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Thankyou! That's perfect

Oh ok   Thanks  

I have not seen capital pi used like that before.    

Is Geno's answer helpful ?

$$\\y'=(x+1)* 8(2x-3)^3+(2x-3)^4\\
y'=8(x+1)* (2x-3)^3+(2x-3)^3*(2x-3)\\
y'=(8x-8)* (2x-3)^3+(2x-3)^3*(2x-3)\\
y'=(2x-3)^3[(8x-8)+(2x-3)]\\
y'=(2x-3)^3[8x+2x-8-3]\\
y'=(2x-3)^3[10x-11]\\
y'=(10x-11)(2x-3)^3\\$$

Thankyou I have the right answer for the first path but I am struggling with the second one. I end up with something really long, I definitely used the right product rule: f(x)g'(x)+g(x)f'(x), which I know is the same as your uv one but I get stuck after I end up with: 

(x+1)8(2x-3)^3 + (2x-3)^4

Thankyou!

Refer to http://web2.0calc.com/questions/what-is-the-percentage-error-in-3-14-as-an-approximation-for

Using the method Percentage error = |(V_true - V_used)/V_true| X 100.

I got a percentage error of -0.0507%.

Of course here are the numbers I used.

Observed Value: 3.14

True Value: 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458700660631558817488152092096282925409171536436789259036001133053054882046652138414695194151160943305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664308602139494639522473719070217986094370277053921717629317675238467481846766940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859502445945534690830264252230825334468503526193118817101000313783875288658753320838142061717766914730359825349042875546873115956286388235378759375195778185778053217122680661300192787661119590921642

$$\frac{1}{7} \div 8$$ is essentially $$\frac{1}{7} \times \frac{1}{8}$$

Using the fraction multiplication method just multiply the two Numerators and the two Denominators. Using this method you should get: $$\frac{1 \times 1}{7 \times 8}$$ = $$\frac{1}{56}$$

The answer is $$\frac{1}{56}$$

Thanks for all the answers!

I used the capital pi notation, which workes the same way as sigma notation, except that one uses products, not sums:

∏(i=1 to 5) 2i = 2*4*6*8*10 = 3840

So, after some trying and failing:

$$P(A) =\displaystyle \;\;\frac{Pi \limits_{n=1}^X\ ^{n{_i}}C_{k_i}}{^nC_k}$$

I'm really not able to write ∏ in the formula :(

Anyway, I really hope it's written correctly :)

$$y = (x + 1)(2x-3)^4$$

for this one you will use the product rule

y=uv 

y'=uv'+vu'

u=x+1                      u'=1

v=(2x-3)^4               v'=[4(2x-3)^3]*2=8(2x-3)^3

and from here you should be able to finish it    

$$\\y=\frac{ln(x+2)}{x+2}$$

use quotient rule

$$\\If \;\;y=\frac{u}{v}\\\\
y'=\frac{vu'-uv'}{v^2}\\\\\\
u=ln(x+2)\\
u'=\frac{1}{x+2}\\\\
v=x+2\\
v'=1\\\\\\$$

Can you finish it?

What kind of a graph do you want to make?

@@ End of Day Wrap       Sun 23/11/14       Sydney,   Australia     Time 7:00 pm          ♬

Hello everybody,

It was a beautifully quiet day today.  It was a great opportunity for the answerers to take a breather.  

I only remember seeing answers by CPhill and geno3141.  Thanks guys.  

Thanks for answering the left over questions from yesterday too:)

I would really like this forum to be a 'teaching site' not an 'answer site'.

Too often people are being supplied with full answers. I am also guity of this.  When the forum gets very busy it is often easier just to give answers (even with working) than it is to offer something different.

There are times when giving full answers is totally appropriate but sometimes there a better options.  

There are many wonderful teaching resources available on the net.   If we bring the asker together with one or more of these resources this will often result in the best learning.  Partial answers and 'hints' can also be very appropriate. The asker is always encouraged to ask for clarification or for further help.  All answerers really like it when the askers (or other people) interact with us.

Of course some askers really only want the answer but I don't think any of us want to offer an answer only service.  The forum gets much too busy at times.  It would be better if 'question dumpers' were discouraged.

Please, before you answer in full, think about whether that is best way to promote learning.  

For example: Today I was asked "What is 15% of $12.80"

Mon 23/11/14

1) Probability   Thanks Geno and anon.

In two dimensions, two lines are paralel if they are running in exactly the same direction.  Yes 
Tahlalia is correct.  The two lines will never touch.  If they are drawn in a number plane they will have exactly the same gradient.

They are like the tracks of a train on a very straight part of the railway line.

It means never touching

Yeah....I  kinda' thought it would either be a "decimal issue" or that you possibly didn't include all the terms in your summation......glad we could help ya'.......

Ohh I know what I did wrong. I forgot to move the decimal places in 7x2x0.3^6

Thanks ya'll!

Here is the onsite calculator output.....

2^7+7*(2^6)*(.3)+21(2^5)*(.3)^2+35(2^4)*(.3)^3+35(2^3)*(.3^4)+21*(2^2)*(.3)^5+7*(2)*(.3^6)+(.3^7)

=  340.4825447     

I didn't seem to have any problems with it...... (???)

(2.3)^7=(2+0.3)^7

$${{\mathtt{2}}}^{{\mathtt{7}}} = {\mathtt{128}}$$

$${\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{6}}}{\mathtt{\,\times\,}}{\mathtt{0.3}} = {\frac{{\mathtt{672}}}{{\mathtt{5}}}} = {\mathtt{134.4}}$$

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{0.3}}}^{{\mathtt{2}}} = {\frac{{\mathtt{1\,512}}}{{\mathtt{25}}}} = {\mathtt{60.48}}$$

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{{\mathtt{0.3}}}^{{\mathtt{3}}} = {\frac{{\mathtt{378}}}{{\mathtt{25}}}} = {\mathtt{15.12}}$$

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.3}}}^{{\mathtt{4}}} = {\frac{{\mathtt{567}}}{{\mathtt{250}}}} = {\mathtt{2.268}}$$

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{0.3}}}^{{\mathtt{5}}} = {\mathtt{0.204\: \!12}}$$

$${\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{{\mathtt{0.3}}}^{{\mathtt{6}}} = {\mathtt{0.010\: \!206}}$$

$${{\mathtt{0.3}}}^{{\mathtt{7}}} = {\mathtt{0.000\: \!218\: \!7}}$$

total = $${\mathtt{128}}{\mathtt{\,\small\textbf+\,}}{\mathtt{134.4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{60.48}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15.12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.268}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.204\: \!12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.010\: \!206}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.000\: \!218\: \!7}} = {\mathtt{340.482\: \!544\: \!7}}$$

I keep getting 340.47212 on my calculator though when I added all the numbers up. Wouldn't it be 340.472? Am I missing a number or step? 

Yes that sounds good Chris :))

They just rounded the last answer (in red) from the blue one before it....."+smaller terms" means they just didn't list all of the evaluated terms..... if you evaluate all the first line of stuff......I'm thinking it should = that "blue" answer

put it into the site calc on the home page as

975000*(1+1/2)

The binomial expansion of  

$$\\(a+b)^7= \sum \limits_{n=0}^7} \;^7C_n\;b^n\;a^{(7-n)}\\\\\\
so\\\\
2.3^7 \;= \;(2+0.3)^7= \sum \limits_{n=0}^7} \;^7C_n\;(0.3)^n\;2^{(7-n)}\\\\$$

Does that help?