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Set the two givens =

(x-2a)(x-b) = x^2 +ax+b

x^2 -2ax-bx+2ab = x^2+ax+b

-2ax-bx+2ab = ax+b

so    -2ax-bx = ax      and   2ab = b  

        -2a-b = a          and    2 ab = b

         -b = 3a             and   2a = 1

         -b = 3a             and  a = 1/2

        -b = 3/2

         b = -3/2                  You can add them together...I hope !  

THANK YOU

THANK YOU SO MUCH

Now I get it

THANK YOU SO MUCH

Now I get it

I haven't been active here for a long time, so I will post an answer :)

\(\begin{array}{rcl} f(x) &=& xe^{-x} + 3x\\ f'(x) &=& x\dfrac{d}{dx} \left(e^{-x}\right)+e^{-x}\dfrac{d}{dx} \left(x\right) + 3\dfrac{d}{dx} x\\ &=& x\left(-e^{-x}\right) + e^{-x} + 3\\ &=& (1 - x)e^{-x} + 3 \end{array}\)

....or factorals are just multiplication    6!  =   6 * 5 * 4 * 3 *2  

Notice that Guest used the word asap in his title. This is one of the things that most answerers don't like.

Thanks guest. I had forgotten how to do that.

So at the very least you have refreshed my memory.

3 \nabla 6 = \dfrac{3 + 6}{1 + 3 \times 6} = \dfrac{9}{19}.

Asker guest,

I repeat CalculatorUser's question.

Was this meant to be solved with the aid of a computer program OR were you meant to devise your own mathermatical technique.

Answering guest, I am pleased that you answered and that you showed your code. Thanks for that

Just sub in the numbers and then do the addition.

I will start you off

\(\displaystyle{f(x)=x^{(x+1)}(x+2)^{(x+3)}}\\ \displaystyle{f(0)=0^{(0+1)}(0+2)^{(0+3)}}\\ \displaystyle{f(0)=0^{1}*2^3}\\ \displaystyle{f(0)=0}\)

i 100% agree with you Melody... in fact sometimes i feel like people don't even want to learn and that is disrespectful to the answerer considering they took their valuable and precious time to answer one's questions,,,so it's unfair to not be appreciative...like at least say thanks or something don't just copy down the answers. 
like first of all it does not AFFECT the answerer's learning but theirs,,,,instead of taking answers they could use that question to learn how to do that topic and then apply their knowledge to other problems similar to the one they were struggling on. In my opinion, people should learn and not just like you know what i mean LOL :') 

Yes I suppose there is an ettique to foruming that is slowly learned. 

But I think people need to learn to use words and phrases like 'please' and 'thank you' 

and

'Thanks but I do not understand this bit of your answer .....'

And they need to learn not to use terms  like  'quick', 'urgent',  'anwer now' etc.

Coupon 1:

        First sweater cost = $18

        Second sweater cost = 18* .6 = $10.8

         Total: $28.8

Coupon 2:

       Cost of both: $36

       25% off: $36*.75 = $27

Coupon 2 is a better deal

i agree with CU and Melody!! 

but i used the word URGENT before too lol but it was for math,,,,i wasn't looking for any answers tho D: i was just asking a question that was on a practice worksheet i printed out for extra practice ( and turns out my answer was right, and the answer key was wrong ) 

yeah, and some of these include but not limited to:
 

-help

-help please

-help due tommorow

-help?

-triangles help

-alegebra help

-urgent due tommrow

-urgent answer asap

-asap urgent

I myself had used the word 'urgent' before, but it wasn't for math, it was for other reasons.. lol.

yes I think melody's is correct!
 

What the guest is trying to ask is a combination question.

Do you really think so? It can easily be done as well, since the code I wrote can easily sum them up.

And the sum of 50,412 lucky numbers =27,336,542,310 mod 13 =12

CU: Here is the computer code in question:

a=1;b=0;c=0;d=0;e=0;f=0;p=0; cycle:if(a+b+c==d+e+f, goto loop, goto next); loop:print a*100000+b*10000+c*1000+d*100+e*10+f,", ",;p=p+1; next:f++;if(f<10, goto cycle, 0);f=0;e++;if(e<10, goto cycle, 0);f=0;e=0;d++;if(d<10, goto cycle,0);f=0;e=0;d=0;c++;if(c<10, goto cycle,0);f=0;e=0;d=0;c=0;b++;if(b<10, goto cycle,0);f=0;e=0;d=0;c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

thanks calc user

\(11C8\\ = \frac{11! }{8!(11-8)!}\\ = \frac{8!*9*10*11 }{8!(3)!}\\ = \frac{9*10*11 }{(3)!}\\ = \frac{9*10*11 }{2*3}\\ = \frac{3*5*11 }{1}\\ =165\)

I think there are only 2 reasons why people would use the word 'urgent' it their titile.

1) They are cheating and need the answer very quickly. (in an exam or tomorrows homework etc)

2) They think they are more important than everyone else and should get priority.

Most times it is probably both of those put together.

Is that a coding problem? If so, Guest it would be pleasing if you posted the code you wrote

LOL why does this website sensor the K word but not the F word

its not bad though, its not like it affects anyone negatively.

lowkey this website has some bad designs and i have login problems.

Killersteve may annoy others, but to me he is a sign that this website needs improvement, as users have to manually create features like welcoming new users.

This is just my opinion

Agree.

But just trying:

Factorial can be defined as

n!=n×(n−1)×(n−2)...

But what is strange is that if I plug -3! into the calculator,

(-3)! = 1

So.... kind of weird.

Let's just say its somehow 1.

11!/8! = 

11 * 10 * 9 = 

990