the square root of 64009 is - simplification
the square root of 64009 is
253 explanation:
so finding the square root by vedic method, four digits are in the number, so the digits in it’s square root will be 2. the quotient is 46, so the square root is 46.hint –i. four digits in the number so two digits in the square root.ii. first square root digit = 4iii. 21 - = 5 = remainderiv. pair of digits 16 was descended before 5. so new dividend = 516.v. divisor = 4 2 = 8, i.e., double of square root digit.vi. 86 divides 516 by 6, so write 6 ahead digit 4 in the quotient.vii. 516 – 86 6 = 516 – 516 = 0remainder = 0, square root = quotient = 46. solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))` on sapling its

oxidation:
cu --> cu^2+ + 2e^-

reduction:
o\_2 + 4e^- -->2o^2- no, it is smaller because if you rounded 0.09 to the nearest 0.1, you would get 0.1, and another reason you can say it is smaller is because if you put it in fractional terms, 0.09 would look like 9/100, but with 0.1, it would be 10/100, so from that, you can hopefully and clearly see that 0.09 is smaller that 0.1 solution: `1 + omega + omega^2 = 0`   

`z, z omega , z + z omega`   

`z , z omega , z(1 + omega)`   

`z , z omega , - z omega^2 `   

area `/\_ abc= 1/2 xx b xxh `   

`1/2 xx ab xx od`   

`1/2 xx 1 xx sqrt3/2 = sqrt3/4`   

so `1, omega , -omega^2`   

`z, z omega , - z omega^2 `   

area = `|z|^2 xx sqrt3/4 = 4/sqrt3`   

`|z|^2 = 16/3`   

`|z| = 4/ sqrt3`   

answer 65 ÷ ( – 13) = **-5** given (pq – prq) = (pq×1 – pq×r)taking pq common,(pq – prq) = pq(1 – r) answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a sr2+ → strontium cl- → chloride[readmore]therefore the chemical name will appear as strontium chloride in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a]