Mark M. answered • 05/25/23
Tutor
5.0
(278)
Mathematics Teacher - NCLB Highly Qualified
Coefficient is 7! / (3!(7 - 3)! = (7*6*5*4) / (3*2*1) = 140
Term is (2x)4(-5y)3, expand this, then multiply for 140 or leave it as
140(2x)4(-5y)3
Asma W.
Thank you!
Report
05/25/23
AJ L.
Small correction, 7!/(3!(7-3)!) = 7!/(3!4!) = (7*6*5*4*3*2*1)/(3*2*1*4*3*2*1)=(7*6*5)/(3*2*1)=35, so the term would be 35(2x)^4(-5y)^3.05/25/23