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You didn't answer my question:)

 should it be 1.6x  ?

I'll assume it should

x^2+1.6x-4.2=0

How about times both sides by 10 and see if that helps.

$$10x^2+16x-42=0$$

I think I will divide everthing by 2

$$5x^2+8x-21=0$$

I am looking for 2 numbers that multiply to -5*21 that is -105 and add to 8

Since the multply to a negative, one is neg and the other is pos.

Since they add to a positive, the bigger one will be the positive one.

some of The factors of 105 are  5,7,3,      and 15-7=8

So the numbers are  +15 and -7

Now i am going to replace 8x with 15x-7x

$$5x^2+15x-7x-21=0$$

Now factorise in pairs

$$\\5x(x+3)-7(x+3)=0\\
(5x-7)(x+3)=0$$

So either   5x-7=0    or       x+3=0

so              5x   =7     or       x=-3

so                 x=7/5     or      x=-3

do you know how to finish it?

It would have been easier just to use the quadratic formula with the original equation LOL   :)

I thought you were suppose to do it by factorising - that is why I did it this 'hard' way.

You know solving this is like cutting a piece of double chocolate brownie and stuffing in your mouth!

4x + 3 +5x + y 

So to solve this you just have to first bring the alikes together!

= 4x + 5x +3+ y

Now add the like terms 

= 9x + 3 + y ANSWER

So see , easy as I said!

btw......isnt the brownie yummy?

It isn't  that hard to beat me melody .........especially when I'm giving my best! ;)

oh yh lol haha

area of normal circle divided by two, so

$${\frac{{{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{2}}}}$$ 

x^2+1.6-4.2=0

Is it meant to be 1.6x  Tassy?

Yes I suppose I thought that you might have been a bit grumpy (or at least I would have if I had thought about it)  but I was having a 'down' day and you made me laugh (or at least smile), and then you disappeared all of a sudden and left me all by myself.  :(

It is alright you have been back now but I did not see you here this time.  I read this after you had left again.

Oh well, see you next time I guess.  

Were you going to find that error for me?  The one where we got different answers.  I don't know which of us made the mistake :(

Tetration has already arbitrated - looks like I won.  lol

Thanks Tetration. :)

All these things except my answer , are flying in the air ..............above my head!

There are 4 , 16's in 75.

75 has a big stomach but if It had a five more inch big , it could take 5 , 16's , It would be a sad story seeing 75 executed for eating 4 , 16's , terribly painful for Mr.Einstein, seeing his dear fella executed!

It is something hard to explain terrible to tell painful to write boring to read sad to know depressing to think and useless to understand.

Well now becoz I am in a cool mood and will soon go and pour a cool bucket on my head and an reading  that again ....I just  laughed  ;/

I just answered this in another place!

What is goin on in here??

Wallah! I was angry sad that time when I answered those. that 2+2 one etc , I didn't know those made your day!i left becoz I thought most would " shout" at me for bing rude or just telling the truth!lol!

btw how did I do that crime to u melody?

Only one 18 in 25, you know 25 doesn't have a big stomach so....

Yes it is totally confusing!     Thanks Alan,  I will try and work out what you are saying.    

This can be expanded to get 3p +p², noting that the two negative signs cancel each other out.

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The 10' can be represented as (10/60)° = (1/6)° so:

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{75}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)} = {\mathtt{0.256\: \!008\: \!189\: \!722}}$$

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

Let   $$y1=(1-3x)^4$$   and   $$y2=1+x$$

$$\frac{dy1}{dx}=4(1-3x)\times(-3)=-12(1-3x)$$

$$\frac{dy2}{dx}=1$$

From the product rule:  $$\frac{dy}{dx}=y1\times \frac{dy2}{dx}+y2\times \frac{dy1}{dx}$$

So:

$$\frac{dy}{dx}=(1-3x)^4\times 1+(1+x)\times (-12(1-3x))$$

or

$$\frac{dy}{dx}=(1-3x)^4-12(1-3x)(1+x)$$

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Here's a more detailed solution:

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*** Thx :) ***

Thanks for the answers, and the link :-)

Yes, CPhill, it was something like that I had in mind.

I just don't have a "feel" for modulo, the same way that I have for, say, regular division :)

The correct answer is: -9x²+8x+7   (checked by wolfram alpha)

I think the wrong answer originated from here:

 "... 2x + 8 - (1- 3x)^2

now here we will use an identity for the expression in brackets 

Identity is (a-b)^2= a^2 - 2ab + b^2 

now we will use this identity to solve the bracketed expression!

= 2x + 8 - 1 - 2x1x3x + (3x)^2 ..."

I'll solve it like this:

2x + 8 - (1 - 3x)² =

2x + 8 - (1 - 6x + 9x²) =

2x + 8 - 1 + 6x - 9x² 

So it's mistakes made based on the minus before the paranthesis. 

I was enjoying interacting with the Rosala fish tonight.  I have had a heavy day.  

But she swam away and left me all alone.