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Byju's Answer
Standard VII
Mathematics
Exponents with Like Bases
Simplify:i 23...
Question
Simplify:
(i)
2
3
×
5
44
×
33
35
(ii)
12
25
×
15
28
×
35
36
(iii)
10
27
×
28
65
×
39
56
(iv)
1
4
7
×
1
13
22
×
1
1
15
(v)
2
2
17
×
7
2
9
×
1
33
52
(vi)
3
1
16
×
7
3
7
×
1
25
39
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Solution
We have the following:
(i)
2
3
×
5
44
×
33
35
=
2
×
5
×
33
3
×
44
×
35
=
1
×
1
×
11
1
×
22
×
7
=
1
×
1
×
1
1
×
2
×
7
=
1
14
(ii)
12
25
×
15
28
×
35
36
=
1
×
3
×
5
5
×
4
×
3
=
1
×
1
×
1
1
×
4
×
1
=
1
4
(iii)
10
27
×
28
65
×
39
56
=
10
×
1
×
3
27
×
5
×
2
=
1
×
1
×
3
27
×
1
×
1
=
3
27
=
1
9
(iv)
1
4
7
×
1
13
22
×
1
1
15
=
11
7
×
35
22
×
16
15
=
11
×
35
×
16
7
×
22
×
15
=
1
×
5
×
16
1
×
2
×
15
=
1
×
1
×
8
1
×
1
×
3
=
8
3
=
2
2
3
(v)
2
2
17
×
7
2
9
×
1
33
52
=
36
17
×
65
9
×
85
52
=
36
×
65
×
85
17
×
9
×
52
=
4
×
5
×
5
1
×
1
×
4
=
1
×
5
×
5
1
×
1
×
1
=
25
(vi)
3
1
16
×
7
3
7
×
1
25
39
=
49
16
×
52
7
×
64
39
=
7
×
4
×
4
1
×
1
×
3
=
112
3
=
37
1
3
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Similar questions
Q.
Simplify:
(i)
12
25
×
15
28
×
35
36
(ii)
10
27
×
39
56
×
28
65
(iii)
2
2
17
×
7
2
9
×
1
33
52
Q.
Simplify:
(i) (
−24
)
× (68) + (−24) × 32
(ii) (
−9
)
×
18
−
(
−9
)
× 8
(iii) (
−147
)
÷ (−21)
(iv) 16 ÷ (−1)
Q.
The digit at unit's place in the number
(
13
)
1225
+
(
11
)
1915
−
(
23
)
1225
is equal to
Q.
The digit at unit's place in the number
(
13
)
1225
+
(
11
)
1915
−
(
23
)
1225
is equal to
Q.
Simplify:
(i) (− 4)
3
(ii) (− 3) × (− 2)
3
(iii) (− 3)
2
× (− 5)
2
(iv)(− 2)
3
× (−10)
3
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