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Let’s check out your problem:
Use the following function rule to find
f
(
144
)
f(144)
f
(
144
)
.
\newline
f
(
x
)
=
1
+
5
x
f(x) = 1 + 5\sqrt{x}
f
(
x
)
=
1
+
5
x
\newline
f
(
144
)
=
f(144) =
f
(
144
)
=
_____
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Math Problems
Precalculus
Evaluate functions
Full solution
Q.
Use the following function rule to find
f
(
144
)
f(144)
f
(
144
)
.
\newline
f
(
x
)
=
1
+
5
x
f(x) = 1 + 5\sqrt{x}
f
(
x
)
=
1
+
5
x
\newline
f
(
144
)
=
f(144) =
f
(
144
)
=
_____
Identify Function and Input:
Identify the function and the input value.
\newline
We are given the function
f
(
x
)
=
1
+
5
x
f(x) = 1 + 5\sqrt{x}
f
(
x
)
=
1
+
5
x
and we need to find the value of
f
(
144
)
f(144)
f
(
144
)
.
Substitute Input into Function:
Substitute the input value into the function.
\newline
To find
f
(
144
)
f(144)
f
(
144
)
, we substitute
144
144
144
for
x
x
x
in the function:
\newline
f
(
144
)
=
1
+
5
144
f(144) = 1 + 5\sqrt{144}
f
(
144
)
=
1
+
5
144
Calculate Square Root:
Calculate the
square root
of
144
144
144
.
\newline
The square root of
144
144
144
is
12
12
12
, since
12
×
12
=
144
12 \times 12 = 144
12
×
12
=
144
.
Continue Calculation:
Continue the calculation with the square root result.
\newline
Now that we know
144
=
12
\sqrt{144} = 12
144
=
12
, we can continue the calculation:
\newline
f
(
144
)
=
1
+
5
×
12
f(144) = 1 + 5 \times 12
f
(
144
)
=
1
+
5
×
12
Multiply and Calculate:
Multiply
5
5
5
by
12
12
12
.
5
5
5
times
12
12
12
is
60
60
60
.
f
(
144
)
=
1
+
60
f(144) = 1 + 60
f
(
144
)
=
1
+
60
Add to Get Final Result:
Add
1
1
1
to
60
60
60
.
\newline
Adding
1
1
1
to
60
60
60
gives us
61
61
61
.
\newline
f(
144
144
144
) =
61
61
61
More problems from Evaluate functions
Question
What is the domain of this function?
\newline
(
9
,
–
2
)
(
6
,
–
10
)
(
–
3
,
9
)
(
5
,
2
)
(9,–2)(6,–10)(–3,9)(5,2)
(
9
,
–2
)
(
6
,
–10
)
(
–3
,
9
)
(
5
,
2
)
\newline
Choices:
\newline
(
A
)
{
–
3
,
5
,
6
,
9
}
(
B
)
{
–
10
,
–
2
,
2
,
9
}
(
C
)
{
2
,
5
,
6
,
9
}
(
D
)
{
–
6
,
–
3
,
5
,
9
}
(A)\{–3,5,6,9\}\newline(B)\{–10,–2,2,9\}\newline(C)\{2,5,6,9\}\newline(D)\{–6,–3,5,9\}
(
A
)
{
–3
,
5
,
6
,
9
}
(
B
)
{
–10
,
–2
,
2
,
9
}
(
C
)
{
2
,
5
,
6
,
9
}
(
D
)
{
–6
,
–3
,
5
,
9
}
Get tutor help
Posted 11 months ago
Question
Look at this set of ordered pairs:
\newline
(
2
,
14
)
(2, 14)
(
2
,
14
)
\newline
(
10
,
12
)
(10, 12)
(
10
,
12
)
\newline
(
3
,
−
19
)
(3, -19)
(
3
,
−
19
)
\newline
Is this relation a function?
\newline
Choices:
\newline
[[yes]
\text{[[yes]}
[[yes]
[no]]
\text{[no]]}
[no]]
Get tutor help
Posted 11 months ago
Question
Use the following function rule to find
f
(
11
)
f(11)
f
(
11
)
.
\newline
f
(
x
)
=
−
10
−
7
x
f(x) = -10 - 7x
f
(
x
)
=
−
10
−
7
x
\newline
f
(
11
)
=
f(11) =
f
(
11
)
=
_____
Get tutor help
Posted 11 months ago
Question
Find the slope of the line
y
=
−
3
x
−
2
5
y = -3x - \frac{2}{5}
y
=
−
3
x
−
5
2
.
\newline
Simplify your answer and write it as a proper fraction, improper fraction, or integer.
\newline
_
_
_
_
_
\_\_\_\_\_
_____
Get tutor help
Posted 11 months ago
Question
A line has a slope of
3
3
3
and passes through the point
(
−
2
,
−
10
)
(-2,-10)
(
−
2
,
−
10
)
. Write its equation in slope-intercept form.
\newline
Write your answer using integers, proper fractions, and improper fractions in simplest form.
Get tutor help
Posted 11 months ago
Question
What is the range of this function?
\newline
y = |x|
\newline
Choices:
\newline
all real numbers
\text{all real numbers}
all real numbers
\newline
{
y
∣
y
≥
0
}
\{y \mid y \geq 0\}
{
y
∣
y
≥
0
}
\newline
{
y
∣
y
≤
0
}
\{y \mid y \leq 0\}
{
y
∣
y
≤
0
}
\newline
{
y
∣
y
>
0
}
\{y \mid y > 0\}
{
y
∣
y
>
0
}
Get tutor help
Posted 1 year ago
Question
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the translation
1
1
1
unit down of
f
(
x
)
=
∣
x
∣
f(x) = |x|
f
(
x
)
=
∣
x
∣
.
\newline
Write your answer in the form
a
∣
x
−
h
∣
+
k
a|x - h| + k
a
∣
x
−
h
∣
+
k
, where
a
a
a
,
h
h
h
, and
k
k
k
are integers.
\newline
g
(
x
)
=
g(x) =
g
(
x
)
=
______
\newline
Get tutor help
Posted 1 year ago
Question
What is the domain of this function?
\newline
(
8
,
–
1
)
(
7
,
–
11
)
(
–
4
,
10
)
(
4
,
3
)
(8,–1)(7,–11)(–4,10)(4,3)
(
8
,
–1
)
(
7
,
–11
)
(
–4
,
10
)
(
4
,
3
)
\newline
Choices:
\newline
(
A
)
{
–
4
,
4
,
7
,
8
}
(A)\{–4,4,7,8\}\newline
(
A
)
{
–4
,
4
,
7
,
8
}
(
B
)
{
–
11
,
–
1
,
3
,
10
}
(B)\{–11,–1,3,10\}\newline
(
B
)
{
–11
,
–1
,
3
,
10
}
(
C
)
{
3
,
4
,
7
,
8
}
(C)\{3,4,7,8\}\newline
(
C
)
{
3
,
4
,
7
,
8
}
(
D
)
{
–
7
,
–
4
,
4
,
8
}
(D)\{–7,–4,4,8\}
(
D
)
{
–7
,
–4
,
4
,
8
}
Get tutor help
Posted 1 year ago
Question
What is the domain of this function?
\newline
(
10
,
–
3
)
(
2
,
–
9
)
(
–
5
,
8
)
(
3
,
1
)
(10,\,–3)(2,\,–9)(–5,\,8)(3,\,1)
(
10
,
–3
)
(
2
,
–9
)
(
–5
,
8
)
(
3
,
1
)
\newline
Choices:
\newline
[
{
–
5
,
2
,
3
,
10
}
]
[
{
–
9
,
–
3
,
1
,
8
}
]
[
{
1
,
2
,
3
,
10
}
]
[
{
–
10
,
–
5
,
2
,
3
}
]
[\{–5,2,3,10\}][\{–9,\,–3,1,8\}][\{1,2,3,10\}][\{–10,\,–5,2,3\}]
[{
–5
,
2
,
3
,
10
}]
[{
–9
,
–3
,
1
,
8
}]
[{
1
,
2
,
3
,
10
}]
[{
–10
,
–5
,
2
,
3
}]
Get tutor help
Posted 11 months ago
Question
What is the domain of this function?
\newline
(
1
,
–
4
)
(
9
,
–
12
)
(
–
6
,
11
)
(
7
,
5
)
(1,–4)(9,–12)(–6,11)(7,5)
(
1
,
–4
)
(
9
,
–12
)
(
–6
,
11
)
(
7
,
5
)
\newline
Choices:
(
A
)
{
–
6
,
1
,
7
,
9
}
\newline\\(A)\{–6,1,7,9\}\newline
(
A
)
{
–6
,
1
,
7
,
9
}
(
B
)
{
–
12
,
–
4
,
5
,
11
}
(B)\{–12,–4,5,11\}\newline
(
B
)
{
–12
,
–4
,
5
,
11
}
(
C
{
5
,
1
,
7
,
9
}
(C\{5,1,7,9\}\newline
(
C
{
5
,
1
,
7
,
9
}
(
D
)
{
–
9
,
–
6
,
1
,
7
}
(D)\{–9,–6,1,7\}
(
D
)
{
–9
,
–6
,
1
,
7
}
Get tutor help
Posted 11 months ago
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