Find dy/dx by implicit differentiation"
xe^y-10x+3y=0
i really do need !

Implicit differentation means you assume y is a function of x:
Dx(xe^y) -Dx(10x) +Dx(3y) = Dx(0) Dx(xe^y) use the product rule:
 (dx/dx)e^y + (dy/dx)xe^y = e^y + y'xe^y
Dx(10x) = (dx/dx)(10) = 10
Dx(3y) = 3(dy/dx) = 3y'
Dx(0) = (d0/dx)0 = 0

Get all your y' on one side and everything else to the other:
e^y + y'xe^y -10 +3y' = 0
y'xe^y + 3y' = 10 -e^y
y'(xe^y +3) = 10 -e^y
(dy/dx) =y' =(10-e^y) / (xe^y +3)

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post