# k-4 divided by 9=3 help me?

k-4

----- = 3

9

Relevance

k-4/9=3

you want to isolate k to find the answer, so you have to do the operations in reverse. so

k-4=3x9

k-4=27

k=27+4

k=31

• 3 years ago

hi, in no way ever use the "." character for multiplication between 2 digits: what's "3.4" ? 3.4 = 34/10 = 17/5 ? 3.4 = 3×4 = 12 ? This ambiguity can, via its presence on my own, detract an answerer from wishing to assist you. because of the fact human beings do no longer rather choose to situation explaining the paradox in such fairly some words... So consistent with hazard could you care to rephrase your expression as to make it unambiguous? and in no way overlook the formulation that sparkling up a great deal of problems in case you surely situation to apply them: "hi", "Please" and "thank you". = = = = = = = = = = = = = = = = = = = = = = = = = f(n) = 10? + 3×4??² + 5 permit's calculate some values: f(0) = a million + 3×4² + 5 = fifty 4 f(a million) = 10 + 3×4³ + 5 = 207 f(2) = one hundred + 3×4² + 5 = 873 So the biggest integer p that be a divisor of all f(n), could divide f(0)=fifty 4. fifty 4 = a million×2×3³ All divisors of fifty 4 are: a million, 2, 3, 6, 9, 18, 27 and fifty 4. permit's examine them from appropriate to smallest = = = = = = = = = = = = = = = = = = = = = = = = = Even DIVISORS: = = = = = = = = = = = = = = = = = = = = = = = = = f(a million) = 207 that's even. So all even divisors won't be able to divide f(a million). = = = = = = = = = = = = = = = = = = = = = = = = = DIVISOR = 27: = = = = = = = = = = = = = = = = = = = = = = = = = f(a million) = 207 which isn't divisible via 27 So 27 isn't the terrific divisors of all f(n) = = = = = = = = = = = = = = = = = = = = = = = = = DIVISOR = 9: = = = = = = = = = = = = = = = = = = = = = = = = = 10? = a million = a million 10¹ = 10 = a million + 9 = a million + 9×a million 10² = one hundred = a million + ninety 9 = a million + 9×11 10³ = 1000 = a million + 999 = a million + 9×111 ... 10? = a million + 9×1111...a million ? n cases digit "a million" Or greater precise shown: 10? = (a million + 9)?    = 1? + ?(n-a million;n)×9×1??¹ + ?(n-2;n)×9²×1??² + ... + 9?         (the place the ? are the binomial coefficients: ?(p;n)=n!/[p!(n-p)!]  ) So 10? can continuously be expressed as: 10? = a million + 9×p(n) with p(n) being an integer Likewise: 4² = a million + 15 = a million + 3×5 4³ = a million + sixty 3 = a million + 3×21 ... The binomial data additionally works because of the fact 4=a million+3... And 4??² is often expressed as 4??² = a million + 3×q(n+2) with q(n+2) being an integer as a effect: f(n) = 10? + 3×4??² + 5   = [a million + 9×p(n)] + 3×[a million + 3×q(n+2)] + 5   = 9×p(n) + 9×q(n+2) + 9   = 9[p(n) + q(n+2) + a million] as a effect f(n) is divisible via 9 for any integer n. And 9 is as a effect the needed appropriate divisor of all f(n) for any integer n. Regards, Dragon.Jade :-)