What is the answer to solvefor j,j(n)=(9n^4+3n)(2n-4) ?

The answer to solvefor j,j(n)=(9n^4+3n)(2n-4) is j=6(3n^3+1)(n-2);\quad n\ne 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor j,j(n)=(9n^4+3n)(2n-4)

solve for

j, j (n) = (9 n^{4} + 3 n) (2 n - 4)

j = 6 (3 n^{3} + 1) (n - 2); n \neq = 0

Solution steps

j (n) = (9 n^{4} + 3 n) (2 n - 4)

Divide both sides by

n; n \neq = 0

j = 6 (3 n^{3} + 1) (n - 2); n \neq = 0

\frac{x - a}{b + c} + \frac{x - b}{a + c} + \frac{x - c}{a + b} = 3

11 = x - 6

p, r = q + p + s

(\frac{x}{4}) + 3 = (\frac{2 x}{6}) - (\frac{1}{16})

ka (- 3) = a

What is the answer to solvefor j,j(n)=(9n^4+3n)(2n-4) ?
The answer to solvefor j,j(n)=(9n^4+3n)(2n-4) is j=6(3n^3+1)(n-2);\quad n\ne 0