SOLUTION: Simplify \frac{(mn^3)*2*1/m*1/n^2}{(2^{-1} m^{-1})*(-3)*(mn)^2}.

SOLUTION: Simplify \frac{(mn^3)21/m1/n^2}{(2^{-1} m^{-1})(-3)*(mn)^2}.

Question 1209029: Simplify \frac{(mn^3)*2*1/m*1/n^2}{(2^{-1} m^{-1})*(-3)*(mn)^2}.
Answer by yurtman(42) (Show Source): You can put this solution on YOUR website!
Let's simplify the expression:
$$\frac{(mn^3)\cdot 2\cdot \frac{1}{m}\cdot \frac{1}{n^2}}{(2^{-1}m^{-1})\cdot (-3)\cdot (mn)^2}$$
We can simplify the expression by canceling terms that are in both the numerator and denominator, multiplying numbers, combining multiplied terms into a single fraction, and evaluating exponents.
Steps to solve:
**1. Cancel multiplied terms that are in the denominator:**
$$\frac{n^{3} \cdot 2 \cdot 1 \cdot \frac{1}{n^{2}}}{\left(2^{-1}m^{-1}\right)(-3)(mn)^{2}}$$
**2. Multiply the numbers:**
$$\frac{2n^{3} \cdot \frac{1}{n^{2}}}{\left(2^{-1}m^{-1}\right)(-3)(mn)^{2}}$$
**3. Combine multiplied terms into a single fraction:**
$$\frac{\frac{2n^{3} \cdot 1}{n^{2}}}{\left(2^{-1}m^{-1}\right)(-3)(mn)^{2}}$$
**4. Multiply the numbers:**
$$\frac{\frac{2n^{3}}{n^{2}}}{\left(2^{-1}m^{-1}\right)(-3)(mn)^{2}}$$
**5. Cancel terms that are in both the numerator and denominator:**
$$\frac{\frac{2n}{1}}{\left(2^{-1}m^{-1}\right)(-3)(mn)^{2}}$$
**6. Divide by 1:**
$$\frac{2n}{\left(2^{-1}m^{-1}\right)(-3)(mn)^{2}}$$
**7. Evaluate the exponent:**
$$\frac{2n}{\left(\frac{1}{2}m^{-1}\right)(-3)(mn)^{2}}$$
**8. Combine multiplied terms into a single fraction:**
$$\frac{2n}{\left(\frac{1m^{-1}}{2}\right)(-3)(mn)^{2}}$$
**9. Multiply by 1:**
$$\frac{2n}{\left(\frac{m^{-1}}{2}\right)(-3)(mn)^{2}}$$
**10. Distribute exponent:**
$$\frac{2n}{\frac{m^{-1}}{2} \cdot (-3)m^{2}n^{2}}$$
**11. Re-order terms so constants are on the left:**
$$\frac{2n}{-3 \cdot \frac{m^{-1}}{2} \cdot m^{2}n^{2}}$$
**12. Combine multiplied terms into a single fraction:**
$$\frac{2n}{\frac{-3m^{-1}m^{2}n^{2}}{2}}$$
**13. Combine exponents:**
$$\frac{2n}{\frac{-3m^{1}n^{2}}{2}}$$
**14. The first power of a number is the number itself:**
$$\frac{2n}{\frac{-3mn^{2}}{2}}$$
**15. Eliminate quotient in denominator:**
$$\frac{2n \cdot 2}{-3mn^{2}}$$
**16. Multiply the numbers:**
$$\frac{4n}{-3mn^{2}}$$
**17. Cancel terms that are in both the numerator and denominator:**
$$\frac{4}{-3nm}$$
**The final answer is:**
$$\frac{4}{-3mn}$$

RELATED QUESTIONS
please help to simplify by adding like terms : - 4mn^2/m^-1 + 2mn^2m^-1 -... (answered by tanimachatterjee)

If m and n are consecutive even integers and m > n > 0, how many integers are greater... (answered by Fombitz)

The number of all pairs (m,n) are positive integers such that 1/m + 1/n + 1/mn = 2/5... (answered by MathLover1,ikleyn)

If the random variable X has Hypergeometric distribution with parameters n, N and M,... (answered by textot)

Evaluate the expression for mn where m=2/5 and... (answered by rfer)

if tanx+sinx=m and tanx-sinx=n then prove that m^2-n^2=4(mn)^1/2 (answered by robertb)

If {{{a^m*a^n=a^mn}}},then m(n-2)+n(m-2)is?options are (a)-1,(b)1 (c)0... (answered by jim_thompson5910)

(((m^2-n^2�m^2-2mn+n^2)�(n^2-mn�m^2+mn))) (answered by solver91311)

Where did I go wrong? Please giveme the corrected answer. Given m=n Multiply... (answered by kusha)