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\(400 = 20^2\), so we have \((2^4 + 2^2)^2 = (2^4)^2 + 2(2^4)(2^2)+(2^2)^2 = 2^8 + 2^7 + 2^4\). \(8+7+4=19\).

- Daisy

It says in the question that x^x^x^x.... =2 

so

x^(x^x^x^x...)  must be  x^2

Wow, here's is my try!

In the left column, it's a steadily increasing sum, +2, +3, +4, so 10 down will be 55, we can also use the arithmetic series formula, to get 55!

Now, we just add 10, then 11, and finally end up with \(\boxed{181}\) .

If \(|x|=3x+1\), find \((64x^2+48x+9)^{2018}\).

Rewriting expression:

\((64x^2+48x+9)^{2018},\\ =[(8x+3)^2]^{2018},\\ =(8x+3)^{4036}.\)

Solving for \(x\):

\(|x|=3x+1\\ x^2=(3x+1)^2\\ x^2=9x^2+6x+1\\ 8x^2+6x+1=0\\ (2x+1)(4x+1)=0\\ x_1=-0.5, x_2 = -0.25\)

After plugging our values into the equation to check if they work, we will see that only \(-0.25 \) satisfies the equation. 

\((8\cdot(-0.25)+3)^{4036}=(-2+3)^{4036}=1^{4036}=\boxed1\)

If \(x^2 + y^2 +\frac12 = x + y\), find \(x^y+y^x\).

Now plugging our values into the expression: 

\(x^y+y^x\Rightarrow \frac12^\frac12+\frac12^\frac12\\ =\sqrt{\frac12}+\sqrt{\frac12}\\ =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\\ =\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\boxed{\sqrt2}\)

(a) You're correct it is 6, because of \(a^2+b^2=c^2, a^2+8^2=10^2, a^2=10^2-8^2, a^2=6^2, a=6.\)

(b) \(a^2+b^2=c^2, a^2+24^2=25^2, a^2=25^2-24^2, a^2=25+24, a^2=49, a=7.\)

(c) \(a^2+b^2=c^2, a^2+8^2=17^2, a^2=17^2-8^2, a^2=225, a=15.\)

(d) \(a^2+b^=c^2, a^2+15^2=25^2, a^2=25^2-15^2, a^2=400, a=20.\)

what is the question?

See the answer here:  https://web2.0calc.com/questions/help_15984

{10! / [8! * 2!]} / 2^10 =45 / 1,024 =0.0439453125.

OR: 10C2 / 2^10 =45 / 1,024            =0.0439453125.

I'm sorry, I am confused on x^(x^x^x^x..) = x^2. How do you decide this?

From the \(10\) to \(99\), there are \(90\) numbers. Out of these numbers, \(9\) have the same digits(\(11, 22, 33, 44, 55, 66, 77, 88, 99\)). So your probability would be \(\frac{81}{90} = \frac{9}{10}\). 

- Daisy

We have ₁₀C₈ \(* (\frac{1}{2})^2 ≈ 0.044\)

- Daisy

We can make a chart for all the sums:

You can see that the \(9\) has \(4\) possible ways, and there are \(36\) total ways to get all the sums. Therefore, your probability would be \(\frac{4}{36} = \frac{1}{9}\).

- Daisy

\(4, 0\)

\(3,1\)

\(2,2\)

Since both are indistinguishable, the order of the balls does not matter.

- Daisy

If the same person can't hold both offices, we have \(20\) people availible for the first office, and \(19\) for the second. \(20*19=380\).

- Daisy

There are \(5\) letters to choose from and \(4\) letters make up a word. We can get the permutation ₅P₄ = \(120\).

- Daisy

I think you are missing some of your question!

- Daisy

thanks!

.jjf

Aha i see, Thanks for your help

because x^x^x^x .... =2

so

x^(x^x^x^x..) = x^2

How do you know it is correct?

I assume you are comparing it to an answer that you have, like in an answer sheet.

That does not necessarily mean it is correct, it just means that they are the same.

They might both be wrong    

The answer is fine though   

I know I have not done it your method. I always do them like this.

I think Heureka does it your way. I don't know about other people. 

My way is easy though. ..

How does \({x}^{{x}^{{x}^{{x}^{...}}}} = x^2\)?

Oh yeah that makes more sense! I was kind of confused on the question. I should of used permutations instead, haha. Thanks!

- Daisy