Question

omg this is so fraustrating, everyones ignoreing me:( please help! medals awarded:

(n^2+3n+2)/(2)÷(n+1)/(2(n+2)

Answer 1

When you divide a fraction by a fraction, you are essentially multiplying by the recipocal of that fraction.

Eg:

(a/b)/(c/d)= (a/b)*(d/c)

Do you need further clarification?

Answer 2

i still dont get what u mean?

Answer 3

(a/b)/(c/d) = (a*d)/(b*c)

Answer 4

he means that\[\huge\frac{\frac ab}{ \frac cd}=\frac ab\cdot \frac cd=\frac{ac}{bd}\]so in your problem we have\[\huge{{n^2+3n+2\over2}\over{n+1\over2(n+2)}}={n^2+3n+2\over2}\cdot{2(n+2)\over n+1}\]see anything that can cancel yet?

Answer 5

oh i got it, so basicall anything that is division swutches to multiplication?

Answer 6

yes, but you must be sure to take the \(reciprocal\)

Answer 7

my answer I got was (n+1)^2/4 is this right?

Answer 8

what do you mean by take the reciprical, flip it around right?

Answer 9

Not quite.. I don't think that Turing Test is right. He/she hasn't taken the reciprocal in the initial algebraic working out, although in your working out he/she has.

Answer 10

Yes, just swap the numerator and denominator. (number on top and bottom)

Answer 11

I don't see the /4 offhand...
yes that's what I mean, flip it=reciprocal

Answer 12

did I fail to take the reciprocal?
I don't think so... where?

Answer 13

is the answer (n+1)^2/4 right?

Answer 14

no I'm pretty sure I flipped the denominator, unless I'm misreading the question
I get just (n+2)^2... but maybe my eyes are crossed... let me check

Answer 15

\[\huge{{n^2+3n+2\over2}\over{n+1\over2(n+2)}}={n^2+3n+2\over2}\cdot{2(n+2)\over n+1}\]so far we agree? or am I misreading something?

Answer 16

(n^2+3n+2)/(2)÷(n+1)/(2(n+2)
This is a little unclear on the the order of operations...
Is it
[(n^2+3n+2)/(2)]÷[(n+1)/(2(n+2)]?

Answer 17

i got the answer (n+1)^2/4

Answer 18

is this answer correct or incorrect?

Answer 19

Turing, assuming I'm reading the OoO correctly, (n+2)^2 is the correct answer.

Answer 20

@SmoothMath I assumed the parentheses implicitly...
so yeah you must have failed to get the reciprocal maybe @katlin95 ?

Answer 21

yea lol

Answer 22

Yeah, the division symbol seemed to imply that the parenthesis should be where you put them.

Answer 23

maybe if I write it like this it is more cear...\[\large{{n^2+3n+2\over2}\div{n+1\over2(n+2)}}={n^2+3n+2\over2}\cdot{2(n+2)\over n+1}\]so work it from there and you should get my answer

Answer 24

can u guys help me solve this?


subtract and write the answer in simpleast form: (9x)/(3x-4)-(12)/(3x-4)

Answer 25

What he said. Specifically, take notice that
\[\div (n+1)/(2(n+2)) becomes * (2(n+2))/(n+1)\]

Answer 26

^^^what he said
@katlin95 I see you have already posted that problem separately, which is good
I have to go, you should wait for an answer there please
thanks, later :)