hectictar

11 apples - 1 apple  =  10 apples

11 thirds - 1 third  =  10 thirds

\(\dfrac{11}{3}\ -\ \dfrac13\ =\ \dfrac{10}{3}\)

1. We will have to fit the pieces into 3 compartments like this:

If we try to place a piece between these compartments, it will create an unfillable area.

Within each of these compartments, there are only 2 ways to fit the pieces:

So the total number of possibilities   =   2 * 2 * 2   =   2³  =  8

They are:

1.  AAA

2.  AAB

3.  ABA

4.  ABB

5.  BAA

6.  BAB

7.  BBA

8.  BBB

And here is a picture of them:

What is this? Mad Libs?

\(x^{400}\ =\ 9^{1000}\)

                              Take the  200th root of both sides of the equation.

\(x^{\frac{400}{200}}\ =\ 9^{\frac{1000}{200}}\)

\(x^{2}\ =\ 9^{5}\)

                              Take the ± sqrt of both sides. (We could have taken the 400th root of both sides to start with.)

\(x\ =\ \pm\sqrt{9^5}\)

                              Rewrite  9⁵  as  3¹⁰

\(x\ =\ \pm\sqrt{3^{10}}\)

\(x\ =\ \pm3^5\)

\(x\ =\ \pm243\)

area of shaded region  =  area of sector ACD - area of blue segment - area of △ACD

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AB  and  DB  are both radiuses of circle B so they are the same length.

AD  =  DB  =  AB  =  20

So △ABD  is an equilateral triangle.

m∠ABD  =  60°

m∠ACD  =  2( m∠ABD )   =   2( 60° )  =  120°

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m∠GCD  =  (1/2)(m∠ACD)  =  (1/2)(120°)  =  60°

△CGD is a 30-60-90 triangle.

GD  =  (1/2)(AD)  =  (1/2)(20)  =  10

CG  =  10 / sqrt(3)

CD =   20 / sqrt(3)

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area of sector ABD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot(20)^2\cdot\big(\frac{60^\circ}{360^\circ}\big)\ =\ \frac{400\pi}{6}\)

area of △ABD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(10\sqrt3)\ =\ 100\sqrt3\)

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area of blue segment  =  area of sector ABD - area of △ABD =  \(\frac{400\pi}{6}-100\sqrt3\)

area of sector ACD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot\Big(\frac{20}{\sqrt3}\Big)^2\cdot\Big(\frac{120^\circ}{360^\circ}\Big)\ =\ \frac{400\pi}{9}\)

area of △ACD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(\frac{10}{\sqrt3})\ =\ \frac{100}{\sqrt3}\)

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area of shaded region  =  \(\frac{400\pi}{9}-(\frac{400\pi}{6}-100\sqrt3)-\frac{100}{\sqrt3}\ \approx\ 46\) square units

In order for ∠A to be the largest angle, It must be that the side formed by ∠A is the largest side. So...

But by the triangle inequality theorem,

So in order for ∠A to be the largest angle, it must be that   \(\frac53
 

The least possible value of  n - m  is  \(\frac92-\frac53\ =\ \frac{27}{6}-\frac{10}{6}\ =\ \boxed{\frac{17}{6}}\)

\(\sqrt[2]{3}\cdot\sqrt[3]{2}\ =\ 3^{\frac12}\cdot2^{\frac13}\ =\ 3^{\frac36}\cdot2^{\frac26}\ =\ (3^3\cdot2^2)^{\frac16}\ =\ (27\cdot4)^{\frac16}\ =\ (108)^{\frac16}\ =\ \sqrt[6]{108}\)

I found three different possible values of  d  using this graph: