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(0.72×(-5)×pi/36)/(1.38×(-17)×pi/90)

=150/391

Deleted

You are right Mellie, at least one is beyond rude! 

Melody has been on here for over three years. More than a skilled mathematician, she’s a professional teacher. That means she spent four or more years in college learning to fluently read, write, and translate mòron, idiòt, and stùpid so dumbfùcks like them might learn something.

Sometimes pìgs can learn to sing –never on key, though.   

That is a good answer last guest :))

YES try constructing it.  Start wit the longest side and then try to get a triangle using the two short lengths.

See what happens.     

What is the derivative of

\(f(x)=x^3*(5x+1)^{-(2/3)}\\ \mbox{use product rule and composite function rule.}\\ u=x^3\\ u'=3x^2\\ v=(5x+1)^{-2/3}\\ v'=(-2/3)(5x+1)^{(-5/3)}\times 5\\ f'(x)=uv'+vu'\)

You should be able to finish it  :))

          lnx,x>=1

           /

f(x)={

           \

              e^(x-1)+βx-β-1,x<1

f has a derivative at x=1

find b

If there is a derivative at x=1 then the curve must be continuous and smooth at x=1

\(f(1) = ln(1)=0\\ f(x) = lnx \qquad for \;\;x\ge1\\ f ' (x) = \frac{1}{x} \qquad for \;\;x\ge1\\ f ' (1) = \frac{1}{1}=1 \)

Now I am going to ignore the fact that the following function is not defined for x=1

but it needs to have the same f(1) and f ' (1) as    f(x)=ln x does

\(f(x)=e^{(x-1)}+\beta x-\beta-1\\ f(x)=\frac{e^x}{e}+\beta x-\beta-1\\ f'(x)=e^x+\beta\\ NOW\\ f(1)=0\\ f(1)=e^{(1-1)}+\beta* 1-\beta-1\\ f(1)=1+\beta-\beta-1\\ f(1)=0 \qquad good \\also\\ f'(1)=1\\ f'(1)=e^1+\beta\\ 1=e+\beta\\ \beta=1-e \)

I do not think I have displayed this very well but  beta = 1-e

It's not rude!! We're just curious.

Thanks Guest.

\(\frac{x}{-4}=\frac{-4}{3}\\ \mbox{multiply both sides by -4}\\ \frac{x}{-4}\times -4=\frac{-4}{3}\times -4\\ x=\frac{-4*-4}{3}\\ x=\frac{16}{3}\\ x=5\frac{1}{3}\\\)

You could also just use the quadratic equation :))

Oh dear,

I am so sorry guest.

I have stuffed up your answer.  I am very sorry.    

I must have clicked 'edit' in yours instead of in mine by mistake.  I did not even know that I could do that!

I went back to try and fix it but that didn't help.  I don't think I tried to edit it 4 times, not even by mistake.  I do not know what has happened.  I do not know why it has changed it like that either.

I am SOOOOO  SORRY    

Simplify x(40-2x)(40-2x) over -4xsquared + 1600

If what I have done is what you meant then this would have been the best way to present it.

Simplify         [ x(40-2x)(40-2x) ] / [ -4x^2 + 1600 ]               

I have used difference of 2 squares to factoise the bottom, that is     \(a^2-b^2=(a-b)(a+b)\)

\(\frac{x(40-2x)(40-2x)}{-4x^2+1600}\\ =\frac{x(40-2x)(40-2x)}{1600-4x^2}\\ =\frac{x(40-2x)(40-2x)}{(40)^2-(2x)^2}\\ =\frac{x(40-2x)(40-2x)}{(40-2x)(40+2x)}\\ =\frac{x(40-2x)}{(40+2x)}\\ =\frac{x\times 2(20-x)}{2(20+x)}\\ =\frac{x(20-x)}{20+x}\\ or\\ =\frac{20x-x^2}{20+x}\\\)

Thanks guest, I have only just seen you answer, it was not there when I started :))

NOTE:  The 2 answers are the same. 

Simplify the following: {nl} (x (40-2 x) (40-2 x))/(1600-4 x^2) {nl} {nl} Combine powers. (x (40-2 x) (40-2 x))/(1600-4 x^2) = (x (40-2 x)^(1+1))/(1600-4 x^2): {nl} (x (40-2 x)^1+1)/(1600-4 x^2) {nl} {nl} 1+1 = 2: {nl} (x (40-2 x)^2)/(1600-4 x^2) {nl} {nl} Factor -4 out of 1600-4 x^2: {nl} (x (40-2 x)^2)/-4 (x^2-400) {nl} {nl} x^2-400 = x^2-20^2: {nl} (x (40-2 x)^2)/(-4 (x^2-400)) {nl} {nl} Factor the difference of two squares. x^2-20^2 = (x-20) (x+20): {nl} (x (40-2 x)^2)/(-4(x-20) (x+20)) {nl} {nl} Factor 2 out of 40-2 x: {nl} (x 2 (20-x)^2)/(-4 (x-20) (x+20)) {nl} {nl} Multiply each exponent in 2 (20-x) by 2: {nl} (x×2^2 (20-x)^2)/(-4 (x-20) (x+20)) {nl} {nl} 2^2 = 4: {nl} (4 x (20-x)^2)/(-4 (x-20) (x+20)) {nl} {nl} 4/(-4) = 4/(4 (-1)) = 1/(-1): {nl} (x (20-x)^2)/(-1 (x-20) (x+20)) {nl} {nl} Multiply numerator and denominator of (x (20-x)^2)/(-(x-20) (x+20)) by -1: {nl} -(x (20-x)^2)/((x-20) (x+20)) {nl} {nl} A common factor of (20-x)^2 and x-20 is x-20, so -(x (20-x)^2)/((x-20) (x+20)) = (-x (x-20) (x-20))/((x+20) (x-20)): {nl} -(x (x-20)^2)/((x+20) (x-20)) {nl} {nl} -(x (x-20) (x-20))/((x+20) (x-20)) = (x-20)/(x-20)×-(x (x-20))/(x+20) = -(x (x-20))/(x+20): {nl} Answer: |  {nl} | -(x(x-20))/(x+20)

2,5=1,05^n

n=Log(2.5)/Log(1.05)

n=18.780(rounded)

Solve for X over the real numbers:
X^2+3 X-1 = 0

Add 1 to both sides:
X^2+3 X = 1

Add 9/4 to both sides:
X^2+3 X+9/4 = 13/4

Write the left hand side as a square:
(X+3/2)^2 = 13/4

Take the square root of both sides:
X+3/2 = sqrt(13)/2 or X+3/2 = -sqrt(13)/2

Subtract 3/2 from both sides:
X = sqrt(13)/2-3/2 or X+3/2 = -sqrt(13)/2

Subtract 3/2 from both sides:
Answer: | 
| X = sqrt(13)/2-3/2 or X = -3/2-sqrt(13)/2

Hi Mellie,

We could never copy and paste pictures into our posts.

The system of doing it is very similar to what there was before but it does 'look' a little different.

It is only different at a very superficial level.

If you could do it before, then you should be able to work it out ok.

I think you need to go through the not so pleasant task of re-remembering what to do.

I think Alan sent you the page on how to do it.  That was written for an old version of the forum but it has only changed very superficially.  

You have to click into the 'insert a image' on the ribbon when you are making a post.

This takes you to an 'image info' page.

JUST IGNORE THAT.

There is a second tab at the top that says 'upload'

click into that ans you should get a screen that you are familiar with.  It is the same as before.

Then it may be self explanatory or maybe you will need to look at the page Alan sent you.

F=-9

To find the mean, add all the numbers together and divide by the number of numbers.

To find the median, rearrange the numbers so they are listed from smallest to largest, and take the middle number as the median.

The mode is the number that is used the most often.

If x and y are in the ratio a:b, then x:y  =  a:b  or  x/y  =  a/b.

Since  x + y  =  c,  then  y  =  c - x..

Since  x/y  =  a/b  and  y  =  c - x, then, by substituting:  x/(c - x)  =  a/b.

Cross multiplying:  x(b)  =  a(c - x)   --->   bx  =  ac - ax   --->   ax + bx  =  ac   --->   x(a + b)  =  ac

Dividing both sides by  (a + b)   --->   x  =  ac / (a + b)

Go online to this link and enter your numbers in their calculator, which will give you your answers:

http://www.calculatorsoup.com/calculators/statistics/mean-median-mode.php

LancelotLink turns into a really POd chimp when I hire away his de/coders–especially if I don’t cross his palm with a few bananas. He’s almost as POed when he catches Sir LaTex gold-bricking –he’s very skilled at it. Lancelot throws banana peels at him after catching him perusing his stamp collection on company time.

Along with roasted peanuts, Sir La-Tex likes the occasional mouse, but I am uncertain if he likes shrimp/prawns. I’ve not had BBQed prawns before, but I often have BBQed crawdads –or eat them raw – they are a troll staple and easily found near troll toll bridges. I may try some BBQed prawns the next time I BBQ a steak for my dogs. We dogs (and trolls) really like our steaks. 

Just try to construct the triangle, the question answers itself.

Solve for f over the integers:
abs(f+10) = 1

Split the equation into two possible cases:
f+10 = 1 or f+10 = -1

Subtract 10 from both sides:
f = -9 or f+10 = -1

Subtract 10 from both sides:
Answer: | 
| f = -9 or f = -11

Simplify the following:
-1--50

-(-50)  =  50:
-1+50

-1+50 = 49:
Answer: | 
| 49

what over -4 = -4/3

-4 X -4/3=5 1/3, so that( 51/3)/-4=-4/3