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:
Use this 3 following steps:
1)At first, we should find the roots :
3x+8=0 => x=-8/3
x-4=0 => x=4
2)Using a chart for the limits:
photo
3)choose your solution:
[-8/3,-4)==> Sentence 3 is correct
Multiply all terms in the first equation by 2 and all terms in the second by 3.
You should obtain:
6x + 16y = 34
-6x + 27y = 9
------------------
43y =43, and so y = 1. Subbing 1 for y in the first eq'n, we get
3x + 8(1) = 17, or 3x = 9, or x = 3.
The solution is (3, 1).
Answer is D
sum of constants = 8 + 4 = 12
sum of coefficients = 7 + 3 = 10
12 > 10
Answer:
7z
Step-by-step explanation:
4z-(-3z).
Two negatives equal a positive only <u>when they are right next to each other.</u>
4z+3z=7z. Hope this helps :D
