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i got this.

There are five odd digits and four even digits to be used. Because the digits must alternate between odd and even, this means that there is only one possible way to distribute odds (O) and evens (E): OEOEOEOEO. Now, there are  5! =120 ways to arrange the odd numbers, as there are five choices for the first slot, four for the second, and so on. Similarly, there are  4!=24 ways to arrange the even numbers. Our final answer is the product of  120 and 24, which is 2880.

And i find this logic clearer, Why?

It is a nice example of LaTex formating too.

Try plugging in 1/5 for y in the given equation and then try solving for x.

Yes, I got that.  thanks :)

The better question would be......

Given.

Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.  
When he averages 40 miles per hour, he arrives at his workplace five minutes late.  
When he averages 60 miles per hour, he arrives fifteen minutes early.

Given the simplest version of this data. (no breakfast breaks, no breakdowns etc) 

Prove that this set of data cannot be valid.

OK then.   You can still use my technique for this problem.
Note that \[\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}} =
\dfrac{x^2+1}{x^2-1}.\]
So
\begin{eqnarray*}
    \dfrac{x^2+1}{x^2-1} &=& \dfrac{5}{4} \\
    \dfrac{(x^2+1)+(x^2-1)}{(x^2+1)-(x^2-1)} &=& \dfrac{5+4}{5-4} \\
    \dfrac{2x^2}{2} &=& \dfrac{9}{1} \\
    x^2 &=& 9 \\
    x &=& \pm 3.
\end{eqnarray*}

Thanks guest but next time please answer on the original post and just leave a little note here saying that you have done so.

Another link

EP has answered it.  Thanks EP.

In general:

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{n} \\\\ \mathbf{x+y } &\ge& \mathbf{4n} \\ \hline \end{array}\)

\(\text{The smallest possible value for $x+y$ is $4n$}\)

[ABCD] = X          [AEF] = 1/4(X)

[ADF] = 1/8(X)

[ADFE] = 3/8(X)

[ADGE] = 3/10(X)  

Since z = 2x     put that into the equation

5x + 10 + 2x -10

= 7x    

https://web2.0calc.com/questions/help-geo_9

Given positive integers \(x\) and \(y\) such that \(x \ne y\) and \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{18}\),
what is the smallest possible value for \(x+y\)?

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{18} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{18} \\\\ \mathbf{xy} &=& \mathbf{18*(x+y)} \\ \hline \end{array}\)

\(\large{AM\ge GM}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 18*(x+y) \\ (x+y)^2 &\ge& 4*18*(x+y) \\ x+y &\ge& 4*18 \\ \mathbf{x+y } &\ge& \mathbf{72} \\ \hline \end{array}\)

\(\text{The smallest possible value for $x+y$ is $72$} \)

Find all positive integers n, less than 17, for which n! is an integral multiple of 49.

\(\boxed{\dfrac{n}{7}=2 \\ \mathbf{n}=\mathbf{14} }\)

I will avoid expanding any of the cubed expression.  Define a polynomial $p$ to be $p(x)=(x+4)^3 + (x+5)^3 - (x+7)^3 - (x-2)^3$.  Note that $p$ is at most quadratic.  Thus, $p(x)=ax^2+bx+c$ for some real $a,b,c$.  (In fact, $a,b,c$ are integers, but it's enough to say they are real.) We have
\begin{eqnarray*}
    p(0) = c &=& 4^3+5^3-7^3 - (-2)^3 = -146 \\
    p(1) = a+b+c &=& 5^3+6^3-8^3-(-1)^3 = -170 \\
    p(-1) = a-b+c &=& 3^3+4^3-6^3-(-3)^3 = -98
\end{eqnarray*}
Solving for $a$, $b$, and $c$, we have $a=12$, $b=-36$, and $c=-146$, giving $p(x)= 12x^2-36x-146$.   So
\[
    x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{3}{2}\pm\dfrac{1}{6\sqrt{519}}.
\]

I failed to account for the fact that  (x-1)   might be negative or positive !   My bad !   (,.....as they say)

A' is the set of all elements that are in the universal set U, but not in set A.

So, the answer is c) {2, 4, 6}

Area  between  = 

pi   (10^2  - 8^2)   =

36 pi  ≈   113.1    units^2  

Area of a circle = pi r^2        do this for the larger circle  and the smaller circles  then subtract the area of the smaller from the larger.

There is no need to delete this.

Though it may not be a direct solution to the posted problem it’s still a good presentation of mathematical problem solving.  

--. .-

15  20   25    30

Possible  x  values              Mean     Median

37                                         25.4       25

22                                         22.4       22 

12                                         20.4       20

Sum  of  x  values  =   71

1, 1, 4, 6

1, 2, 2, 6

1, 2, 3, 4

2, 2, 2, 3

4/6^4 = 1/324

=^._.^=

I think the best way to solve this is to just start counting.

Sides:

1, 1, sqrt(2)

1, 2, sqrt(5)

1, 3, sqrt(10)

sqrt(2), sqrt(2), 2

sqrt(2), sqrt(5), 4

That's 5. :))

=^._.^=