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There are 10 prime numbers between 0 and 30 which are: 2,3,5,7,11,13,17,19,23,29
Irrational number between 2 and 2.5 = √2*2.5 = √5
As it is clear now that, 2 < √5 < 2.5
We know that √5 is an irrational number.
So, √5 is the required irrational number between 2 and 2.5.
(√2×√3)½ = (2×3)½×½ = (6)¼
As the bar is under two digits multiply with 100
100x = 2.7435*100
100x = 274.35
now subtract with x terms
that gives
99x = 274.35  2.7435
99 x = 271.60
x = 271.60/99
x = 2716/9900
√4 = 2, √9 = 3, √16 = 4 but √2 = 1.414 it is a nonterminating and nonrepeating number and such numbers are called irrational numbers. So, √2 is not a rational number.
(2√3 +√3) = 3√3 ( By addition of like terms )
So option C is correct answer.
If this number is rationalised then 3√3×3√3/3+√3×3√3
=(3√3)^{2/3}^2+√3^{2}
=9+36√3/6
=126√3/6
=2√3 this is an irrational number.
63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder
The sum of the digits of a number is subtracted from the number, the resulting number is always divisible by which of the following numbers?
The resulting number is always divisible by 9. e.g.,
Consider 47.
Sum of its digits is 4+7=11
So,47−11=36 which is exactly divisible by 9
(1  10)/(9 + (3))
=9/6 = 3/2
If 2^{x} = 4^{y} = 8^{z} and = 4, then the value of x is :
If a = 1/ (32√2) & b = 1/ ( 3+2√2) then the value of a^{2} +b^{2} is :
a = 1/ (32√2)
b = 1/ ( 3+2√2)
∴ ab = 1/(32√2) * 1/ (3+2√2)
=> 1/ (98)
=> 1
We know;
(a+b)^{2} = a^{2} + b^{2 }+ 2ab
And, a^{2}+ b^{2} = (a+b)^{2}  2ab
=> [ 1/(32√2) + 1/(3+2√2) ]^{2} 2*1
=> [3+2√2+32√2 / 1]^2 2
=> (6)^{2}  2
=> 36  2
=> 34
[(x)^2]^{1/3]1/4}
= [(x^{2})^{1/12}]
= x^{1/6}
4√6*3√24=4√6*3√2*√2*√*2*√3
=4√6*6√6
4*6*√6*√6=4*6*6=24*6=144
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