Question

(18 points)

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Q.1. For what value of n,2n × 5n ends in 5

Q.2. If the prime factorization of a natural number n is 2^3 × 3^2 × 5^2 × 7 , write the number of consecutive zeroes in m.

Q.3. Write whether 2√45 + 3√20 ÷ 2√5 on simplification gives rational or irrational?

Q.3. Given that HCF(2520,6600) , LCM(2520,6600)=252 × x then the value of x is

Q.4. (2+√3)(2-√3) is

Answer 1

que no... 4
(2)^2- (root3)^2
4-3
1

Answer 2

1
${2}^{n} \times {5}^{n}$
this alwaws ends with zero for any value of n. so this never ends with 5
2
${2}^{3} \times {3}^{2} \times {5}^{2} \times 7 = {2}^{2} \times {5}^{2} \times 2 \times {3}^{2} \times 7$
we here have
${5}^{2} \times {2}^{2}$
as one of factors so no of consecutive zeres is 2 and the number is 12600
3
$2 \sqrt{45} + 3 \sqrt{20} \div 2 \sqrt{ 5} \\ = 4 \sqrt{45} + 3 \sqrt{4} \div 2 \\ = 4 \sqrt{45 } + 3(2) \div 2 \\ = 4 \sqrt{45} + 3$
which is irrational

4 is rational as
$(2 + \sqrt{3} )(2 - \sqrt{3} ) \\ = {2}^{2} + { \sqrt{3} }^{2} = 4 + 3 = 7$