All Answers 358179 Answers

the real question is.... r u black??

Let the numbers be represented by n. Then

3n - 12 < 21

Add 12 to both sides:

3n < 33

Divide both sides by 3:

n < 11

So any number less than 11 satisfies the condition.

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To simplify this:

(x-y)² → x² - 2xy + y² 

y(2x - y) → 2xy - y²

Add these up and you get (x-y)² + y(2x - y) = x²

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The normal interpretation of "inverse function" doesn't mean simply replace the x's by y's and the y's by x's.  It means if, say, f(x) = y, then f^-1(y) = x.

So, to get the (real) inverse to y = x³/4 + x - 1 we need to solve for x in terms of y (taking the only real solution), and then switch x and y values.  We get y = the expression given by Chris.

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Looks good Melody.

I think that you need to recount geno.

Start with some construction.

Draw a line through A parallel to DB and EC and a line through C to meet this line in a right angle. Call the point of intersection F. Extend DB to meet the new line through C at G.

Since AB:BC = 4:3, by similar triangles FG:GC = 4:3, so let FG = 4a and GC = 3a, for some constant a.

Let the length of CE = p and the length of BD =q.

The area of the parallelogram is 60, so (p + q).3a/2 = 60, so,    pa + qa = 40................(1)

The area of the triangle ABD is 24, so,  q.4a/2 = 24, so                     qa =  12................(2)

From (1) and (2), pa = 28.

The area of the big triangle ACE = p.7a/2 = 28*7/2 = 98,

in which case the area of the triangle ADE = 98 -24 - 60 = 14.

the answer is 21

@@ End of Day Wrap    Mon 23/2/15     Sydney, Australia   Time 1:35 am  (Really Tues morning)    ♪ ♫

Hi all,

Our wonderful answerers today were Silver27, Tetration, CPhill, Anima, geno3141, abhanda, Alan, Rosala, Eddierose and Heureka.   Thank you.   

Interest threads:

24/2/15

Thank you Heueka, you are a gem. 

Thanks Alan  

Using the search function.

Finding anonymous user posts can be very difficult.

They are really only findable if they have some unique words in them and then you can find them with the search function.

for example you could search for  'Magick of the Quadratic"  just by typing it (without the quotation marks) into the search box at the top of the page.  I like that post. 

If I wanted to search for Rosala's posts I would type in

by:rosala

If you become a member you get a private watchlist.  That is where I store posts that I am likely to want to find again. 

$$\\\mathbf{How\ to\ calcultate\ the\ \textit{Euler phi function} \ \phi(n):}\\
$ We have the prime factorization of n = p_1\cdot p_2\cdot p_3 \cdots\\
\phi(n) = n \cdot (1-\frac{1}{p_1})\cdot (1-\frac{1}{p_2}) \cdot (1-\frac{1}{p_3}) \cdots$$

$$\\\mathbf{Example\ 1: n = 6 } $\\
The prime factorization of 6 = 2 * 3 = p_1*p_2 \\
\phi(6) = 6 \cdot (1-\frac{1}{2}) \cdot (1-\frac{1}{3}) \\
\phi(6) = 6 \cdot \frac{1}{2} \cdot \frac{2}{3} \\
\phi(6) = \frac{6}{3} \\
\phi(6) = 2$$

$$\\\mathbf{Example\ 2: n = 9 } $\\
The prime factorization of 9 = 3^2 = p_1^2 \\
\phi(9) = 9 \cdot (1-\frac{1}{3}) \\
\phi(9) = 9 \cdot \frac{2}{3} \\
\phi(6) = 3 \cdot 2 \\
\phi(6) = 6$$

$$\\\mathbf{Example\ 3: n = 7 } $\\
The prime factorization of 7 = 7 = p_1 \qquad 7 $ is a prime number!$ \\
\phi(7) = 7 \cdot (1-\frac{1}{7}) \\
\phi(7) = 7 \cdot \frac{6}{7} \\
\phi(7) = 6$$

$$\\\mathbf{Example\ 4: n = 11 } $\\
The prime factorization of 11 = 11 = p_1 \qquad 11 $ is a prime number!$ \\
\phi(11) = 11 \cdot (1-\frac{1}{11}) \\
\phi(11) = 11 \cdot \frac{10}{11} \\
\phi(11) = 10$$

$$\boxed{\text{ In general $ \phi(p) = p-1 $, if p is a prime number }}\\\\
\begin{array}{lr}
p = 2: &\phi(2) = 1 \qquad =(2-1)\\
p = 3: &\phi(3) = 2 \qquad =(3-1)\\
p = 5: &\phi(5) = 4 \qquad =(5-1)\\
p = 7: &\phi(7) = 6 \qquad =(7-1)\\
p = 11: &\phi(11) = 10 \qquad =(11-1)\\
p = 13: &\phi(13) = 12 \qquad =(13-1)\\
\cdots & \phi(p) = p-1
\end{array}$$

The first 99 values of the Phi function are:

	+0	+1	+2	+3	+4	+5	+6	+7	+8	+9
0+		1	1	2	2	4	2	6	4	6
10+	4	10	4	12	6	8	8	16	6	18
20+	8	12	10	22	8	20	12	18	12	28
30+	8	30	16	20	16	24	12	36	18	24
40+	16	40	12	42	20	24	22	46	16	42
50+	20	32	24	52	18	40	24	36	28	58
60+	16	60	30	36	32	48	20	66	32	44
70+	24	70	24	72	36	40	36	60	24	78
80+	32	54	40	82	24	64	42	56	40	88
90+	24	72	44	60	46	72	32	96	42	60

$$\frac{2}{5}\times b = \frac{2}{5}\times \frac{b}{1}=\frac{2b}{5}$$

I doubt that this calc doest that though I could be wrong.

Here is one that does.

Sin50•6.5 = 4.98 

OR maybe the asker may have meant

$$\\log_{256}x = 4\\\\
x=256^4$$

$${{\mathtt{256}}}^{{\mathtt{4}}} = {\mathtt{4\,294\,967\,296}}$$

Well Chris, what you have done looks all very impressive but I'm going to have a go at it myself.

I am treading on shaky ground here so if another mathematicians wants to correct me I shall not be too surprise.

let f(x)= $$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}$$, solve for (f^-1)'(3).

$$\\y=(0.25)x^3+x-1\\\\
$inverse function$\\\\
x=(0.25)y^3+y-1\\\\
\frac{dx}{dy^{-1}}=0.75y^2+1\\\\
\frac{dy^{-1}}{dx}=\frac{1}{0.75y^2+1}\\\\$$

----------------------------------------------

$$\\x=(0.25)y^3+y-1\\\\
When\;x=3\; \\\\
3=0.25*y^3+y-1\\\\
0=0.25y^3+y-4\\\\$$

Are you showing us what you have learned.  If so that is really good.

OR

Is there are question here?    

$$\\\frac{2}{x^{3/7}}+\frac{5}{x^{2/7}}-\frac{3}{x^{1/7}}=0\\\\
x^{3/7}*\left(\frac{2}{x^{3/7}}+\frac{5}{x^{2/7}}-\frac{3}{x^{1/7}}\right)=x^{3/7}*0\\\\
2+5x^{1/7}-3x^{2/7}=0\\\\
let\;y=x^{1/7}\\\\
2+5y-3y^2=0\\\\
3y^2-5y-1=0\\\\
y=\frac{5\pm\sqrt{25+12}}{6}\\\\
y=\frac{5\pm\sqrt{37}}{6}\\\\
x=y^{1/7}=\left(\frac{5\pm\sqrt{37}}{6}\right)^{1/7}$$

I haven't checked it but I think that is somewhere near correct.