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Seems basically correct, assuming the first "x" in your a x (x^

**k**) is supposed to be a multiplication symbol, not a variable. You could divide through by 3 to have .. log(y) = (**k**/3)*log(x) +log(a)/3 .. Y = log(y) .. m =**k**/3 .. X = log(x) ..**c**= log(a)/3...1 Answers · Science & Mathematics · 30/01/2018

(x + y)^n: Coefficient of term x^(n-k) y^

**k**=**C**(n, k) x^97: Coefficient of x^(100-3) (-1)^3 is C(100...3 Answers · Science & Mathematics · 12/01/2018

... AB = BA, then (A + B)^k = Σ(j = 0 to

**k**)**C**(k, j) A^j B^(k-j). For instance, look at the case when k = 2...1 Answers · Science & Mathematics · 21/12/2017

y' + x*y = 130x This is linear differential equation therefore the answer is y = y_h + y_p; solve first for y_h y' + x*y = 0 -> y' = -x*y -> y'/y = -x -> ln(y) =

**C**- ((x^2)/2) -> y = e^[**C**- ((x^2)/2)] -> y =**C***exp(-(x^2)/2); Now solve for y_h by...2 Answers · Science & Mathematics · 14/12/2017

If a = nc + k and b=nd + k (where

**k**,**c**, and d are integers), then a - b = (nc + k) - (nd + k) = n(c - d), which...1 Answers · Science & Mathematics · 26/11/2017

Binomial probability formula: P(X=

**k**) =**C**(n,k) * p^k * q^(n-k) n : number of trials (15) k : number...2 Answers · Science & Mathematics · 01/11/2017

...achievable There is equality in C-S when (x(k)−x(k−1))/√

**k**=**c**√k for k=1,2…9 and c is a constant ⟹ x(k)−x(k−1) = ck for k=1...3 Answers · Science & Mathematics · 31/10/2017

Use the binomial probability formula: P(X =

**k**) =**C**(n,k) * p^k * q^(n-k) n : number of trials (11) k : number...1 Answers · Science & Mathematics · 29/09/2017

Use the binomial probability formula: P(X =

**k**) =**C**(n,k) * p^k * q^(n-k) n : number of "trials" (6) k : ...3 Answers · Science & Mathematics · 04/09/2017

. let second number = x first number is twice second number: =========>first number = 2x third number is 15 less than the second number: =>third number = x - 15 if (x - 15...

5 Answers · Science & Mathematics · 17/08/2017