All Answers 358087 Answers

Ha Ha Ha

Geno, you make a great troll!

.....And it was all due to rosala!!!!!!

Look who won the Nobel Peace Prize !

Here's one way........

5 6 7 8

(56) 7 8

And (56) is what we need !!!

been busy with mid-semester exams.

Must have been the exams from h**l!

Very nice, Alan.....I added that last line after you posted your first answer.....I meant for the all the digits to be different.......

Here's a closer look at a "split" graph of this......I have chosen to let "k" = 1, but it's value really doesn't have any effect on the end behavior.....

Note that because of the one denominator of "x" and the other of (x+1) that this graph will have vertical asymptotes at x =0  and x = -1  (as expected).  When x approaches -1 from the left, the graph tends toward +infinity. When x approaches -1 from the right, the graph approaches -infinity. When it approaches 0 from both sides, the graph approaches + infinity. (Just as Alan has said.) Note that as x moves toward larger positive and negative values, it approaches y = 0. Thus, this graph has no numerical mins or max's.

You've lost a sign in there Anonymous!

$$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,-\,}}{\mathtt{a}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,-\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$    |*2
$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}} = -{\mathtt{1}}$$$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{a}} = {\frac{\left({i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\right)\\
{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\\
\end{array} \right\}$$

just one xtra thing to solve this without you need

 standart quadratic formula enjoy

The minimum can only be -∞ and the maximum +∞

(I assumed k was positive in my original answer, but positive or negative the max/min are still +∞/-∞)

but what the answer in this qu ? !!! 
#Alan

Percent change is given by 100*(new number - original number)/original number

$${\frac{{\mathtt{100}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\mathtt{1\,739\,000}}{\mathtt{\,-\,}}{\mathtt{2\,018\,000}}\right)}{{\mathtt{2\,018\,000}}}} = -{\mathtt{186.174\: \!430\: \!128\: \!840\: \!436\: \!1}}$$

or ≈ -186%   (the negative sign indicates a reduction)

When x = 0 the function goes to +∞ (the maximum).  When x tends to -1 from x>-1 the function goes to -∞ (the minimum). (When x tends to -1 from x<-1 the function goes to +∞).

2 * 5 = 10 - 4 = 6 * 3 = 18

Sorry for the late response, been busy with mid-semester exams.

Heres the official answer, hope it helps.

Book value of existing X-ray machine

= $7200

Book loss on sale of existing X-ray machine

= 7,200 - 5000 = $2,200

Tax shield from book loss on sale of existing X-ray machine

= 0.3(2,200) = $660

Hence:

After-tax proceeds from sale of old asset

= 5,000 + 660 = $5,660

Initial Investment

= 65,000 – 5,660 + 2,000 = $61,340

Here's a way of generating more sets of values:

Choose a Pythagorean triple, say 3, 4, 5.  Let a = 3m, b = 4m, c = 3n, d = 4n, and e = 5√(m² + n²) 

a² + b² + c² + d² = (3² + 4²)m² + (3² + 4²)n² = (3² + 4²)(m² + n²) = 5²(m² + n²) = e²

Now choose m and n to be part of another Pythagorean triple, say m = 5 and n = 12, and we have e as an integer (=5*13).

So a = 3*5 = 15;  b = 4*5 = 20;  c = 3*12 = 36;  d = 4*12 = 48; e = 5*13 = 65;

$${{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{36}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{48}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

$${{\mathtt{65}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

(I don't remember seeing Chris's last line when I gave my previous answer.  Hmm!)

.

Write this as 

1/4 - a - 1/4 + a² = -1/2   remembering that two minuses make a plus, which is why the a² term is + not -.

Simplify:

a² - a = -1/2

Add 1/2 to both sides

a² - a + 1/2 = 0

Solve using the quadratic formula:

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{a}} = {\frac{\left({i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\right)\\
{\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\\
\end{array} \right\}$$

You can see that the solutions are complex (that is, they involve i, the square root of -1); there are no real- number solutions.

. 

No!  It's such a well known question I must have seen it a hundred times.

A googol is 10^100 so a googol times a googol is (10^100)*(10^100) = 10^200.

What I did was to divide both sides by ncr(20,5) and then take the fifth root of both sides.  This leaves a quartic in p.  There are four solutions to this quartic; two real and two complex.  The 0.0414 was one of the real solutions (the other was around 0.6, so I doubt this was the desired result!).

(I used Mathcad to do the calculations).

Ding!!  Ding!! Ding!!!......we have a winner!!

kitty<3  !!!!

I came up with........

1 , 2 , 8, 10, 13

Note, that any positive integer multiples of these answers "work," too. (Just like in the Pythagorean "Multiples")

[Mmmm...I wonder if there is a general "formula" for generating the "correct" values??? ]

Here's a pretty good "non-technical" explanation of how it works and how it's derived.

Hm...

a = 4, b = 8, c = 10, b = 12, and e = 18 :)