Question

simplify. assume that no variable equals 0.\\(\frac{-20(m^{2}v)(-v)^{3}}{5(-v)^{2}(-m^{4})}\\)\\(\bigcirc \frac{4v^{2}}{m^{2}}\\)\\(\bigcirc -4m^{2}v^{2}\\)\\(\bigcirc -\frac{4v}{m}\\)\\(\bigcirc -\frac{4v^{2}}{m^{2}}\\)\
question 11\
1 pts\
simplify. assume that no variable equals 0.\\((3n^{2} + 1) + (8n^{2} - 8)\\)\\(\bigcirc 4n^{2}\\)\\(\bigcirc 11n - 7\\)\\(\bigcirc 11n^{2} - 7\\)\\(\bigcirc 11n^{2} + 7\\)

Explanation:

First Problem (Simplifying the Rational Expression)

Step1: Simplify the coefficients and signs

First, handle the coefficients: \(\frac{-20}{5} = -4\). Then, handle the signs of the variables. Recall that \((-v)^3 = -v^3\) and \((-v)^2 = v^2\), and \((-m^4)= -m^4\). So the expression becomes:
\[
\frac{-20(m^2v)(-v^3)}{5(v^2)(-m^4)}=\frac{(-20)\times(-1)m^2v^4}{5\times(-1)m^4v^2}
\]
Simplify the signs: \((-20)\times(-1) = 20\) and \(20\div(-5)= -4\). So now we have \(\frac{-4m^2v^4}{m^4v^2}\)

Step2: Simplify the variables using exponent rules

For the \(m\) terms: use the rule \(\frac{a^m}{a^n}=a^{m - n}\), so \(\frac{m^2}{m^4}=m^{2-4}=m^{-2}=\frac{1}{m^2}\) (since \(a^{-n}=\frac{1}{a^n}\)). For the \(v\) terms: \(\frac{v^4}{v^2}=v^{4 - 2}=v^2\). Multiply these with the coefficient \(-4\):
\[
-4\times\frac{1}{m^2}\times v^2=-\frac{4v^2}{m^2}
\]
Wait, wait, let's re - check the sign calculation. Let's do the sign part again. The original numerator: \(-20\times(m^2v)\times(-v)^3=-20m^2v\times(-v^3)=20m^2v^4\) (because \((-v)^3=-v^3\), so \(-20\times(-v^3)=20v^3\), then times \(m^2v\) gives \(20m^2v^4\)). The denominator: \(5\times(-v)^2\times(-m^4)=5\times v^2\times(-m^4)= - 5m^4v^2\). So the fraction is \(\frac{20m^2v^4}{-5m^4v^2}\). Now, \(\frac{20}{-5}=-4\), \(\frac{m^2}{m^4}=m^{2 - 4}=m^{-2}=\frac{1}{m^2}\), \(\frac{v^4}{v^2}=v^{4-2}=v^2\). So multiplying these together: \(-4\times\frac{1}{m^2}\times v^2 =-\frac{4v^2}{m^2}\)

Wait, but let's check the options. The fourth option is \(-\frac{4v^2}{m^2}\). Wait, maybe I made a mistake in the first sign calculation. Let's start over.

Original expression: \(\frac{-20(m^2v)(-v)^3}{5(-v)^2(-m^4)}\)

First, expand the powers:

\((-v)^3=-v^3\), \((-v)^2 = v^2\), \((-m^4)=-m^4\)

So numerator: \(-20\times m^2v\times(-v^3)=(-20)\times(-1)\times m^2\times v\times v^3 = 20m^2v^{1 + 3}=20m^2v^4\)

Denominator: \(5\times v^2\times(-m^4)=-5m^4v^2\)

Now, divide numerator by denominator: \(\frac{20m^2v^4}{-5m^4v^2}=\frac{20}{-5}\times\frac{m^2}{m^4}\times\frac{v^4}{v^2}\)

\(\frac{20}{-5}=-4\), \(\frac{m^2}{m^4}=m^{2-4}=m^{-2}=\frac{1}{m^2}\), \(\frac{v^4}{v^2}=v^{4 - 2}=v^2\)

So \(-4\times\frac{1}{m^2}\times v^2=-\frac{4v^2}{m^2}\)

Step1: Remove the parentheses

Since we are adding two polynomials \((3n^2 + 1)+(8n^2-8)\), the parentheses can be removed directly because we are adding (there is no negative sign in front of the second parentheses to distribute). So we get \(3n^2+1 + 8n^2-8\)

Step2: Combine like terms

For the \(n^2\) terms: \(3n^2+8n^2=(3 + 8)n^2 = 11n^2\). For the constant terms: \(1-8=-7\). So the simplified form is \(11n^2-7\)

Answer:

\(\boldsymbol{-\frac{4v^2}{m^2}}\) (which corresponds to the fourth option: \(-\frac{4v^2}{m^2}\))

Second Problem (Simplifying the Polynomial)