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here is the decimal \(0.9 \overline{7}\)

This is written as \(9\frac{7}{9} \cdot \frac{1}{10}\)

which equals to \(\frac{44}{45}\)

as the question asks

\(44-45={\boxed{-1}}\)

niceeee.

here's a way

\(240=2^4\cdot3\cdot5=2^3(2\cdot3\cdot5)\). For \(240k\) to be a perfect cube (and not a perfect square), \(k\) must be at least \(2^2\cdot3^2\cdot5^2=\boxed{900}\).

(a) See the graph below.

(b) g(x) = 1/3*f(2x + 2).

(c) We stretch the graph horiztonally by a factor of 2, then stretch the graph vertically by a factor of 3, then shift down 2 units.

240 0= 24 * 10 = 2^3 * 3 * 10 = 2^4 * 3 * 5

We need 2 2s, 2 3s, and 2 5s. This gives us:

2^2 * 3^2 * 5^2 = 900. 

240x60=14400, which is not a perfect cube

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i dont think they want to cheat by searching it up...

a^3 + b^3 = (a + b)((a + b)^2 - 3ab)

Use vieta's formulas!

x^2 - 2x + 6 = 0 

a + b = 2

ab = 6

Plugging this in gives us

(2)(4 - 18) =  -28.

That's very interesting! I know another strange fact that i^i is even real!

i = e^{pi/2i}

\(e^{\pi i/2 \cdot e^{\pi i/2}}\) = \(e^{\pi i^2 /2 }\)

The i^2 cancels out as negative and we see that \(e^{-\pi/2}\) = i ^i which is actually real!

Cool, isn't it! :)

6 hours

Pythagorean theorem

Let the side of the triangle be \(2s\)

The area of "success" is a \(60 ^ \circ\) sector of a circle centered at Point A, and the radius is half the side length of the triangle. 

The total area is the area of an equilateral triangle. 

Can you take it from here?

It simplifies to \(a + b - 2\), but note that \(a + b = {- b \over a}= {7 \over 2}\). 

Can you take it from here?

There are 4 8's, so the probability is \({4 \over 52} \times {3 \over 51}\)

Expand it, and substitute \(i^2\) for -1...

The quadratic is \((x-7)(x -4)\). 

Now, just expand it...

1 = 797
2 = 2369
3 = 3941
 

There are a total of 15 partitions with 4 parts as follows: 

9 + 1 + 1 + 1 = 12
8 + 2 + 1 + 1 = 12
7 + 3 + 1 + 1 = 12
7 + 2 + 2 + 1 = 12
6 + 4 + 1 + 1 = 12
6 + 3 + 2 + 1 = 12
6 + 2 + 2 + 2 = 12
5 + 5 + 1 + 1 = 12
5 + 4 + 2 + 1 = 12
5 + 3 + 3 + 1 = 12
5 + 3 + 2 + 2 = 12
4 + 4 + 3 + 1 = 12
4 + 4 + 2 + 2 = 12
4 + 3 + 3 + 2 = 12
3 + 3 + 3 + 3 = 12

Note: Simply pick the ones that meet your conditions. 

28 and 11

\(\frac{i\sqrt{1}}{i\sqrt{3}}=\frac{1}{\sqrt{3}}\) which is a real number.

So, this number can be represented on a real number line. (No need for complex plane (argand diagram).)

Yes, we first "simplify" the number then we put it. 

And the number is not imaginary; it is real. 

4+3+2+1=10  so only 2 more can be added

1,2,3,  6

1,2,4,  5

1,3,3,  5

2,2,3,  5

1,2,  5,4

1,3,  4,4

2,2,  4,4

1,  4,3,4

2,  3,3,4

3,2,3,4

That is all of them but some are copies.  You can count how many are originals

There are 8 partitions that work.

The answer is 61.

The side length is 15/7.