It will be whichever equation you see that has the "c" = +3, in the format:
F(x) = y = ax^2 + bx + c, where a, b, and c are all integers, such that plugging a "0" into the x's will give:
y = a•0^2 + b•0 + 3 = 0 + 0 + 3 = 3
Answer:
y
=
−
2
x
+
9
Step-by-step explanation:
Write in slope-intercept form, y
=
m
x
+
b
.
Answer:
c, e, f, and i
Step-by-step explanation:
a. is bounded above at y = 1 and below at y = 0
b. is unbounded both above and below
c. is bounded below at y = 0 and unbounded above
d. is unbounded both above and below
e. is bounded below at y = 0 and unbounded above
f. is bounded below at y = 0 and unbounded above
g. is bounded below at y = 0 and bounded above at y = 1
h. is unbounded both above and below
i. is bounded below at y = 0 and unbounded above
j. is unbounded both above and below
Answer:
(P, Q) = (-75, 57)
Step-by-step explanation:
The equation will have infinitely many solutions when it is a tautology.
Subtract the right side from the equation:
Px +57 -(-75x +Q) = 0
x(P+75) +(57 -Q) = 0
This will be a tautology (0=0) when ...
P+75 = 0
P = -75
and
57-Q = 0
57 = Q
_____
These values in the original equation make it ...
-75x +57 = -75x +57 . . . . . a tautology, always true
Answer: 4 por 8
Step-by-step explanation: