what is the product of $(3y^{-4})(2y^{-4})$?\n$\frac{6}{y^{8}}$\n$\frac{1}{6y^{8}}$\n$\frac{6}{y^{16}}$\n$\fr…
what is the product of $(3y^{-4})(2y^{-4})$?\n$\frac{6}{y^{8}}$\n$\frac{1}{6y^{8}}$\n$\frac{6}{y^{16}}$\n$\frac{1}{6y^{16}}$
Answer
Explanation:
Step1: Multiply the coefficients
$3\times2 = 6$
Step2: Multiply the variables using exponent rule
When multiplying variables with the same base ($y$ in this case), we add the exponents. So $y^{-4}\times y^{-4}=y^{-4+( - 4)}=y^{-8}=\frac{1}{y^{8}}$
Step3: Combine the results
The product is $6\times\frac{1}{y^{8}}=\frac{6}{y^{8}}$
Answer:
A. $\frac{6}{y^{8}}$