Find the product
-$\left(3 t-\frac{11}{10}\right)\left(3 t+\frac{11}{10}\right)$

A) $9 t^{2}-\frac{121}{100}$
B) $9 t^{2}-\frac{33}{5} t-\frac{121}{100}$
C) $9 t^{2}+\frac{33}{5} t$
D) $9 t^{2}+\frac{121}{100}$

Question

Find the product
-$\left(3 t-\frac{11}{10}\right)\left(3 t+\frac{11}{10}\right)$

A) $9 t^{2}-\frac{121}{100}$
B) $9 t^{2}-\frac{33}{5} t-\frac{121}{100}$
C) $9 t^{2}+\frac{33}{5} t$
D) $9 t^{2}+\frac{121}{100}$

Quizplus · Accepted Answer

This is an application of the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$, where $a = 3t$ and $b = \frac{11}{10}$. The product is $9t^2 - \left(\frac{11}{10}\right)^2 = 9t^2 - \frac{121}{100}$.

Find the Product - (3t−1110)(3t+1110)\left(3 t-\frac{11}{10}\right)\left(3 t+\frac{11}{10}\right)(3t−1011​)(3t+1011​) A) 9t2−1211009 t^{2}-\frac{121}{100}9t2−100121​ B) 9t2−335t−1211009 t^{2}-\frac{33}{5} t-\frac{121}{100}9t2−533​t−100121​

Find the Product
- $\left(3 t-\frac{11}{10}\right)\left(3 t+\frac{11}{10}\right)$ A) $9 t^{2}-\frac{121}{100}$ B) $9 t^{2}-\frac{33}{5} t-\frac{121}{100}$