Test: Number Systems - 2 - Class 9 MCQ

There are 10 prime numbers between 0 and 30 which are: 2,3,5,7,11,13,17,19,23,29. Number one is not a prime number.
Now, for 1, the number of positive divisors or factors is only one i.e. 1 itself.  

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.

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 = 271.61

x = 271.61/99

x = 27161/9900

√4 = 2, √9 = 3, √16 = 4 but √2 = 1.414 it is a non-terminating and non-repeating 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. 

63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder 

find remainder when 99/4 ( by 4 because to check periodicity)

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

a = 1/ (3-2√2) 

b =  1/ ( 3+2√2) 

∴ ab =  1/(3-2√2) * 1/ (3+2√2) 

=>  1/ (9-8) 

=> 1 

We know; 

(a+b)² = a² + b² + 2ab

And, a²+ b² = (a+b)² - 2ab

=> [ 1/(3-2√2) + 1/(3+2√2) ]² -2*1

=> [3+2√2+3-2√2 / 1]^2 -2 

=> (6)² - 2 

=>  36 - 2
=>  34

2ⁿ(2⁴-2) + 2^-3
2ⁿx2x2³
14 + 2^-3
2⁴
7/8+1/8
1

a - a^-1 = a - 1/a = a² - 1
                               a
(x^m)ⁿ = x^mn just use this

[(x)^2]^1/3]1/4

= [(x²)^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