Asked by ronda
are these correct? and simplifeid correctly?
ax2b/6 x 4/5a = 4ab/15a
xy^2/7 x 7x^2/y = xy^2
3/17rs x 17s^2/4r = 51s/68r^2
2v/3u x 4/6vu = 4/9u^2
31m^2n/41 x 21/mn x 2n^2/3m = 1302n^n 123
3tr/s x 2rt/s^2 = 6t^2r^2/s^3
ax2b/6 x 4/5a = 4ab/15a
xy^2/7 x 7x^2/y = xy^2
3/17rs x 17s^2/4r = 51s/68r^2
2v/3u x 4/6vu = 4/9u^2
31m^2n/41 x 21/mn x 2n^2/3m = 1302n^n 123
3tr/s x 2rt/s^2 = 6t^2r^2/s^3
Answers
Answered by
drwls
Use * for multiplicatrion when you also have x symbols present
a * 2b/6 * 4/(5a) = 8 b/5
(I assumed the "5a" was the denominator)
(xy^2)/7 * (7x^2)/y = x^3y
Try the others again. You aren't doing it right.
a * 2b/6 * 4/(5a) = 8 b/5
(I assumed the "5a" was the denominator)
(xy^2)/7 * (7x^2)/y = x^3y
Try the others again. You aren't doing it right.
Answered by
ronda
for the second one wouldnt it just be x^3
Answered by
drwls
No. You have a y^2 in the numerator and a y in the denominator. They do not cancel out
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