Question

simplify the expression \\(\frac{-15x^{-4}y^{-9}}{3x^{-10}y^{-2}}\\) and choose the correct option from the given choices (the options are shown with radio buttons and include expressions like \\(\frac{x^6}{5y^7}\\), \\(\frac{-5y^7}{x^6}\\), \\(\frac{-5x^6}{y^7}\\), \\(\frac{y^7}{5x^5}\\))

Explanation:

Step1: Simplify the coefficient

Divide the coefficient of the numerator (-15) by the coefficient of the denominator (3): $\frac{-15}{3} = -5$.

Step2: Simplify the \(x\)-terms

Use the rule of exponents for division \(a^m\div a^n=a^{m - n}\). For \(x\)-terms: \(x^{-4}\div x^{-10}=x^{-4-(-10)} = x^{6}\).

Step3: Simplify the \(y\)-terms

For \(y\)-terms: \(y^{-9}\div y^{-2}=y^{-9-(-2)} = y^{-7}=\frac{1}{y^{7}}\) (but we can also use the rule \(a^m\div a^n=a^{m - n}\) directly in the fraction). Combining with the previous steps, we have \(-5\times x^{6}\times\frac{1}{y^{7}}=\frac{-5x^{6}}{y^{7}}\) (Wait, let's re - check the exponent calculation for \(y\): The exponent of \(y\) in the numerator is \(-9\), in the denominator is \(-2\). So \(y^{-9}\div y^{-2}=y^{-9 + 2}=y^{-7}\)? No, wait, the rule is \(a^m\div a^n=a^{m - n}\), so \(y^{-9}\div y^{-2}=y^{-9-(-2)}=y^{-9 + 2}=y^{-7}\)? Wait, no, \(-9-(-2)=-9 + 2=-7\), but when we move a negative exponent from numerator to denominator or vice - versa, \(y^{-7}=\frac{1}{y^{7}}\), but let's do it again. Wait, the original expression is \(\frac{-15x^{-4}y^{-9}}{3x^{-10}y^{-2}}\). So for \(y\): \(\frac{y^{-9}}{y^{-2}}=y^{-9-(-2)}=y^{-7}\), and for \(x\): \(\frac{x^{-4}}{x^{-10}}=x^{-4 + 10}=x^{6}\). The coefficient: \(\frac{-15}{3}=-5\). So putting it together: \(-5\times x^{6}\times y^{-7}\). But \(y^{-7}=\frac{1}{y^{7}}\), so \(-5x^{6}\times\frac{1}{y^{7}}=\frac{-5x^{6}}{y^{7}}\)? Wait, no, wait I think I made a mistake in the sign of the exponent for \(y\). Wait, the formula is \(a^m\div a^n=a^{m - n}\). So \(y^{-9}\div y^{-2}=y^{-9-(-2)}=y^{-9 + 2}=y^{-7}\)? No, \(-9-(-2)=-9 + 2=-7\), yes. But if we rewrite the fraction as \(\frac{-15}{3}\times x^{-4-(-10)}\times y^{-9-(-2)}\), which is \(-5\times x^{6}\times y^{-7}\). But \(y^{-7}=\frac{1}{y^{7}}\), so \(-5x^{6}\times\frac{1}{y^{7}}=\frac{-5x^{6}}{y^{7}}\). Wait, but let's check the options. One of the options is \(\frac{-5x^{6}}{y^{7}}\) (the third option from the left? Wait, the options are: first (from left, after Clear All) \(\frac{y^{7}}{5x^{5}}\), second \(\frac{-5x^{6}}{y^{7}}\), third \(\frac{-5y^{7}}{x^{6}}\), fourth \(\frac{x^{6}}{5y^{7}}\). Wait, my previous calculation for \(y\) was wrong. Let's recalculate the exponent of \(y\): The exponent of \(y\) in the numerator is \(-9\), in the denominator is \(-2\). So \(y^{-9}\div y^{-2}=y^{-9-(-2)}=y^{-9 + 2}=y^{-7}\)? No, that's not right. Wait, the rule is when you divide exponents with the same base, you subtract the exponents. So \(a^m\div a^n=a^{m - n}\). So \(y^{-9}\div y^{-2}=y^{-9-(-2)}=y^{-9 + 2}=y^{-7}\)? But \(y^{-7}=\frac{1}{y^{7}}\), but if we have \(y^{-9}\) in the numerator and \(y^{-2}\) in the denominator, it's equivalent to \(y^{-9}\times y^{2}=y^{-7}\) (because dividing by \(y^{-2}\) is multiplying by \(y^{2}\)). Wait, no, dividing by \(y^{-2}\) is multiplying by \(y^{2}\), so \(y^{-9}\times y^{2}=y^{-7}\). But maybe I mixed up the direction. Wait, let's start over. The expression is \(\frac{-15x^{-4}y^{-9}}{3x^{-10}y^{-2}}\).

First, coefficient: \(\frac{-15}{3}=-5\).

For \(x\): \(x^{-4}\div x^{-10}=x^{-4-(-10)}=x^{6}\) (because \(-4 + 10 = 6\)).

For \(y\): \(y^{-9}\div y^{-2}=y^{-9-(-2)}=y^{-7}\)? No, wait, \(-9-(-2)=-9 + 2=-7\), so \(y^{-7}\). But \(y^{-7}=\frac{1}{y^{7}}\), so the term with \(y\) is \(\frac{1}{y^{7}}\).

So putting it all together: \(-5\times x^{6}\times\frac{1}{y^{7}}=\frac{-5x^{6}}{y^{7}}\), which is the second option (the one with \(\frac{-5x^{6}}{y^{7}}\)). Wait, no, the second option (from left, after Clear All) is \(\frac{-5x^{6}}{y^{7}}\)? Wait, the options are:

1. \(\frac{y^{7}}{5x^{5}}\)

2. \(\frac{-5x^{6}}{y^{7}}\)

3. \(\frac{-5y^{7}}{x^{6}}\)

4. \(\frac{x^{6}}{5y^{7}}\)

Wait, my mistake in the exponent of \(x\): \(x^{-4}\div x^{-10}=x^{-4-(-10)}=x^{6}\), correct. For \(y\): \(y^{-9}\div y^{-2}=y^{-9-(-2)}=y^{-7}\), but \(y^{-7}=\frac{1}{y^{7}}\), so the fraction is \(-5\times x^{6}\times\frac{1}{y^{7}}=\frac{-5x^{6}}{y^{7}}\), which is the second option. Wait, but let's check again. Wait, maybe I had the exponent of \(y\) reversed. Let's use the rule for negative exponents: \(a^{-n}=\frac{1}{a^{n}}\) and \(\frac{1}{a^{-n}}=a^{n}\). So \(\frac{y^{-9}}{y^{-2}}=y^{-9}\times y^{2}=y^{-7}\) (since \(\frac{1}{y^{-2}} = y^{2}\)). So that's correct. And \(\frac{x^{-4}}{x^{-10}}=x^{-4}\times x^{10}=x^{6}\) (since \(\frac{1}{x^{-10}}=x^{10}\)). And \(\frac{-15}{3}=-5\). So multiplying them together: \(-5\times x^{6}\times y^{-7}=\frac{-5x^{6}}{y^{7}}\), which is the second option.

Answer:

\(\boldsymbol{\frac{-5x^{6}}{y^{7}}}\) (the second option from the left, with the label \(\frac{-5x^{6}}{y^{7}}\))