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:
Simplify:
−3=12y−5(2y−7)
−3=12y+(−5)(2y)+(−5)(−7)(Distribute)
−3=12y+−10y+35
−3=(12y+−10y)+(35)(Combine Like Terms)
−3=2y+35
Flip the equation.
2y+35=−3
Subtract 35 from both sides.
2y+35−35=−3−35
2y=−38
Divide both sides by 2.
2y/2=−38/2
y=−19
Total value: $144
Explanation: each book cost $4.50
Answer:
Yes, all equilateral triangles are acute.
Step-by-step explanation:
yea
hope this helps :)))
