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Byju's Answer
Standard VI
Mathematics
Exponents with Unlike Bases and Same Exponent
Question 6Lik...
Question
Question 6
Like term as
4
m
3
n
2
is
a)
4
m
2
n
2
b)
−
6
m
3
n
2
c)
6
p
m
3
n
2
d)
4
m
3
n
2
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Solution
b)
−
6
m
3
n
2
→
We know that, the like terms contain the same literal factor. So, the like term as
4
m
3
n
2
is
−
6
m
3
n
2
, as it contains the same literal factor
m
3
n
2
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4
Similar questions
Q.
Question 6
Like term as
4
m
3
n
2
is
a)
4
m
2
n
2
b)
−
6
m
3
n
2
c)
6
p
m
3
n
2
d)
4
m
3
n
2
Q.
Factorise
4
m
3
n
2
+
12
m
2
n
2
+
18
m
4
n
3
Q.
Which of the following is the like term of
6
m
2
n
3
?
Q.
Question 6
The
21
s
t
term of an AP whose first
two terms are – 3
and 4, is
A) 17
B) 137
C) 143
D) -143
Q.
Question 4
The
11
t
h
term of the AP
−
5
,
−
5
2
,
0
,
5
2
,
⋯
⋯
is
A) -20
B) 20
C) -30
D) 30
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Exponents with Unlike Bases and Same Exponent
Standard VI Mathematics
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