@rationaltiger Correct!

=1+11+111 times 68
=12+548
=560

24+3/4(24)(2)+(3/4^2)(24)(2)+(3/4^3)(24)(2)+.....
(The x2 is because the ball drops the same distance up and down)

This is an infinite arithmetic sequence.

Let S = 3/4(24)(2)+(3/4^2)(24)(2)+... = 36+(3/4)(36)+(3/4^2)(36)+....
So 4/3S=48+36+(3/4)(36)+....
So 4/3S-S=1/3S=48
So S=144
So S+24=168?

MISSPELL has 8 letters with 2 S’s and two L’s. 8!/(4!2!) = 10,080 arrangements, but one of these is the correct spelling, so there are 10,079 misspellings in total.

@quacker88 Thanks for your help!

@quacker88 Alright! Thanks for the answer!

@energizedpanda XD

@fabulouscamel

@energizedpanda basically covered it! Essentially, to find something like 12^(3.7), you could first turn the exponent into a fraction in simplest form-- i.e. 12^(37/10), and then we know that x^(a/b) = x^(a * 1/b) = (x^a)^(1/b) (this is an important exponent law), so 12^(37/10) = (12^37)^(1/10). Then, you would first calculate the giant number that is 12^(37), and take the 10th root of that. (Not so great to do by hand)

Hope that answered your question! 😄

Thanks @RZ923

See, if I have 45 hamburgers and I need to put them in 0 boxes how many hamburgers are in each box?
It does not make sense right? You'll never run out of hamburgers.

The square root of 16 is 4
4^2 = 16

@the-blade-dancer said in Is it possible to...:

Find the Mean Absolute Deviation without having to calculate the mean and the difference of each piece of data from the mean? In other words, can MAD problems be solved simply, or is it just something you have to grind through?

calm the problem down first or else your bones may break

@victorioussheep Here's a diagram that I drew:
b86da18e-ca84-496c-bf32-02b656a5dc8b-image.png
(only drew half of it because that's all we need)
So, first off, the length of \(\overline{AD}\) comes from the Pythagorean Theorem. Or, by chance, if you happen to remember that \(119-120-169\) is a Pythagorean triple, \(1428-1440-2028\) is exactly the same triangle, except all the dimensions are multiplied by \(12\). What the problem states is that \(\triangle BOP\) happens to be similar to a \(5-12-13\) triangle. The solution took a huge shortcut, but the idea is that you'll eventually find out that the dimensions of \(\triangle BOP\) are \(420-1008-1092\), which is exactly \(84\) times the dimensions of a \(5-12-13\) triangle. I know that the solution says to use angle bisector, but I can't think of a way to do it that isn't tedious, maybe the other mods could weigh in on that.

Here's how I did it though-- as long as you get that \(\overline{AD}\) is \(1440\), recall that two tangents of a circle that meet have the same length.
Let's redraw the diagram:
faea46b1-cba8-4928-b9ed-e1ec265a8712-image.png
So, \(\overline{BP_1} \cong \overline{BP_2}, \overline{AP_2} \cong \overline{AP_3}, \overline{DP_1} \cong \overline{DP_3},\) and I labeled them \(a, b, c\).
If you want to prove them, here's how:
To prove that \(\overline{BP_1} \cong \overline{BP_2}\), draw line \(\overline{BO}\) and prove that triangles \(\triangle{BOP_1}\) and \(\triangle{BOP_2}\) are congruent (hint: they're both right triangles!).
You can do the same with all of the other lines.

Anyways, knowing this, we can set up a system of equations, like the one I did to the right of the triangle. Then, notice that the radius of the circle is equivalent to \(c\) because quadrilateral \(OP_1DP_3\) is a square. Solving for \(c\) should give you \(\boxed{420}\).

Hope this makes sense! I know that it isn't the method that the solution used, but hopefully at least the part about multiplying the dimensions is clearer.
Also, if the other mods want to weigh in on the angle bisector method, that'd be great 🙂