4 years ago

Find the area of the rectangle polynomial

Mathematics

1 answer:

Andru [333]4 years ago

8 0

hello,

$Area =(2x + 4) \times ( 4x - 2) \\ Area =8 {x}^{2} - 4x + 16x - 8 \\ Area = {8x}^{2} + 12x - 8$

$\boxed{crsjr}$

You might be interested in

You start a new business. The location costs $100,000 to rent for one year. The items cost you $12 to make. You sell the items f

worty [1.4K]

Answer:my stuff is gliched

Step-by-step explanation:

4 0

3 years ago

lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.

zaharov [31]

Answer with Step-by-step explanation:

Given

$f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1$