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Math Problems
Grade 8
Evaluate a nonlinear function
Use the following function rule to find
f
(
4
)
f(4)
f
(
4
)
.
\newline
f
(
x
)
=
−
5
∣
x
−
7
∣
f(x) = -5|x - 7|
f
(
x
)
=
−
5∣
x
−
7∣
\newline
f
(
4
)
=
f(4) =
f
(
4
)
=
_____
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Use the following function rule to find
f
(
2
)
f(2)
f
(
2
)
.
\newline
f
(
x
)
=
2
x
2
f(x) = 2x^2
f
(
x
)
=
2
x
2
\newline
f
(
2
)
=
f(2) =
f
(
2
)
=
_____
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Use the following function rule to find
f
(
6
)
f(6)
f
(
6
)
.
\newline
f
(
x
)
=
8
x
+
x
2
f(x) = 8x + x^2
f
(
x
)
=
8
x
+
x
2
\newline
f
(
6
)
=
f(6) =
f
(
6
)
=
_____
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Use the following function rule to find
f
(
4
)
f(4)
f
(
4
)
.
\newline
f
(
x
)
=
∣
7
−
x
∣
f(x) = |7 - x|
f
(
x
)
=
∣7
−
x
∣
\newline
f
(
4
)
=
f(4) =
f
(
4
)
=
_____
Get tutor help
Use the following function rule to find
f
(
−
6
)
f(-6)
f
(
−
6
)
.
\newline
f
(
x
)
=
5
∣
x
∣
−
2
f(x) = 5|x| - 2
f
(
x
)
=
5∣
x
∣
−
2
\newline
f
(
−
6
)
=
f(-6) =
f
(
−
6
)
=
_____
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Use the following function rule to find
f
(
−
4
)
f(-4)
f
(
−
4
)
.
\newline
f
(
x
)
=
∣
x
+
4
∣
f(x) = |x + 4|
f
(
x
)
=
∣
x
+
4∣
\newline
f
(
−
4
)
=
f(-4) =
f
(
−
4
)
=
_____
Get tutor help
Use the following function rule to find
f
(
12
)
f(12)
f
(
12
)
.
\newline
f
(
x
)
=
3
x
2
f(x) = 3x^2
f
(
x
)
=
3
x
2
\newline
f
(
12
)
=
f(12) =
f
(
12
)
=
_____
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Use the following function rule to find
f
(
5
)
f(5)
f
(
5
)
.
\newline
f
(
x
)
=
−
3
−
∣
x
∣
f(x) = -3 - |x|
f
(
x
)
=
−
3
−
∣
x
∣
\newline
f
(
5
)
=
f(5) =
f
(
5
)
=
_____
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Use the following function rule to find
f
(
6
)
f(6)
f
(
6
)
.
\newline
f
(
x
)
=
(
–
1
−
x
)
2
f(x) = (–1 - x)^2
f
(
x
)
=
(
–1
−
x
)
2
\newline
f
(
6
)
=
f(6) =
f
(
6
)
=
_____
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Use the following function rule to find
f
(
11
)
f(11)
f
(
11
)
.
\newline
f
(
x
)
=
5
∣
–
2
−
x
∣
f(x) = 5|–2 − x|
f
(
x
)
=
5∣–2−
x
∣
\newline
f
(
11
)
=
f(11) =
f
(
11
)
=
_____
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Use the following function rule to find
f
(
2
)
f(2)
f
(
2
)
.
\newline
f
(
x
)
=
12
∣
x
∣
+
9
f(x) = 12|x| + 9
f
(
x
)
=
12∣
x
∣
+
9
\newline
f
(
2
)
=
f(2) =
f
(
2
)
=
_____
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Use the following function rule to find
f
(
7
)
f(7)
f
(
7
)
.
\newline
f
(
x
)
=
9
+
∣
x
∣
f(x) = 9 + |x|
f
(
x
)
=
9
+
∣
x
∣
\newline
f
(
7
)
=
f(7) =
f
(
7
)
=
_____
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Use the following function rule to find
f
(
12
)
f(12)
f
(
12
)
.
\newline
f
(
x
)
=
(
–
9
+
x
)
2
f(x) = (–9 + x)^2
f
(
x
)
=
(
–9
+
x
)
2
\newline
f
(
12
)
=
f(12) =
f
(
12
)
=
_____
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Use the following function rule to find
f
(
10
)
f(10)
f
(
10
)
.
\newline
f
(
x
)
=
2
∣
x
∣
−
9
f(x) = 2|x| - 9
f
(
x
)
=
2∣
x
∣
−
9
\newline
f
(
10
)
=
f(10) =
f
(
10
)
=
_____
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Use the following function rule to find
f
(
1
)
f(1)
f
(
1
)
.
\newline
f
(
x
)
=
(
x
+
4
)
2
f(x) = (x + 4)^2
f
(
x
)
=
(
x
+
4
)
2
\newline
f
(
1
)
=
f(1) =
f
(
1
)
=
_____
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Use the following function rule to find
f
(
9
)
f(9)
f
(
9
)
.
\newline
f
(
x
)
=
(
x
−
11
)
2
f(x) = (x - 11)^2
f
(
x
)
=
(
x
−
11
)
2
\newline
f
(
9
)
=
f(9) =
f
(
9
)
=
_____
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Use the following function rule to find
f
(
4
)
f(4)
f
(
4
)
.
\newline
f
(
x
)
=
(
x
−
6
)
2
f(x) = (x - 6)^2
f
(
x
)
=
(
x
−
6
)
2
\newline
f
(
4
)
=
f(4) =
f
(
4
)
=
_____
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Use the following function rule to find
f
(
4
)
f(4)
f
(
4
)
.
\newline
f
(
x
)
=
∣
x
+
10
∣
f(x) = |x + 10|
f
(
x
)
=
∣
x
+
10∣
\newline
f
(
4
)
=
f(4) =
f
(
4
)
=
_____
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Use the following function rule to find
f
(
5
)
f(5)
f
(
5
)
.
\newline
f
(
x
)
=
(
x
+
11
)
2
f(x) = (x + 11)^2
f
(
x
)
=
(
x
+
11
)
2
\newline
f
(
5
)
=
f(5) =
f
(
5
)
=
_____
Get tutor help
Use the following function rule to find
f
(
−
6
)
f(-6)
f
(
−
6
)
.
\newline
f
(
x
)
=
−
10
∣
x
∣
+
4
f(x) = -10|x| + 4
f
(
x
)
=
−
10∣
x
∣
+
4
\newline
f
(
−
6
)
=
f(-6) =
f
(
−
6
)
=
_____
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∫
1
f
(
x
)
d
x
=
\int_{1} f(x) d x=
∫
1
f
(
x
)
d
x
=
\newline
F
(
x
)
=
∣
4
x
−
20
∣
F(x)=|4 x-20|
F
(
x
)
=
∣4
x
−
20∣
\newline
f
(
x
)
=
F
′
(
x
)
f(x)=F^{\prime}(x)
f
(
x
)
=
F
′
(
x
)
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F
(
x
)
=
∣
4
x
−
20
∣
F(x)=|4 x-20|
F
(
x
)
=
∣4
x
−
20∣
\newline
f
(
x
)
=
F
′
(
x
)
f(x)=F^{\prime}(x)
f
(
x
)
=
F
′
(
x
)
\newline
∫
−
5
5
f
(
x
)
d
x
=
\int_{-5}^{5} f(x) d x=
∫
−
5
5
f
(
x
)
d
x
=
Get tutor help
rite a rule for this pattern, Use
n
n
n
for the term number in
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Use the following function rule to find
f
(
1
)
f(1)
f
(
1
)
.
\newline
f
(
x
)
=
∣
12
+
x
∣
f(x) = |12 + x|
f
(
x
)
=
∣12
+
x
∣
\newline
f
(
1
)
=
f(1) =
f
(
1
)
=
___
Get tutor help
Use the following function rule to find
f
(
12
)
f(12)
f
(
12
)
.
\newline
f
(
x
)
=
∣
−
1
+
x
∣
f(x)=|-1+x|
f
(
x
)
=
∣
−
1
+
x
∣
\newline
f
(
12
)
=
□
f(12) = \square
f
(
12
)
=
□
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DeltaMath
\newline
Khan Academy
\newline
Weekly Bellwork Ha
\newline
My IXL
\newline
DD.
4
4
4
Evaluate a nonlinear function HKG
\newline
Use the following function rule to find
f
(
10
)
f(10)
f
(
10
)
\newline
f
(
x
)
=
∣
x
∣
−
4
f
(
10
)
=
□
\begin{array}{l} f(x)=|x|-4 \\ f(10)=\square \end{array}
f
(
x
)
=
∣
x
∣
−
4
f
(
10
)
=
□
\newline
Submit
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Use the following function rule to find
f
(
7
)
f(7)
f
(
7
)
.
\newline
f
(
x
)
=
−
10
∣
x
∣
f
(
7
)
=
\begin{array}{l} f(x)=-10|x| \\ f(7)= \end{array}
f
(
x
)
=
−
10∣
x
∣
f
(
7
)
=
\newline
Submit
Get tutor help
Which sign makes the statement true?
\newline
7.9
×
1
0
4
7.9 \times 10^4
7.9
×
1
0
4
?
\text{?}
?
7
,
900
7,900
7
,
900
\newline
Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
Function
g
g
g
can be thought of as a scaled version of
f
(
x
)
=
∣
x
∣
f(x)=|x|
f
(
x
)
=
∣
x
∣
.
\newline
What is the equation for
g
(
x
)
g(x)
g
(
x
)
?
\newline
Choose
1
1
1
answer:
\newline
g
(
x
)
=
∣
x
∣
3
g(x)=\frac{|x|}{3}
g
(
x
)
=
3
∣
x
∣
\newline
g
(
x
)
=
∣
x
+
4
∣
g(x)=|x+4|
g
(
x
)
=
∣
x
+
4∣
\newline
g
(
x
)
=
∣
x
∣
+
4
g(x)=|x|+4
g
(
x
)
=
∣
x
∣
+
4
\newline
g
(
x
)
=
3
∣
x
∣
g(x)=3|x|
g
(
x
)
=
3∣
x
∣
Get tutor help
GIVEN:
\newline
Q
T
V
W
QTVW
QT
VW
is a rectangle,
\newline
Q
R
‾
≅
T
S
‾
\overline{QR} \cong \overline{TS}
QR
≅
TS
\newline
PROVE:
\newline
△
S
W
Q
≅
△
R
V
T
\triangle SWQ \cong \triangle RVT
△
S
W
Q
≅
△
R
V
T
Get tutor help
Exercise
1
1
1
.
1
1
1
.
2
2
2
\newline
List all elements of the following sets:
\newline
a.
{
1
n
∣
n
∈
{
3
,
4
,
5
,
6
}
}
\left\{\frac{1}{n}\mid n \in \{3,4,5,6\}\right\}
{
n
1
∣
n
∈
{
3
,
4
,
5
,
6
}
}
Get tutor help
Given
f
(
x
)
=
2
sec
(
2
x
)
f(x)=2 \sec (2 x)
f
(
x
)
=
2
sec
(
2
x
)
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
Given
y
=
−
2
sin
(
2
x
)
y=-2 \sin (2 x)
y
=
−
2
sin
(
2
x
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
x
2
=
−
5
y
2
+
y
3
-x^{2}=-5 y^{2}+y^{3}
−
x
2
=
−
5
y
2
+
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
2
,
1
)
(2,1)
(
2
,
1
)
.
\newline
Answer:
d
y
d
x
∣
(
2
,
1
)
=
\left.\frac{d y}{d x}\right|_{(2,1)}=
d
x
d
y
∣
∣
(
2
,
1
)
=
Get tutor help
If
−
y
3
−
y
+
2
+
y
2
=
x
2
-y^{3}-y+2+y^{2}=x^{2}
−
y
3
−
y
+
2
+
y
2
=
x
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
4
,
−
2
)
(4,-2)
(
4
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
4
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(4,-2)}=
d
x
d
y
∣
∣
(
4
,
−
2
)
=
Get tutor help
If
0
=
4
+
5
x
3
−
y
2
0=4+5 x^{3}-y^{2}
0
=
4
+
5
x
3
−
y
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
−
3
)
(1,-3)
(
1
,
−
3
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
−
3
)
=
\left.\frac{d y}{d x}\right|_{(1,-3)}=
d
x
d
y
∣
∣
(
1
,
−
3
)
=
Get tutor help
If
4
x
−
4
−
4
x
2
=
y
2
−
y
3
4 x-4-4 x^{2}=y^{2}-y^{3}
4
x
−
4
−
4
x
2
=
y
2
−
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
2
)
=
\left.\frac{d y}{d x}\right|_{(1,2)}=
d
x
d
y
∣
∣
(
1
,
2
)
=
Get tutor help
If
y
2
+
3
x
3
=
2
y
y^{2}+3 x^{3}=2 y
y
2
+
3
x
3
=
2
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
1
,
3
)
(-1,3)
(
−
1
,
3
)
.
\newline
Answer:
d
y
d
x
∣
(
−
1
,
3
)
=
\left.\frac{d y}{d x}\right|_{(-1,3)}=
d
x
d
y
∣
∣
(
−
1
,
3
)
=
Get tutor help
If
−
2
x
2
+
y
2
+
y
3
+
3
x
=
0
-2 x^{2}+y^{2}+y^{3}+3 x=0
−
2
x
2
+
y
2
+
y
3
+
3
x
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
2
,
1
)
(2,1)
(
2
,
1
)
.
\newline
Answer:
d
y
d
x
∣
(
2
,
1
)
=
\left.\frac{d y}{d x}\right|_{(2,1)}=
d
x
d
y
∣
∣
(
2
,
1
)
=
Get tutor help
If
5
x
2
−
5
−
3
y
2
=
0
5 x^{2}-5-3 y^{2}=0
5
x
2
−
5
−
3
y
2
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
4
,
5
)
(4,5)
(
4
,
5
)
.
\newline
Answer:
d
y
d
x
∣
(
4
,
5
)
=
\left.\frac{d y}{d x}\right|_{(4,5)}=
d
x
d
y
∣
∣
(
4
,
5
)
=
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If
−
3
−
3
x
2
−
5
y
=
−
y
3
−
y
2
-3-3 x^{2}-5 y=-y^{3}-y^{2}
−
3
−
3
x
2
−
5
y
=
−
y
3
−
y
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
1
,
−
2
)
(-1,-2)
(
−
1
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
−
1
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(-1,-2)}=
d
x
d
y
∣
∣
(
−
1
,
−
2
)
=
Get tutor help
If
y
2
=
2
x
2
+
y
y^{2}=2 x^{2}+y
y
2
=
2
x
2
+
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
2
)
=
\left.\frac{d y}{d x}\right|_{(1,2)}=
d
x
d
y
∣
∣
(
1
,
2
)
=
Get tutor help
