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Math Problems
Grade 8
Write and solve direct variation equations
If
f
(
1
)
=
1
,
f
(
2
)
=
4
f(1)=1, f(2)=4
f
(
1
)
=
1
,
f
(
2
)
=
4
and
f
(
n
)
=
f
(
n
−
1
)
−
3
f
(
n
−
2
)
f(n)=f(n-1)-3 f(n-2)
f
(
n
)
=
f
(
n
−
1
)
−
3
f
(
n
−
2
)
then find the value of
f
(
6
)
f(6)
f
(
6
)
.
\newline
Answer:
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If
f
(
1
)
=
5
,
f
(
2
)
=
4
f(1)=5, f(2)=4
f
(
1
)
=
5
,
f
(
2
)
=
4
and
f
(
n
)
=
3
f
(
n
−
1
)
−
f
(
n
−
2
)
f(n)=3 f(n-1)-f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
−
f
(
n
−
2
)
then find the value of
f
(
5
)
f(5)
f
(
5
)
.
\newline
Answer:
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If
f
(
1
)
=
5
,
f
(
2
)
=
1
f(1)=5, f(2)=1
f
(
1
)
=
5
,
f
(
2
)
=
1
and
f
(
n
)
=
3
f
(
n
−
1
)
+
f
(
n
−
2
)
f(n)=3 f(n-1)+f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
+
f
(
n
−
2
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
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If
f
(
1
)
=
2
,
f
(
2
)
=
5
f(1)=2, f(2)=5
f
(
1
)
=
2
,
f
(
2
)
=
5
and
f
(
n
)
=
f
(
n
−
1
)
+
3
f
(
n
−
2
)
f(n)=f(n-1)+3 f(n-2)
f
(
n
)
=
f
(
n
−
1
)
+
3
f
(
n
−
2
)
then find the value of
f
(
5
)
f(5)
f
(
5
)
.
\newline
Answer:
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If
f
(
1
)
=
4
,
f
(
2
)
=
1
f(1)=4, f(2)=1
f
(
1
)
=
4
,
f
(
2
)
=
1
and
f
(
n
)
=
3
f
(
n
−
1
)
+
f
(
n
−
2
)
f(n)=3 f(n-1)+f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
+
f
(
n
−
2
)
then find the value of
f
(
5
)
f(5)
f
(
5
)
.
\newline
Answer:
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If
f
(
1
)
=
9
f(1)=9
f
(
1
)
=
9
and
f
(
n
)
=
3
f
(
n
−
1
)
−
n
f(n)=3 f(n-1)-n
f
(
n
)
=
3
f
(
n
−
1
)
−
n
then find the value of
f
(
3
)
f(3)
f
(
3
)
.
\newline
Answer:
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Given that
f
(
x
)
=
3
x
,
g
(
x
)
=
x
+
3
f(x)=3 x, g(x)=x+3
f
(
x
)
=
3
x
,
g
(
x
)
=
x
+
3
and
h
(
x
)
=
2
f
(
x
+
1
)
+
3
g
(
x
)
h(x)=2 f(x+1)+3 g(x)
h
(
x
)
=
2
f
(
x
+
1
)
+
3
g
(
x
)
, then what is the value of
h
(
6
)
h(6)
h
(
6
)
?
\newline
Answer:
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Given that
f
(
x
)
=
3
x
,
g
(
x
)
=
x
+
4
f(x)=3 x, g(x)=x+4
f
(
x
)
=
3
x
,
g
(
x
)
=
x
+
4
and
h
(
x
)
=
−
2
f
(
x
+
3
)
−
2
g
(
x
−
3
)
h(x)=-2 f(x+3)-2 g(x-3)
h
(
x
)
=
−
2
f
(
x
+
3
)
−
2
g
(
x
−
3
)
, then what is the value of
h
(
6
)
h(6)
h
(
6
)
?
\newline
Answer:
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If
f
(
1
)
=
10
f(1)=10
f
(
1
)
=
10
and
f
(
n
)
=
−
5
f
(
n
−
1
)
−
n
f(n)=-5 f(n-1)-n
f
(
n
)
=
−
5
f
(
n
−
1
)
−
n
then find the value of
f
(
3
)
f(3)
f
(
3
)
.
\newline
Answer:
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Given that
f
(
x
)
=
4
x
,
g
(
x
)
=
x
−
1
f(x)=4 x, g(x)=x-1
f
(
x
)
=
4
x
,
g
(
x
)
=
x
−
1
and
h
(
x
)
=
2
f
(
x
+
3
)
−
2
g
(
x
)
h(x)=2 f(x+3)-2 g(x)
h
(
x
)
=
2
f
(
x
+
3
)
−
2
g
(
x
)
, then what is the value of
h
(
6
)
h(6)
h
(
6
)
?
\newline
Answer:
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Given
g
(
x
)
=
−
x
−
1
g(x)=-x-1
g
(
x
)
=
−
x
−
1
, solve for
x
x
x
when
g
(
x
)
=
0
g(x)=0
g
(
x
)
=
0
.
\newline
Answer:
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Given
f
(
x
)
=
−
x
+
3
f(x)=-x+3
f
(
x
)
=
−
x
+
3
, solve for
x
x
x
when
f
(
x
)
=
0
f(x)=0
f
(
x
)
=
0
.
\newline
Answer:
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Given
h
(
x
)
=
−
x
+
3
h(x)=-x+3
h
(
x
)
=
−
x
+
3
, solve for
x
x
x
when
h
(
x
)
=
0
h(x)=0
h
(
x
)
=
0
.
\newline
Answer:
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−
2
x
+
b
y
=
59
-2x+by=59
−
2
x
+
b
y
=
59
\newline
5
x
−
y
=
1
5x-y=1
5
x
−
y
=
1
\newline
If the system of equations has exactly one solution
(
x
,
y
)
(x,y)
(
x
,
y
)
and
x
=
2
x=2
x
=
2
, what is the value of
b
b
b
?
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If
a
1
=
5
a_{1}=5
a
1
=
5
and
a
n
=
−
2
a
n
−
1
+
1
a_{n}=-2 a_{n-1}+1
a
n
=
−
2
a
n
−
1
+
1
then find the value of
a
4
a_{4}
a
4
.
\newline
Answer:
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If
a
1
=
9
a_{1}=9
a
1
=
9
and
a
n
=
n
a
n
−
1
−
2
a_{n}=n a_{n-1}-2
a
n
=
n
a
n
−
1
−
2
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
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If
a
1
=
4
a_{1}=4
a
1
=
4
and
a
n
=
5
a
n
−
1
−
n
a_{n}=5 a_{n-1}-n
a
n
=
5
a
n
−
1
−
n
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
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If
a
1
=
1
a_{1}=1
a
1
=
1
and
a
n
=
n
a
n
−
1
+
4
a_{n}=n a_{n-1}+4
a
n
=
n
a
n
−
1
+
4
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
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If
a
1
=
9
a_{1}=9
a
1
=
9
and
a
n
+
1
=
2
a
n
+
2
a_{n+1}=2 a_{n}+2
a
n
+
1
=
2
a
n
+
2
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
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4
x
−
32
k
x
=
53
4 x-32 k x=53
4
x
−
32
k
x
=
53
\newline
In the given equation,
k
k
k
is a constant. The equation has no solution. What is the value of
k
k
k
?
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evaluate
6
+
4
a
+
b
3
6+\frac{4}{a}+\frac{b}{3}
6
+
a
4
+
3
b
when
a
=
4
a=4
a
=
4
and
b
=
3
b=3
b
=
3
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Find the average value of the function
f
(
x
)
=
2
x
−
11
f(x)=\frac{2}{x-11}
f
(
x
)
=
x
−
11
2
from
x
=
2
x=2
x
=
2
to
x
=
8
x=8
x
=
8
. Express your answer as a constant times
ln
3
\ln 3
ln
3
.
\newline
Answer:
□
ln
3
\square \ln 3
□
ln
3
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a
x
2
+
5
x
+
2
=
0
a x^{2}+5 x+2=0
a
x
2
+
5
x
+
2
=
0
\newline
In the given equation,
a
a
a
is a constant. If the equation has the solutions
x
=
−
2
x=-2
x
=
−
2
and
x
=
−
1
2
x=-\frac{1}{2}
x
=
−
2
1
, what is the value of
a
a
a
?
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3
x
2
+
a
x
−
5
=
0
3 x^{2}+a x-5=0
3
x
2
+
a
x
−
5
=
0
\newline
In the given equation,
a
a
a
is a constant. If the equation has the solutions
x
=
−
5
x=-5
x
=
−
5
and
x
=
1
3
x=\frac{1}{3}
x
=
3
1
, what is the value of
a
a
a
?
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(
x
−
3
)
(
a
x
+
4
)
=
0
(x-3)(a x+4)=0
(
x
−
3
)
(
a
x
+
4
)
=
0
\newline
In the given equation,
a
a
a
is a constant. If the equation has the solutions
x
=
3
x=3
x
=
3
and
x
=
−
2
x=-2
x
=
−
2
, what is the value of
a
a
a
?
Get tutor help
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