giules

Catenary; y= a/2(e^bx +e^-bx) goes through the origin (0,0) and has point (-9,18) & (9,18) solve for a and b.

1st; 1 black from 6 marbles when you have 4 black is 4chances out of 6 total marbles ie is 4/6 2nd; one white when you have a choice of 2 out of a total of 5 marbles; ie 2/5 Therefore the sum will be 4/6 x 2/5 = 8/30 = 4/15

What about if we move the turning point (0,0) to (9,0) and the other points are (0,18) and (18,18, what then?

If we use the above points,I think y= a/2(e^bx +e^-bx) 18 = a/2 (e^b0 + e^-b0) 18 = a/2 (e^0 + e^0) 18 = a/2 (1 + 1) 18 = a/2 (2) 18 = a To find b; then substitute in y= a/2(e^bx +e^-bx) at point (9,0) 0= 18/2(e^b9 +e^-9b) 0= 9(e^9b +e^-9b) 0 = 9e^9b + 9e...