All Answers 352934 Answers

So basically you just have to factor out the common factors on both the denominator or numberstor then cancel them out.

hope this helps!

\(\text{suppose }f(x) \text{ is a well behaved function in that it has a Taylor series}\\ f(x) = \sum \limits_{k=0}^\infty~c_k \dfrac{x^k}{k!}\)

\(\dfrac{f(x)}{x^2} = \sum \limits_{k=0}^\infty~c_k\dfrac{x^{k-2}}{k!}=\dfrac{c_0}{x^2}+\dfrac{c_1}{x}+\dfrac{c_2}{2}+\dots \text{ function of powers of x}\)

\(\text{In order for }\lim \limits_{x\to 0}\dfrac{f(x)}{x^2} = 5 \\ c_0=0,~c_1=0~c_2 = 10 \\ \text{so } f(x) = 5x^2+\text{higher powers of x}\)

a) \(\lim \limits_{x \to 0} f(x) = c_0 = 0\)

b) \(\lim \limits_{x\to 0}\dfrac{f(x)}{x} = c_1 = 0\)

I am not sure

All I can think to do is division which gave me

\(\frac{x^3-2}{x^2+x}\\ =x-1+\frac{x-2}{x^2+2x} \qquad \text{(This has been edited)}\)

I will leave this to another mathematician.  

Continued from earlier

\(\frac{x^3-2}{x^2+x}\\ =x-1+\frac{x-2}{x^2+2x}\\ =x-1+\frac{x-2}{x(x+2)}\\\)

consider just the fraction part

\(\frac{x-2}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2} \qquad \text{for some real A and B}\\ A(x+2)+Bx\equiv x-2\\ Ax+2A+Bx\equiv x-2\\ (A+B)x+2A\equiv x-2\\ so\\ 2A=-2\\ A=-1\\ A+B=1\\ -1+B=1\\ B=2\\ \frac{x-2}{x(x+2)}=\frac{-1}{x}+\frac{2}{x+2} \)

So

\(\frac{x^3-2}{x^2+x} \\ =x-1+\frac{x-2}{x^2+2x} \\ =x-1+\frac{-1}{x}+\frac{2}{x+2}\\ =x-1-\frac{1}{x}+\frac{2}{x+2}\\\)

I think that is right now.   

What do you think Mathbum and guest?

For 3, I got 12/29. Please point out any mistakes I might have made. Thanks,

Simplify the following:
(x^3 - 2)/(x^2 + x)
There isn't much you can do with it !!

Factor common terms out of x^2 + x.
Factor x out of x^2 + x:

(x^3 - 2)/(x (x + 1)), or you can write it like this:x^3/(x^2 + x) - 2/(x^2 + x)

Oh right. Thanks.

Yep, there are others. I listed all of them out and got 12 too.

Hi Curry   

sorry, your question got flagged automatically, I am not sure why.  

Probably the program thought those letters in your equation looked too random and it was flagged as a potential rubbish question.  These questions viewed later by a moderator and shown. 

ab-c=ae+f        solve for a

The idea is to get a by itself in fron to the equal sign.

First get every term WITH an a on the left and everything eles on the right.

\(ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e\)

a=30 and b=15 =30 + (2*15) =60. So, ab =30*15 =450 !!!!!.

End of April shows she has 225....

End of June data point shows she has 275        275 - 225 = 50 dollars more at end of June vs end of April

Yep,   ya got it!  

When you have a hard time, just expand the equation: 

(x^3-2)/ (x^2-2) :   (X)(X)(X)-2 / (x)(x)+X 

now proceed to eliminate which gives:  (X)-2 / X 

X/X= 1 so they cancel out

-2/1 = -2 

:) 

I retried it and got 162. Could someone double check?

a) \(E[X]=\sum \limits_{k=1}^6~\dfrac 1 6 (6k) = \dfrac{6\cdot 7}{2} = 21\)

b) \(E[X] = \dfrac 1 6(6 + 6 + 6 - 6 + 6 - 6) = 2\) (1,2,3,5 are !composite, 4 and 6 are composite)

I'll try .....

12 players total

   each shakes hands with 8 players (total of two other teams) x 3 times      12 x 8 x 3

      each of the 12 shakes hands (once) with 3 of their teammates......                    12 x 3 x 1

          each handshake involves two people so divide by 2                           

(12 x 8 x 3  + 12 x 3 x 1 )   /  2 =   162 handshakes    (I think!) 

                                                 (corrected)

Um...15+2(30)=75, not 60...

55%divided by74%times91

Tangent of an angle in a right triangle =  opposite / adjacent

so for this triangle,   Tangent 20 = x /10

                  or          10 tan(20) = x

                                10 (.36397) = 3.6397 cm ~  3.6 cm  (rounded)

why is this flagged?

thank you very much

Slope-Intercept or Point-Slope?

x=6y for the first one.

x=2y+ 3 for the second.

p=sn

If you want a more in depth answer give us a way to solve the problem with number LOL.

Is (a) 90t ?

oh i see thank you