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Math Problems
Precalculus
Operations with rational exponents
x
2
+
y
2
=
121
?
x^{2}+y^{2}=121 ?
x
2
+
y
2
=
121
?
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Name
\newline
Jaliyan
\newline
Date
\newline
Multiplying Fractions
\newline
2
2
2
. The math club has
96
96
96
members.
1
8
\frac{1}{8}
8
1
of the members are fourth graders and th rest are fifth graders. How many math club members are fifth graders?
\newline
(A)
12
12
12
\newline
C
80
80
80
\newline
B
16
16
16
\newline
D
84
84
84
\newline
4
4
4
. In which equation is the product greater than
3
3
3
?
\newline
3
4
\frac{3}{4}
4
3
\newline
A
1
×
1
3
=
1
12
\text { A } 1 \times \frac{1}{3}=\frac{1}{12}
A
1
×
3
1
=
12
1
\newline
B
3
2
×
3
\frac{3}{2} \times 3
2
3
×
3
\newline
1
12
\frac{1}{12}
12
1
\newline
C
1
12
×
3
1
12
×
3
1
=
3
12
1
12
×
1
3
=
1
36
\begin{array}{l} C \frac{1}{12} \times 3 \\ \frac{1}{12} \times \frac{3}{1}=\frac{3}{12} \\ \frac{1}{12} \times \frac{1}{3}=\frac{1}{36} \end{array}
C
12
1
×
3
12
1
×
1
3
=
12
3
12
1
×
3
1
=
36
1
\newline
D
9
10
×
3
\frac{9}{10} \times 3
10
9
×
3
\newline
3
2
x
\frac{3}{2} x
2
3
x
\newline
9
10
×
1
3
−
9
10
×
3
1
=
27
10
\frac{9}{10} \times \frac{1}{3}-\frac{9}{10} \times \frac{3}{1}=\frac{27}{10}
10
9
×
3
1
−
10
9
×
1
3
=
10
27
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Name Jaliuan
\newline
Date
\newline
Multiplying Fractions
\newline
2
2
2
. The math club has
96
96
96
members.
1
8
\frac{1}{8}
8
1
of the members are fourth graders and t rest are fifth graders. How many math club members are fifth graders?
\newline
(A)
12
12
12
\newline
C
80
80
80
\newline
B
16
16
16
\newline
D
84
84
84
\newline
4
4
4
. In which equation is the product greater than
3
3
3
?
\newline
3
1
=
3
4
\frac{3}{1}=\frac{3}{4}
1
3
=
4
3
\newline
A
1
×
1
3
=
1
12
1 \times \frac{1}{3}=\frac{1}{12}
1
×
3
1
=
12
1
\newline
B
3
2
×
3
\frac{3}{2} \times 3
2
3
×
3
\newline
3
2
\frac{3}{2}
2
3
\newline
=
1
12
=\frac{1}{12}
=
12
1
\newline
4
4
4
\newline
C
1
12
×
3
\frac{1}{12} \times 3
12
1
×
3
\newline
D
9
10
×
3
\frac{9}{10} \times 3
10
9
×
3
\newline
3
2
\frac{3}{2}
2
3
\newline
1
12
×
3
1
=
3
12
=
1
12
=
1
36
\frac{1}{12} \times \frac{3}{1}=\frac{3}{12}=\frac{1}{12}=\frac{1}{36}
12
1
×
1
3
=
12
3
=
12
1
=
36
1
\newline
3
1
=
3
4
\frac{3}{1}=\frac{3}{4}
1
3
=
4
3
0
0
0
\newline
3
1
=
3
4
\frac{3}{1}=\frac{3}{4}
1
3
=
4
3
1
1
1
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(
2
2
2
)
\newline
6
6
6
. In a school contest, each student draws a number from
1
1
1
to
5
5
5
at random out of a large basket. The outcomes are shown in the table below.
\newline
Based on the data in the table, what is the expected probability of drawing a
3
3
3
?
\newline
\begin{tabular}{|l|c|c|c|c|c|}
\newline
\hline Number Drawn &
1
1
1
&
2
2
2
&
3
3
3
&
4
4
4
&
5
5
5
\\
\newline
\hline \begin{tabular}{l}
\newline
Number of \\
\newline
Times Drawn
\newline
\end{tabular} &
83
83
83
&
79
79
79
&
88
88
88
&
72
72
72
&
85
85
85
\\
\newline
\hline
\newline
\end{tabular}
\newline
488
740
\frac{488}{740}
740
488
\newline
5 \longdiv { \frac { 8 1 4 } { 4 0 7 } } \frac { \frac { 2 } { 3 8 8 } } { \sqrt { 4 0 7 } } \quad \frac { - 6 } { 2 8 }
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Aidan ran
3
4
\frac{3}{4}
4
3
mile in a race. Bryan ran
2
5
\frac{2}{5}
5
2
of the distance that Aidan ran. What distance did Bryan run?
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15
15
15
The graph of a linear function is shown on the grid.
\newline
Which equation is best represented by this graph?
\newline
A
y
=
−
7
4
x
+
4
y=-\frac{7}{4} x+4
y
=
−
4
7
x
+
4
\newline
B
y
=
−
7
4
x
+
7
y=-\frac{7}{4} x+7
y
=
−
4
7
x
+
7
\newline
c
y
=
−
4
7
x
+
4
y=-\frac{4}{7} x+4
y
=
−
7
4
x
+
4
\newline
D
y
=
−
4
7
x
+
7
y=-\frac{4}{7} x+7
y
=
−
7
4
x
+
7
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For \#
12
12
12
−
15
-15
−
15
, simplify. You
\newline
12
12
12
.
448
x
21
y
18
5
\sqrt[5]{448 x^{21} y^{18}}
5
448
x
21
y
18
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Simplify the expression completely:
2
x
10
y
15
34
x
5
y
5
\frac{2 x^{10} y^{15}}{34 x^{5} y^{5}}
34
x
5
y
5
2
x
10
y
15
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What is the area of the region bound by the graphs of
f
(
x
)
=
x
−
2
f(x)=\sqrt{x-2}
f
(
x
)
=
x
−
2
,
g
(
x
)
=
14
−
x
g(x)=14-x
g
(
x
)
=
14
−
x
, and
x
=
2
x=2
x
=
2
?
\newline
Choose
1
1
1
answer:
\newline
(A)
19
6
\frac{19}{6}
6
19
\newline
(B)
99
2
\frac{99}{2}
2
99
\newline
(C)
151
2
\frac{151}{2}
2
151
\newline
(D)
45
2
\frac{45}{2}
2
45
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1
−
…
.
=
3
8
4
8
\begin{array}{c}1-\ldots .=\frac{3}{8} \\ \frac{4}{8}\end{array}
1
−
…
.
=
8
3
8
4
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x
5
6
=
32
x^{\frac{5}{6}}=32
x
6
5
=
32
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(
3
3
3
) List the equivalent fractions. Then compare the equivalent fractions and fill in the boxes.
\newline
(a)
3
4
and
4
5
3
4
=
4
5
=
\begin{array}{l} \text { (a) } \frac{3}{4} \text { and } \frac{4}{5} \\ \frac{3}{4}= \\ \frac{4}{5}= \end{array}
(a)
4
3
and
5
4
4
3
=
5
4
=
\newline
So,
□
\square
□
is smaller than
□
\square
□
.
\newline
(b)
5
6
\frac{5}{6}
6
5
and
7
9
\frac{7}{9}
9
7
\newline
5
6
=
7
9
=
\begin{array}{l} \frac{5}{6}= \\ \frac{7}{9}= \end{array}
6
5
=
9
7
=
\newline
So,
□
\square
□
is greater than
□
\square
□
.
\newline
(c)
7
8
\frac{7}{8}
8
7
and
2
3
\frac{2}{3}
3
2
\newline
7
8
=
2
3
=
\begin{array}{l} \frac{7}{8}= \\ \frac{2}{3}= \end{array}
8
7
=
3
2
=
\newline
So, is smaller than
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(
1
,
07
×
1
0
−
1
1
,
56
×
1
0
−
2
)
x
=
(
1
,
32
2
,
89
)
\left(\frac{1,07 \times 10^{-1}}{1,56 \times 10^{-2}}\right)^{x}=\left(\frac{1,32}{2,89}\right)
(
1
,
56
×
1
0
−
2
1
,
07
×
1
0
−
1
)
x
=
(
2
,
89
1
,
32
)
\newline
6
6
6
.
058
058
058
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3
4
x
2
+
1
2
x
=
0
\frac{3}{4} x^{2}+\frac{1}{2} x=0
4
3
x
2
+
2
1
x
=
0
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1
1
1
. Place the following numbers in ascending order:
\newline
\begin{tabular}{|l|l|l|l|}
\newline
\hline
7
4
\frac{7}{4}
4
7
&
−
2
3
-\frac{2}{3}
−
3
2
&
−
8
5
-\frac{8}{5}
−
5
8
&
150
%
150 \%
150%
\\
\newline
\hline
\newline
\end{tabular}
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13
100
÷
11
20
\frac{13}{100} \div \frac{11}{20}
100
13
÷
20
11
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4
10
=
20
=
100
\frac{4}{10}=\frac{20}{}=\frac{}{100}
10
4
=
20
=
100
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x
2
−
3
x
y
+
y
3
=
3
x^{2}-3 x y+y^{3}=3
x
2
−
3
x
y
+
y
3
=
3
\newline
Find the value of
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
4
15
\frac{4}{15}
15
4
\newline
(B)
4
9
\frac{4}{9}
9
4
\newline
(C)
15
4
\frac{15}{4}
4
15
\newline
(D)
9
4
\frac{9}{4}
4
9
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Draw a ring around all of the calculations that are equivalent to
9
16
÷
3
4
\frac{9}{16} \div \frac{3}{4}
16
9
÷
4
3
\newline
16
9
×
3
4
9
16
×
4
3
9
4
×
1
3
16
9
×
4
3
3
4
×
1
1
3
8
×
2
1
\frac{16}{9} \times \frac{3}{4} \quad \frac{9}{16} \times \frac{4}{3} \quad \frac{9}{4} \times \frac{1}{3} \quad \frac{16}{9} \times \frac{4}{3} \quad \frac{3}{4} \times \frac{1}{1} \quad \frac{3}{8} \times \frac{2}{1}
9
16
×
4
3
16
9
×
3
4
4
9
×
3
1
9
16
×
3
4
4
3
×
1
1
8
3
×
1
2
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f
′
(
x
)
=
100
x
3
f^{\prime}(x)=\frac{100}{x^{3}}
f
′
(
x
)
=
x
3
100
and
f
(
5
)
=
−
14
f(5)=-14
f
(
5
)
=
−
14
.
\newline
f
(
1
)
=
f(1)=
f
(
1
)
=
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f
′
(
x
)
=
−
50
x
3
f^{\prime}(x)=-\frac{50}{x^{3}}
f
′
(
x
)
=
−
x
3
50
and
f
(
1
)
=
20
f(1)=20
f
(
1
)
=
20
.
\newline
f
(
5
)
=
f(5)=
f
(
5
)
=
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f
′
(
x
)
=
−
10
x
4
+
8
x
3
f^{\prime}(x)=-10 x^{4}+8 x^{3}
f
′
(
x
)
=
−
10
x
4
+
8
x
3
and
f
(
1
)
=
9
f(1)=9
f
(
1
)
=
9
.
\newline
f
(
0
)
=
f(0)=
f
(
0
)
=
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If
x
3
=
64
x^{3}=64
x
3
=
64
then what is
x
x
x
?
\newline
Step one: How do you undo
x
x
x
to the third
p
p
p
\newline
Take the cubed root of
64
64
64
:
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63
−
(
−
3
)
{
−
2
−
7
−
3
}
÷
3
{
5
+
(
−
2
)
(
−
1
)
}
B
O
D
M
A
S
63 - (-3) \{-2 - 7 - 3\} \div 3 \{5 + (-2) (-1)\} BODMAS
63
−
(
−
3
)
{
−
2
−
7
−
3
}
÷
3
{
5
+
(
−
2
)
(
−
1
)}
BO
D
M
A
S
Get tutor help
(
3
2
3
5
)
−
1
3
=
\left(32^{\frac{3}{5}}\right)^{-\frac{1}{3}}=
(
3
2
5
3
)
−
3
1
=
Get tutor help
w
5
−
w
3
−
64
w
2
<
−
64
w^{5}-w^{3}-64 w^{2}<-64
w
5
−
w
3
−
64
w
2
<
−
64
Get tutor help
Jika
M
=
(
−
2
5
1
−
3
)
M=\left(\begin{array}{cc}-2 & 5 \\ 1 & -3\end{array}\right)
M
=
(
−
2
1
5
−
3
)
dan
k
=
(
0
−
1
−
2
3
)
k=\left(\begin{array}{cc}0 & -1 \\ -2 & 3\end{array}\right)
k
=
(
0
−
2
−
1
3
)
maka determinon dari Matriks KM sama dengan....
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3
16
−
1
8
=
\frac{3}{16}-\frac{1}{8}=
16
3
−
8
1
=
\newline
Submit
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My IXL
\newline
Learning
\newline
Assessment
\newline
G.
9
9
9
Put a mix of decimals and fractions in order MSR
\newline
Put these numbers in order from greatest to least.
\newline
1
9
\frac{1}{9}
9
1
\newline
0
0
0
.
94
94
94
\newline
1
6
\frac{1}{6}
6
1
\newline
7
9
\frac{7}{9}
9
7
\newline
Submit
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(
3
3
3
) 4 \longdiv { 3 2 , 8 6 0 }
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3
3
3
. 4 \longdiv { 3 2 , 8 6 0 }
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№
1
Сравнить
8
0
13
×
1
0
28
\frac{\text { № } 1}{\text { Сравнить } 80^{13} \times 10^{28}}
Сравнить
8
0
13
×
1
0
28
№
1
.
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Express
(
3
7
)
17
÷
(
3
7
)
5
\left(\frac{3}{7}\right)^{17} \div\left(\frac{3}{7}\right)^{5}
(
7
3
)
17
÷
(
7
3
)
5
as a single power.
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Your goal is to make
251
251
251
. You can use
+
,
−
,
÷
,
×
+,-, \div ,\times
+
,
−
,
÷
,
×
Your numbers are
75
,
50
,
100
,
8
,
7
,
4
75, 50, 100, 8, 7, 4
75
,
50
,
100
,
8
,
7
,
4
You’re allowed to use each number once
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(
−
2
)
5
⋅
(
−
2
)
10
(-2)^5 \cdot (-2)^{10}
(
−
2
)
5
⋅
(
−
2
)
10
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7
7
7
.
sin
(
11
π
12
)
\sin \left(\frac{11 \pi}{12}\right)
sin
(
12
11
π
)
Get tutor help
using only positive exponents.
\newline
3
x
4
y
5
15
x
3
y
2
\frac{3 x^{4} y^{5}}{15 x^{3} y^{2}}
15
x
3
y
2
3
x
4
y
5
Get tutor help
(
7
x
4
)
(
5
x
3
)
=
\left(7 x^{4}\right)\left(5 x^{3}\right)=
(
7
x
4
)
(
5
x
3
)
=
Get tutor help
Q
1
1
1
. Find the additive inverse of
\newline
(
1
1
1
)
3
3
3
\newline
(
2
2
2
)
−
11
13
\frac{-11}{13}
13
−
11
\newline
(
3
3
3
)
31
119
\frac{31}{119}
119
31
\newline
(
4
4
4
)
4
−
25
\frac{4}{-25}
−
25
4
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Qus
1
1
1
- Find the additive inverse
θ
\theta
θ
\newline
(
1
1
1
)
3
11
\frac{3}{11}
11
3
\newline
(
2
2
2
)
−
11
13
\frac{-11}{13}
13
−
11
\newline
(
3
3
3
)
31
119
\frac{31}{119}
119
31
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G
89
89
89
b
\newline
(
9
9
9
)
3
2
×
4
3
÷
6
2
=
3^{2} \times 4^{3} \div 6^{2}=
3
2
×
4
3
÷
6
2
=
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3
x
2
+
7
−
3
2
3
−
3
2
=
65
3 x^{2}+7-3^{2} 3-3^{2}=65
3
x
2
+
7
−
3
2
3
−
3
2
=
65
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Найдите значение выражения:
\newline
4
4
9
⋅
1
1
32
⋅
1
1
5
−
(
2
11
14
−
2
2
35
)
⋅
4
2
3
4 \frac{4}{9} \cdot 1 \frac{1}{32} \cdot 1 \frac{1}{5}-\left(2 \frac{11}{14}-2 \frac{2}{35}\right) \cdot 4 \frac{2}{3}
4
9
4
⋅
1
32
1
⋅
1
5
1
−
(
2
14
11
−
2
35
2
)
⋅
4
3
2
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11
11
11
. Given the two functions below:
\newline
f
(
x
)
=
1
8
x
3
+
7
8
x
2
−
15
11
x
−
1
g
(
x
)
=
−
∣
0.5
x
∣
+
2
\begin{array}{c} f(x)=\frac{1}{8} x^{3}+\frac{7}{8} x^{2}-\frac{15}{11} x-1 \\ g(x)=-|0.5 x|+2 \end{array}
f
(
x
)
=
8
1
x
3
+
8
7
x
2
−
11
15
x
−
1
g
(
x
)
=
−
∣0.5
x
∣
+
2
\newline
State the solutions to the equation
f
(
x
)
=
g
(
x
)
f(x)=g(x)
f
(
x
)
=
g
(
x
)
, rounded to the nearest thousandth. \#
6
6
6
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3
9
10
+
4
9
+
7
45
+
4
=
3 \frac{9}{10}+\frac{4}{9}+\frac{7}{45}+4=
3
10
9
+
9
4
+
45
7
+
4
=
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5
8
÷
1
1
3
=
\frac{5}{8} \div 1 \frac{1}{3}=
8
5
÷
1
3
1
=
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11
11
11
.
>
(
7
2
1
/
5
9
1
/
5
)
3
>\left(\frac{72^{1 / 5}}{9^{1 / 5}}\right)^{3}
>
(
9
1/5
7
2
1/5
)
3
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4
4
4
. Resuelve el siguiente ejercicio:
4
+
3
4
−
(
1
2
×
5
)
+
(
5
8
×
2
15
)
−
3
2
[
1
2
+
(
1
2
×
4
)
]
4+\frac{3}{4}-\left(\frac{1}{2} \times 5\right)+\left(\frac{5}{8} \times \frac{2}{15}\right)-\frac{3}{2}\left[\frac{1}{2}+\left(\frac{1}{2} \times 4\right)\right]
4
+
4
3
−
(
2
1
×
5
)
+
(
8
5
×
15
2
)
−
2
3
[
2
1
+
(
2
1
×
4
)
]
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(
2
9
×
3
6
)
+
(
3
4
÷
12
5
)
−
[
1
4
−
(
1
4
×
2
3
)
]
\left(\frac{2}{9} \times \frac{3}{6}\right)+\left(\frac{3}{4} \div \frac{12}{5}\right)-\left[\frac{1}{4}-\left(\frac{1}{4} \times \frac{2}{3}\right)\right]
(
9
2
×
6
3
)
+
(
4
3
÷
5
12
)
−
[
4
1
−
(
4
1
×
3
2
)
]
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Fernando factored
45
y
6
45y^{6}
45
y
6
as
(
9
y
3
)
(
5
y
3
)
(9y^{3})(5y^{3})
(
9
y
3
)
(
5
y
3
)
.
\newline
Salma factored
45
y
6
45y^{6}
45
y
6
as
(
3
y
)
(
15
y
5
)
(3y)(15y^{5})
(
3
y
)
(
15
y
5
)
.
\newline
Which of them factored
45
y
6
45y^{6}
45
y
6
correctly?
\newline
Choose
1
1
1
answer:
\newline
(A) Only Fernando
\newline
(B) Only Salma
\newline
(C) Both Fernando and Salma
\newline
(D) Neither Fernando nor Salma
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