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Math Problems
Grade 7
Divide integers
In the answer, the whole number is the
□
\square
□
and the fraction is the
□
\square
□
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நல்ல கணித ஆசிரியரின் நற்பண்புகள்
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5
÷
5
5\div5
5
÷
5
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9
9
9
divide by
5805
5805
5805
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已知正方形
A
B
C
D
ABCD
A
BC
D
的边长为
2
2
2
,点
Q
Q
Q
为边
B
C
BC
BC
的中点,点
P
P
P
在正方形外部,且满足
∠
A
P
D
=
13
5
∘
\angle APD=135^\circ
∠
A
P
D
=
13
5
∘
,则
P
Q
PQ
PQ
的最大值
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Simplify:
\newline
1
×
5
+
6
1 \times 5 + 6
1
×
5
+
6
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Solve for
x
x
x
:
x
−
8
=
9
x-8=9
x
−
8
=
9
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Add:
\newline
4
+
8
=
4 + 8 =
4
+
8
=
_____
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Add:
\newline
−
6
+
−
1
=
-6 + -1 =
−
6
+
−
1
=
_____
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Multiply:
\newline
−
3
×
−
6
=
-3 \times -6 =
−
3
×
−
6
=
_____
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Divide:
\newline
40
−
5
\frac{40}{-5}
−
5
40
= _____
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Divide:
\newline
−
63
÷
9
=
-63 \div 9 =
−
63
÷
9
=
_____
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Find the value of each of the following using shortcut method
6
5
3
65^3
6
5
3
cube
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Find
m
∠
A
B
C
m \angle A B C
m
∠
A
BC
\newline
107
107
107
\newline
106
106
106
\newline
93
93
93
\newline
121
121
121
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5
5
5
.
b
2
=
50
b^{2}=50
b
2
=
50
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x
2
=
196
x^{2}=196
x
2
=
196
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Evaluate:
\newline
2
+
4
−
93290
2+4-93290
2
+
4
−
93290
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42
×
x
=
x
×
42
=
42
42 \times x = x \times 42 = 42
42
×
x
=
x
×
42
=
42
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1
8
×
2
\frac{1}{8} \times 2
8
1
×
2
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Find
x
\mathrm{x}
x
:
\newline
71
71
71
\newline
21
21
21
\newline
142
142
142
\newline
8
8
8
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?
×
3
6
=
15
×
1
6
? \times \frac{3}{6}=15 \times \frac{1}{6}
?
×
6
3
=
15
×
6
1
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830
x
=
20
100
\frac{830}{x}=\frac{20}{100}
x
830
=
100
20
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7
×
c
×
8
=
42
×
20
×
64
7 \times c \times 8 = 42 \times 20 \times 64
7
×
c
×
8
=
42
×
20
×
64
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−
3
1
2
÷
1
1
4
=
-3 \frac{1}{2} \div 1 \frac{1}{4}=
−
3
2
1
÷
1
4
1
=
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−
3
1
2
÷
4
9
=
-3 \frac{1}{2} \div \frac{4}{9}=
−
3
2
1
÷
9
4
=
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−
2
3
÷
2
1
4
=
\frac{-2}{3} \div 2 \frac{1}{4}=
3
−
2
÷
2
4
1
=
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1
2
3
÷
(
−
2
3
4
)
=
1 \frac{2}{3} \div\left(-2 \frac{3}{4}\right)=
1
3
2
÷
(
−
2
4
3
)
=
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−
1
1
3
÷
2
2
5
=
-1 \frac{1}{3} \div 2 \frac{2}{5}=
−
1
3
1
÷
2
5
2
=
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2
−
3
÷
2
3
4
=
\frac{2}{-3} \div 2 \frac{3}{4}=
−
3
2
÷
2
4
3
=
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5
−
8
÷
1
3
7
=
\frac{5}{-8} \div 1 \frac{3}{7}=
−
8
5
÷
1
7
3
=
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y
=
5
+
6
x
y=5+6 x
y
=
5
+
6
x
\newline
\begin{tabular}{|c|c|}
\newline
\hline
x
x
x
&
y
y
y
\\
\newline
\hline
1
1
1
&
□
\square
□
\\
\newline
\hline
3
3
3
&
□
\square
□
\\
\newline
\hline
4
4
4
&
□
\square
□
\\
\newline
\hline
10
10
10
&
□
\square
□
\\
\newline
\hline
\newline
\end{tabular}
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Transform this equation to solve for
y
y
y
:
y
+
5
=
2
x
y + 5 = 2x
y
+
5
=
2
x
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Rewrite the expression in the form of
4
n
4^n
4
n
.
(
4
2
)
4
(4^2)^4
(
4
2
)
4
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{
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2
×
8
+
8
×
4
2 \times 8+8 \times 4
2
×
8
+
8
×
4
to
\newline
pounds
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810
−
780
810-780
810
−
780
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5000
×
29
5000 \times 29
5000
×
29
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cos
x
=
6
9
\cos x = \frac{6}{9}
cos
x
=
9
6
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100
:
7
×
50
100:7\times50
100
:
7
×
50
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1157
1157
1157
divided by
13
13
13
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ext{ exttt{
−
10
-10
−
10
extdiv extvisiblespace
25
25
25
}}
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Solve for
x
x
x
:
x
=
−
1
x=-1
x
=
−
1
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178
÷
9
178\div9
178
÷
9
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If
x
÷
20
=
0
x \div 20 = 0
x
÷
20
=
0
, then
x
=
x =
x
=
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8
4
8^4
8
4
=
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BODMAS Rule
\newline
Evaluate the following:
\newline
1
1
1
)
6
+
7
×
8
6+7 \times 8
6
+
7
×
8
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11
=
7
V
11=7V
11
=
7
V
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(
7
7
7
)
3
x
:
5
=
6
:
6
3 x: 5=6: 6
3
x
:
5
=
6
:
6
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Solve for
x
x
x
.
\newline
x
−
9
=
1
x-9=1
x
−
9
=
1
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0
=
x
3
−
8
0=x^3-8
0
=
x
3
−
8
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1
2
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