Could you help me to simplify 3pq / 4r to the power of 2 multiplied by 2rs / 6q to the power of 2?

Question

Could you help me to simplify these two equation's
3pq / 4r to the power of 2 multiplied by 2rs / 6q to the power of 2
and
15ab to the power of 2 / 8c to the power of 2 d divided by 3ab / 4cd
If you could set the working out in step's, giving me an explanation at each step

Answer 1

(3pq/4r)^2=9p^2q^2/16r^2
(2rs/6q)^2=4r^2s^2/36q^2
(3pq/4r)^2(2rs/6q)^2
=9p^2q^2*4r^2s^2/16r^2*36q^2
=9p^2s^2/4r^2*36q^2
=p^2s^2/16q^2r^2

2.(225a^2b^2/64c^2)/(3ab/4cd)
=225a^2b^2*4cd/64c^2*3ab
=75abd/16c

Answer 2

1. As both sections are to taken to the power 2 we can perform the simplification first and then raise to the power 2 (ie square it) so we have 3pq/4r times 2rs/6q which simplifies as follows:
3x2 = 6 so the 6 numerator cancells with 6 in the denominator
q numerator cancells with q denominator
r denominator cancells with r numerator
This leaves ps numerator and 4 denominator or ps/4
This is to be squared so the answer is

[p squared x s squared]/16

2. This sum is worded ambiguously eg is it to the power of 2/8c
or is it 15ab to the power of 2 and then divided by 8c
Can't be done

Answer 3

(3pq/4r)^2 x (2rs/6q)^2

Expand the squares gives

[(9p^2 q^2 )/ 16 r^2 ] x [4r^2s^2 /36q^2]

Now multiplying the two sides gives

36 p^2 q^2 r^2 s^2 / 576 r^2 q^2

You can divide top and bottom by 36 q^2 r^2 which would give

p^2 s^2 / 16

which can also be expressed as (ps/4)^2

Hope this helps

Answer 4

The first equation goes like this:

3pq is simplified to p, as it cancels with the 6q^2...leaving 2q for the 6q^2....the 4r^2 cancels to 2r with the 2rs, which turns to just ...s...

thus...you have p/2r multiplied by s over 2q.
this multiplies to get ps/4rq


Im doing the next one....

ok...done:

the next one is:

15ab^2/8c^2 divided by 3ab/4cd

that turns to 15ab^2/8c^2 times by 4cd/3ab

which goes to 5b/2c times by d/1

which equals 5bd/2c...

Answer 5

ok so, you "distribute" the multiplication to each and every of the added or subtracted numbers. Mutliplication over additon is distributive. skill you could distribute the multipliers and multiply each and every addition volume (said as addends) through the mutiplier ourside the parenthesis. volume social gathering 2(3+4) is two x 7 = 14 also works in case you bypass 2 x 3 + 2 x 4 6 + 8 = 14 ok now your concern: Like this: 6(3p+q)-2(2p+3q) (6 x 3p) + (6 x Q) - (2 x 2p) - (2 x 3 p) ^ ^ see how the "minus 2" get accelerated through each and every of the numbers contained in the brackets. So this step is what the different useful human beings disregarded. Now do the standard math of each and every multiplication concern. (6 x 3p) + (6 x Q) - (2 x 2p) - (2 x 3 p) 18p + 6 Q - 4p -6q Rearrange (because including and subtracting might want to be performed in any order) 18p + 6 Q - 4p -6q 18p - 4p + 6q - 6q sounds like you get 14p as your very last answer,

Answer 6

3pq/10to the power of 2 multiplied by 2rs/6q to the power of 2 is 300pq/400r x 100rs/300q

Answer 7

3pq / 4r to the power of 2 multiplied by 2rs / 6q to the power of 2
= 3pq / (4r)^2 x 2rs / (6q)^2
= 3pq / 16r^2 x 2rs / 36q^2
= (3pq/16r^2) x 2rs / 36q^2
= (3pqs/8r) / 36q^2
= ps/96qr

Answer 8

(3pq/4r)^2 x (2rs/6q)^2

= [(9p^2q^2) / (16r^2)] x [(4r^2s^2) / (36q^2)]

=(p^2s^2)/16


[(15ab)^2 / (8c^2)] / [(3ab) / (4cd)]

=[(225a^2b^2) / (64c^2)] * [(4cd) / (3ab)]

=(75abd) / (16c)

Answer 9

1-) ps square divided by 16
I didn't understand the second one

Answer 10

Try a Bacardi and Coke!

Answer 11

You really need to layout the problem better.