I need a little help with this math problem?

Question

Trying to solve this problem but can't quite seem to get it

q^-2 – r^-2/q^-1 –r^-1

Answer 1

charge and radius perhaps.... just = -q + r .....

Answer 2

you want to simplify this expression?

you need to use brackets too in order for us to help you solve it correctly:

do you mean (q^-2 - r^-2) / (q^-1 - r^-1)

or

q^-2 - (r^-2/q^-1) - r^-1

?

Answer 3

q^-3 / r^-3 ?

Answer 4

zxc

Answer 5

But what do you want done with it? solve it means there should be an " = " sign with something on the other side. Like " = 0"

Answer 6

This is reading it as:

(q^-2 – r^-2)/(q^-1 –r^-1)

It MIGHT be:

q^-1 – r^-1

It's a guess, but I assume that you just cancel out.

Answer 7

hmm, i think it's just on the index laws....what does the slash mean..that's the part i don't get, maybe this will help...

basic index rules-

a^b x a^c = a^b+c

eg/ 2^3 + 2^5= 2^8

next index law-

a^b/a^c = a^b-c

i just got that the slash was a fraction, i am thinking that you just need to simplify...

would q^-1 - r^-1  be the correct answer?

Answer 8

the answer is q^-1 - r^-1. You change the signs of the exponents on the bottom and add them to the exponents on top. -2 + 1 = -1.

Answer 9

q^-2 – r^-2/q^-1 –r^-1 = a very confused me  

Answer 10

(q^-2 – r^-2) / (q^-1 –r^-1)

= [(r^2 - q^2) / (qr)^2] / [(r - q) / qr]

= qr(r^2 - q^2) / (r - q)(qr)^2

= (r + q) / qr

= 1/q + 1/r

= q^(-1) + r^(-1)

Answer 11

No idea

Answer 12

If this is an equation, it needs two sides. So, I'll assume the one you gave above became like this:

(q^2 -r^2)/(q^1-r^1)=0

<=> (q-r)*(q+r)/(q-r)=0 (q is different from r)

<=>q=-r

(Not a teacher! )

Answer 13

this is not a problem

I think this is what you mean

[q^-2 – r^-2]/[q^-1 –r^-1}

let p = q^-1 & v= r^-1

(p^2-v^2)/(p-v) = (p+v )(p-v)/(p-v) = p+ v if p<>v

so  q^-2 – r^-2/q^-1 –r^-1 = 1/q + 1/r  q <>r

= (r+q)/qr

Answer 14

I think the answer is Q, but I have not done this in awhile.