Melody

If 

\(a=\sqrt4+\sqrt5+a\\ then\\ 0=\sqrt4+\sqrt5\\ 0=2+\sqrt5\\ 2=-\sqrt5 \)

This is obviously nonsense.

so

This equation has no real solutions.  In fact I do not think it has any solutions at all.

Perhaps you wrote the question incorrrectly?

Here is the graph. From the graph it certainly appears that  3^(x-1) will always be bigger than 2^x-7

Thanks Chris, I believe your answer is correct but I'd like to talk about it a different way.

Let the people form a line and the envelopes are on the table.

The first person has a 1 in 5 chance of selecting the right letter.  (He selects the correct one)

Now there are 4 on the table so the second person has a 1 in 4 chance of getting the right one (He selects the correct one)

Now there are 3on the table so the second person has a 1 in 3 chance of getting the right one (He selects the correct one)

and so on

so 

P(everyone getting the correct letter) = \(\frac{1}{5}\times \frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{1}=\frac{1}{5!}=\frac{1}{120}\)

Suppose f(x)  is a polynomial of degree 4 or greater such that f(1)=2,   f(2)=3, and    f(3)=5. Find the remainder when f(x)    is divided by    (x-1)(x-2)(x-3).

I don't know the answer but I would like to see someone else discuss it :/

Come on guest don't be so mean.

It is good that Steve posts this. There are always new people about and it is good info to be repeated.

Sometimes you get annoyed far too easily.    ;)

Thanks OBT

TRIGONOMETRY

A trick for remembering the exact values of  sin and cos 0, 30, 45, 60, and 90 degrees. 

Many students and teachers could make good use of this!

Thanks XSquaredFactor :)

y = x^2 - 4x + 7

OR

y=(x-2)^2-4+7

y=(x-2)^2+3

x=t+2

y=t^2+3

It seems we are all in agreement :))

Thanks Chris :)