Question

Stuck. Not sure how they got there

Answer 1

which question ?

Answer 2

35. Where the star is

Answer 3

So they took out the common -2 ... but why do they add (inside) didnt they have to subtractÉ Then i dont get how they just got (2) for the second last step

Answer 4

they factored out \(\large -2e^{2x}\)

Answer 5

\[\large a(-\heartsuit)-b(\heartsuit ) = -\heartsuit (a+b)\]

Answer 6

Oh okay, and what about the 2 they got after, and do should I remember that formula you gave me, would that always work ?

Answer 7

its not a formula and you should not memorize it

Answer 8

however it works always as its just a algebraic manipulaiton

Answer 9

its same as saying : a-b = -(-a+b)

Answer 10

Oh okay, I understand now, but how did they get (2) ? for the second last step

Answer 11

remove the brackets and combine like terms

Answer 12

\(\large -2e^{2x}\left[\left(1+e^{2x}\right) + \left(1-e^{2x}\right)\right]\)

Answer 13

is that the numerator ?

Answer 14

yes

Answer 15

\(\large -2e^{2x}\left[\left(1+e^{2x}\right) + \left(1-e^{2x}\right)\right]\)

\(\large -2e^{2x}\left[1+e^{2x} + 1-e^{2x}\right]\)

Answer 16

next, group like terms

Answer 17

\(\large -2e^{2x}\left[\left(1+e^{2x}\right) + \left(1-e^{2x}\right)\right]\)

\(\large -2e^{2x}\left[1+e^{2x} + 1-e^{2x}\right]\)

\(\large -2e^{2x}\left[1+1+e^{2x} -e^{2x}\right]\)

Answer 18

you can cancel +e^2x and -e^2x right ?

Answer 19

\(\large -2e^{2x}\left[\left(1+e^{2x}\right) + \left(1-e^{2x}\right)\right]\)

\(\large -2e^{2x}\left[1+e^{2x} + 1-e^{2x}\right]\)

\(\large -2e^{2x}\left[1+1+e^{2x} -e^{2x}\right]\)

\(\large -2e^{2x}\left[1+1+ 0\right]\)

Answer 20

ohhhhhhhh so when theres an e, it automatcally turns into a 0 ?

Answer 21

that kind of magic almost never happens

Answer 22

whats the value of \(\large a - a\) ?

Answer 23

what do you get when you subtract \(\large a\) from \(\large a\) ?

Answer 24

0

Answer 25

oh sorry I didnt realize the negative

Answer 26

thanks !