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Eyad
 2 years ago
Best ResponseYou've already chosen the best response.1F(x)= x+2m ,x<3 3m(x^2)+k ,3≤x<1 2(x^3)m ,x>1 ___________________________ (this is da multirule) ^^ Where F(x) is continuous at x=3 ,at x=1 Find m,k

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1NO! BAH DOES NOT WAIT!

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1So basically plug in x=3 into x+2m and into 3m(x^2)+k

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.1Alright BAH (i wish if i was a mod >.>) Do BAH KNOW THE ANSWER ?

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1Now equate them because you want them to meet at the same point for the function to be continuous 3+2m=27m+k 2m27m=k+3 25m=k+3 k = (25m+3)

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1Do the same thing for 3m(x^2)+k 2(x^3)m, but at x=1

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1Solve the two equations

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1And everything will be revealed............................................

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.1well that was my answer but i got a diff. value :/

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1k = (25m+3) 3m+k=2m 3m25m3=2m 22m+m=2+3 21m=5 m=5/21

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1k = (125/21 + 3) ... whatever that is
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