You can make 22 batches of chocolate chip cookies. And for the second one you can make about 7 batches of lemon cookies
The function is definately defined at x=0 but not x=1.
But its just one part of the coordinate (x,y).
If the value of y or f(x) is considered, you'll see that it is never possible to attain f(x)=0. In other terms (x,y)= (0,0) is not a defined point in the graph of the function because the graph doesnt pass through that point.
Now I hope you understood what I meant!
Conclusion- The above function is not defined at all points in the space having the abscissa or x=1 in the coordinate and also at ordinate or y=0 in the coordinate.nation:
1600(-400)= f is the y axis and -400 because its taking money from his account
Step-by-step explanation:
Q1 . (f+g)(x) = f(x) + g(x)
=4x-4 +2x^2 -3x
= 2x^2 + x -4
Q2. (f-g)(x) = f(x) - g(x)
= 2x^2−2 - (4x+1)
= 2x^2 -2 -4x -1
= 2x^2 - 4x -3
Q3. h(x)=3x−3 and g(x)=x^2+3
(h.g)(x) = h(x) × g(x)
= (3x-3) × (x^2 + 3)
=3x^3 -3x^2 + 9x -9
Q4.f(x)=x+4 and g(x)=x+6
(f/g)(x) = f(x) ÷ g(x)
= x+4 / x+6
the domain restriction is x>-6
x<-6
x doesn't equal (-6)
