's solve for the other inequality \(t<2t+3+1\):

Of course, all solutions must adhere to both restrictions. Both inequalities show that t must be greater than -4 and less than or equal to -3. -3 is automatically a solution because it is an integer.

We aren't done yet, though. Are there are other solution that adhere to our current restriction that makes \(2t+3\) an integer? We don't even have to worry about the +3 because the sum of any integer and 3 will always yield another integer. Let's set this equal to 0:
 

What I have demonstrate here is that a number divided by 2 will make 2t+3 an integer. The only one that meets our description and our current restriction is -(7/2).

Therefore, there are 2 solutions to this equation

\(t_1=-\frac{7}{2}\)

\(t_2=-3\)

To solve the equation \(0.00000000013284=13284\times10^{-x}\), we must realize that we are trying to get the same base. First, convert \(0.00000000013284\) into a scientific form. 

To do this, move the decimal point of \(0.00000000013284\) such that the decimal point moves in front of the first nonzero digit of the number. The number of times one moves is what the 10 will be raised to. Knowing this, \(0.00000000013284=1.3284\times10^{-10}\). We have now changed the equation to the following:
 

\(1.3284\times10^{-10}=13284\times10^{-x}\)

Now, let's continue moving that decimal of 1.3284 to the right. We'll move it 4 decimal places. This means that \(1.3284\times10^{-10}=13284\times10^{-14}\). Oh look! we have the same base now. Let's solve for x:

And there you go! 

I'm assuming that we are solving for v in the equation \(v^2=\frac{25}{81}\). Let's do that!

Math is a basic necessity. Envisioning a world without it is downright unfeasible. 

There would be no clock or calendar. Therefore, any event that occurs on a schedule is impossible; there is no way to keep track of the time that passed or time in the future. 

Currency would be nonexistent. It only has value if the people accept its worth. But without math, currency would be worthless and useless.

All buildings would collapse, as they are governed by strict mathematical rules.

Technology as a whole would vanish for eternity. 

The world without math couldn't exist.

Of course, if nothing else, you can always do \(x^\frac{1}{4}\)

No one will correct you here! You are 100% right. I did it myself.

I will solve it for you! The expression I will evaluate is \(\left(\frac{-2}{3}\right)^2\):

By the way, if you would like to type it into a calculator, do it like this: (-2/3)^2. The calculator will output 0.444...