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Math Problems
Algebra 1
Write and solve direct variation equations
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
14
14
14
when
x
x
x
is
2
2
2
, find
y
y
y
when
x
x
x
is
9
9
9
.
\newline
Answer:
y
=
y=
y
=
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If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
5
5
5
when
x
x
x
is
15
15
15
, find
y
y
y
when
x
x
x
is
6
6
6
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
9
9
9
when
x
x
x
is
12
12
12
, find
y
y
y
when
x
x
x
is
4
4
4
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
4
4
4
when
x
x
x
is
8
8
8
, find
y
y
y
when
x
x
x
is
10
10
10
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
4
4
4
when
x
x
x
is
2
2
2
, find
y
y
y
when
x
x
x
is
5
5
5
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
5
5
5
when
x
x
x
is
10
10
10
, find
y
y
y
when
x
x
x
is
14
14
14
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
56
56
56
when
x
x
x
is
7
7
7
, find
y
y
y
when
x
x
x
is
6
6
6
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
8
8
8
when
x
x
x
is
12
12
12
, find
y
y
y
when
x
x
x
is
6
6
6
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
9
9
9
when
x
x
x
is
12
12
12
, find
y
y
y
when
x
x
x
is
8
8
8
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
2
2
2
when
x
x
x
is
4
4
4
, find
y
y
y
when
x
x
x
is
6
6
6
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
10
10
10
when
x
x
x
is
15
15
15
, find
y
y
y
when
x
x
x
is
9
9
9
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
7
7
7
when
x
x
x
is
14
14
14
, find
y
y
y
when
x
x
x
is
8
8
8
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
60
60
60
when
x
x
x
is
10
10
10
, find
y
y
y
when
x
x
x
is
15
15
15
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
1
1
1
when
x
x
x
is
4
4
4
, find
y
y
y
when
x
x
x
is
12
12
12
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
24
24
24
when
x
x
x
is
6
6
6
, find
y
y
y
when
x
x
x
is
3
3
3
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
4
4
4
when
x
x
x
is
12
12
12
, find
y
y
y
when
x
x
x
is
15
15
15
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
2
2
2
when
x
x
x
is
8
8
8
, find
y
y
y
when
x
x
x
is
12
12
12
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
are in direct proportion and
y
y
y
is
3
3
3
when
x
x
x
is
12
12
12
, find
y
y
y
when
x
x
x
is
4
4
4
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
25
25
25
when
x
x
x
is
5
5
5
, find
y
y
y
when
x
x
x
is
10
10
10
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
55
55
55
when
x
x
x
is
11
11
11
, find
y
y
y
when
x
x
x
is
13
13
13
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
If
x
x
x
and
y
y
y
vary directly and
y
y
y
is
55
55
55
when
x
x
x
is
11
11
11
, find
y
y
y
when
x
x
x
is
4
4
4
.
\newline
Answer:
y
=
y=
y
=
Get tutor help
The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of fox at a national park is modeled by the following differential equation:
\newline
d
P
d
t
=
3277
16290
P
(
1
−
P
904
)
\frac{d P}{d t}=\frac{3277}{16290} P\left(1-\frac{P}{904}\right)
d
t
d
P
=
16290
3277
P
(
1
−
904
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of fox at the national park is
180
180
180
and is increasing at a rate of
29
29
29
fox per day. Find
lim
t
→
∞
P
′
(
t
)
\lim _{t \rightarrow \infty} P^{\prime}(t)
lim
t
→
∞
P
′
(
t
)
.
\newline
Answer:
Get tutor help
The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people on a beach is modeled by the following differential equation:
\newline
d
P
d
t
=
6154
21945
P
(
1
−
P
724
)
\frac{d P}{d t}=\frac{6154}{21945} P\left(1-\frac{P}{724}\right)
d
t
d
P
=
21945
6154
P
(
1
−
724
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of people on the beach is
154
154
154
and is increasing at a rate of
34
34
34
people per hour. Find
lim
t
→
∞
P
(
t
)
\lim _{t \rightarrow \infty} P(t)
lim
t
→
∞
P
(
t
)
.
\newline
Answer:
Get tutor help
C
=
[
1
4
4
−
1
3
−
2
]
\mathrm{C}=\left[\begin{array}{rr}1 & 4 \\ 4 & -1 \\ 3 & -2\end{array}\right]
C
=
⎣
⎡
1
4
3
4
−
1
−
2
⎦
⎤
and
D
=
[
−
2
2
3
0
]
\mathrm{D}=\left[\begin{array}{rr} -2 & 2 \\ 3 & 0 \end{array}\right]
D
=
[
−
2
3
2
0
]
\newline
Let
H
=
C
D
\mathrm{H}=\mathrm{CD}
H
=
CD
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
D
=
[
−
1
0
2
−
2
1
3
]
\mathrm{D}=\left[\begin{array}{rr}-1 & 0 \\ 2 & -2 \\ 1 & 3\end{array}\right]
D
=
⎣
⎡
−
1
2
1
0
−
2
3
⎦
⎤
and
F
=
[
0
2
3
−
2
]
F=\left[\begin{array}{rr} 0 & 2 \\ 3 & -2 \end{array}\right]
F
=
[
0
3
2
−
2
]
\newline
Let
H
=
D
F
\mathrm{H}=\mathrm{DF}
H
=
DF
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
C
=
[
−
1
−
2
2
]
and
D
=
[
2
1
]
\mathrm{C}=\left[\begin{array}{r} -1 \\ -2 \\ 2 \end{array}\right] \text { and } \mathrm{D}=\left[\begin{array}{ll} 2 & 1 \end{array}\right]
C
=
⎣
⎡
−
1
−
2
2
⎦
⎤
and
D
=
[
2
1
]
\newline
Let
H
=
C
D
\mathrm{H}=\mathrm{CD}
H
=
CD
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
E
=
[
1
−
1
5
5
5
0
]
\mathrm{E}=\left[\begin{array}{rrr}1 & -1 & 5 \\ 5 & 5 & 0\end{array}\right]
E
=
[
1
5
−
1
5
5
0
]
and
F
=
[
−
2
−
1
5
2
5
−
2
]
F=\left[\begin{array}{rr}-2 & -1 \\ 5 & 2 \\ 5 & -2\end{array}\right]
F
=
⎣
⎡
−
2
5
5
−
1
2
−
2
⎦
⎤
\newline
Let
H
=
E
F
\mathrm{H}=\mathrm{EF}
H
=
EF
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
F
=
[
2
3
0
5
−
1
−
1
]
F=\left[\begin{array}{rr}2 & 3 \\ 0 & 5 \\ -1 & -1\end{array}\right]
F
=
⎣
⎡
2
0
−
1
3
5
−
1
⎦
⎤
and
D
=
[
−
1
0
4
2
]
\mathrm{D}=\left[\begin{array}{rr} -1 & 0 \\ 4 & 2 \end{array}\right]
D
=
[
−
1
4
0
2
]
\newline
Let
H
=
F
D
\mathrm{H}=\mathrm{FD}
H
=
FD
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
B
=
[
4
4
1
0
−
2
1
]
and
F
=
[
0
3
0
1
]
B=\left[\begin{array}{rr} 4 & 4 \\ 1 & 0 \\ -2 & 1 \end{array}\right] \text { and } F=\left[\begin{array}{ll} 0 & 3 \\ 0 & 1 \end{array}\right]
B
=
⎣
⎡
4
1
−
2
4
0
1
⎦
⎤
and
F
=
[
0
0
3
1
]
\newline
Let
H
=
B
F
\mathrm{H}=\mathrm{BF}
H
=
BF
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
D
=
[
4
−
1
2
−
1
]
\mathrm{D}=\left[\begin{array}{ll}4 & -1 \\ 2 & -1\end{array}\right]
D
=
[
4
2
−
1
−
1
]
and
A
=
[
3
1
0
2
1
−
2
]
A=\left[\begin{array}{rrr} 3 & 1 & 0 \\ 2 & 1 & -2 \end{array}\right]
A
=
[
3
2
1
1
0
−
2
]
\newline
Let
H
=
D
A
\mathrm{H}=\mathrm{DA}
H
=
DA
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
What happens to the value of the expression
35
+
k
35+k
35
+
k
as
k
k
k
decreases?
\newline
Choose
1
1
1
answer:
\newline
(A) It increases.
\newline
(B) It decreases.
\newline
(C) It stays the same.
Get tutor help
The functions
f
(
x
)
=
−
2.5
(
x
+
2
)
2
+
8
f(x)=-2.5(x+2)^{2}+8
f
(
x
)
=
−
2.5
(
x
+
2
)
2
+
8
and
g
(
x
)
=
−
2.5
(
x
+
2
)
2
+
b
g(x)=-2.5(x+2)^{2}+b
g
(
x
)
=
−
2.5
(
x
+
2
)
2
+
b
are graphed in the
x
y
x y
x
y
-plane as
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
and
y
=
g
(
x
)
y=g(x)
y
=
g
(
x
)
. Let
a
a
a
be the
y
y
y
-coordinate of the vertex of function
f
f
f
and
b
b
b
be the
y
y
y
-coordinate of the vertex of function
g
(
x
)
=
−
2.5
(
x
+
2
)
2
+
b
g(x)=-2.5(x+2)^{2}+b
g
(
x
)
=
−
2.5
(
x
+
2
)
2
+
b
0
0
0
. If
a
a
a
is
5
5
5
less than
b
b
b
, then what is the value of
b
b
b
?
Get tutor help
a
t
2
+
7
2
t
−
4
=
0
a t^{2}+\frac{7}{2} t-4=0
a
t
2
+
2
7
t
−
4
=
0
\newline
The given equation has solutions at
t
=
−
8
t=-8
t
=
−
8
and
t
=
1
t=1
t
=
1
. What is the value of the constant
a
a
a
?
Get tutor help
(
4
x
+
3
)
−
(
5
x
−
7
)
=
a
x
+
b
(4 x+3)-(5 x-7)=a x+b
(
4
x
+
3
)
−
(
5
x
−
7
)
=
a
x
+
b
\newline
In the given equation,
b
b
b
is a constant. What is the value of
b
b
b
?
Get tutor help
(
2
x
+
5
)
(
−
m
x
+
9
)
=
0
(2 x+5)(-m x+9)=0
(
2
x
+
5
)
(
−
m
x
+
9
)
=
0
\newline
In the given equation,
m
m
m
is a constant. If the equation has the solutions
x
=
−
5
2
x=-\frac{5}{2}
x
=
−
2
5
and
x
=
3
2
x=\frac{3}{2}
x
=
2
3
, what is the value of
m
m
m
?
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(
x
+
4
)
(
2
x
−
a
)
=
0
(x+4)(2 x-a)=0
(
x
+
4
)
(
2
x
−
a
)
=
0
\newline
In the given equation,
a
a
a
is a constant. If the equation has the solutions
x
=
4
x=4
x
=
4
and
x
=
−
4
x=-4
x
=
−
4
, what is the value of
a
a
a
?
Get tutor help
What is the value of
x
2
y
4
\frac{x^{2}}{y^{4}}
y
4
x
2
when
x
=
8
x=8
x
=
8
and
y
=
2
y=2
y
=
2
?
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If
y
y
y
varies directly with
x
x
x
and
y
=
12
y = 12
y
=
12
when
x
=
6
x = 6
x
=
6
, find
y
y
y
when
x
=
4
x = 4
x
=
4
.
\newline
Write and solve a direct variation equation to find the answer.
\newline
y
=
y =
y
=
____
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