Help!!!!!!!!! i need this asap!

Mathematics

1 answer:

Nutka1998 [239]3 years ago

4 0

$A=\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]\\\\B=\left[\begin{array}{ccc}2&8\\0.6&3\end{array}\right]\\\\A\cdot B=\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]\cdot\left[\begin{array}{ccc}2&8\\0.6&3\end{array}\right]=\left[\begin{array}{ccc}(3)(2)+(0)(0.6)&(3)(8)+(0)(3)\\(2)(2)+(-1)(0.6)&(2)(8)+(-1)(3)\end{array}\right]\\\\=\left[\begin{array}{ccc}6&24\\3.4&13\end{array}\right] \to\boxed{A.}$

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Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$