Answer: question question for you to do a walk on a walk and jog on a walk on a field trip and jog the field of a walk on the walk in a walk on the field and jog jog the
Step-by-step explanation:
Answer: its x = 3
Step-by-step explanation:
The maximum or minimum of a quadratic function occurs at
x
=
−
b
2
a
. If
a
is negative, the maximum value of the function is
f
(
−
b
2
a
)
. If
a
is positive, the minimum value of the function is
f
(
−
b
2
a
)
.
f
min
x
=
a
x
2
+
b
x
+
c
occurs at
x
=
−
b
2
a
Find the value of
x
equal to
−
b
2
a
.
x
=
−
b
2
a
Substitute in the values of
a
and
b
.
x
=
−
−
12
2
(
2
)
Remove parentheses.
x
=
−
−
12
2
(
2
)
Simplify
−
−
12
2
(
2
)
.
Tap for more steps...
x
=
3
The maximum or minimum of a quadratic function occurs at
x
=
−
b
2
a
. If
a
is negative, the maximum value of the function is
f
(
−
b
2
a
)
. If
a
is positive, the minimum value of the function is
f
(
−
b
2
a
)
.
f
min
x
=
a
x
2
+
b
x
+
c
occurs at
x
=
−
b
2
a
Find the value of
x
equal to
−
b
2
a
.
x
=
−
b
2
a
Substitute in the values of
a
and
b
.
x
=
−
−
12
2
(
2
)
Remove parentheses.
x
=
−
−
12
2
(
2
)
Simplify
−
−
12
2
(
2
)
.
Tap for more steps...
x
=
3
Check the picture below.

since the distance AB cannot be a negative value, thus is not -3-√(130).
Answer:
can you show a link to the graph
Step-by-step explanation:
it will help me answer
This is an arithmetic series, so we know, that:
Sum = (First term + Last term) * Number of terms /2
Here we have:
First term = 33
Last term = 104
Number of terms = 104 - 33 + 1 = 72
and