Question

Find the quotient. 42j4k2 ÷ (-3j3k)

Answer 1

\[\frac{42j^4k^2}{-3j^3k }\]

Isthàt what your problem is supposed to be?

Answer 2

14jk

Answer 3

@essjai's answer is incorrect

Answer 4

lol ^

Answer 5

-14jk if you want it to be right
because you have to first do the denominator because you are dividing you have to subtract the stuff.
j^4 - j^3 = j and k^2 - k = k and 42 / -3 = -14 so it becomes -14jk

Answer 6

@FriedRice while I appreciate the affirmation, it'd be better if expressed correctly :-)

\(j^4-j^3 \ne j\) for most values of \(j\) and
\(k^2-k \ne k\) for most values of \(k\)

However, the rules of exponents say that when dividing quantities with a common base \[\frac{x^n}{x^m}\]that we form the quotient by retaining the common base and using the difference of the exponents as the exponent:
\[\frac{x^n}{x^m} = x^{n-m}\]
\[\frac{j^4}{j^3} = j^{4-3} = j^1 = j\]similarly\[\frac{k^2}{k} = \frac{k^2}{k^1} = k^{2-1} = k^1 = k\]

Taking it all apart and putting it back together, \[42j^4k^3 \div (-j^3k^2) = \frac{42j^4k^3}{-3j^3k^2} = -\frac{42}{3}*\frac{j^4}{j^3}*\frac{k^3}{k^2}\]\[ = -14*j^{4-3}*k^{3-2} = -14*j^1*k^1 \]\[ = -14jk\]