Mathxx
• Member for 6 years, 10 months
• Last seen this week
• Malaysia

I figured it out. $1^2+(\tan \alpha)^2=(\sec \alpha)^2$ So $\left |z \right |=\sec \alpha$ Is it right?

$$\frac{x^2}{4}+\frac{(y-3)^2}{9}=1$$ Let $Y=y-3$ So, \begin{align*} \frac{x^2}{4}+\frac{Y^2}{9} & = 1\\ \frac{x^2}{4} & =1-\frac{Y^2}{9}\\ \frac{x^2}{4} & =\frac{1}{9}(9-Y^2)\\ x^2 &...

By doing long division, you'll get $$-\int \left(-\frac{1}{2(x^2+1)}-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+1\right) dx$$

Found a way to solve from another site. $$S=\frac{1}{8}\left(3\sum_{k=1}^{44}\left(\cos\left(4k\frac{\pi}{180}\right)\right)+221\right)$$ $$\sum_{k=1}^{44} \left( \cos \left(4k \frac{ \pi}{180} \... View answer 1 votes As @AlexR said, P(\text{dog} \cap \neg \text{cat}) = P(\text{dog}) - P(\text{dog} \cap \text{cat}) So you'll get 0.1 for P(Dog and not cat). P(\text{Not cat} \mid \text{Dog})=\frac{0.1}{0.25}=0.4... View answer 1 votes \bar{y}=\frac{\sum y}{n}=\frac{0}{11}=0 \sigma y=\sqrt{\frac{1.2}{11}-0^2} y=x-1506.8 So, \bar{y}=\bar{x}-1506.8 0=\bar{x}-1506.8 \bar{x}=1506.8 \sigma y=\sigma x \sigma x=\sqrt{\... View answer 1 votes I found an alternative way.$$\cos 3x=\cos (2x+x)=\cos (2x)\cos (x)-\sin (2x)\sin (x)=(2\cos ^2x-1)\cos x-(2\sin x\cos x)\sin x=2\cos ^3x-\cos x-2\sin ^2x\cos x=2\cos ^3x-\cos x-2(1-\...

The proposed solution by the author. The common difference of the given sequence is $7$, and there are $21$ terms in the sequence, since $\frac{177-37}{7}=20$. Let $n$ be a sum of five terms in the ...

If you're saying the most basic algebra. Then I would suggest this. Introduction to Algebra

https://artofproblemsolving.com/wiki/index.php?title=Math_books Try this links. Books which specific in every level according to the level.

By using the chain rule. $\frac{d}{dx}\sqrt{\left | x \right |}$ $=\frac{\frac{d}{dx}(\left |x \right |)}{2\sqrt{\left | x \right |}}$ $=\frac{x}{\left |x \right |}\frac{1}{2\sqrt{\left | x \right ... View answer 0 votes I've found the answer by using Law of Cosines.$BC^2=8^2+8^2-2(8)(8)\cos 120=192BC=8\sqrt{3}AB+BC+AC=24+8\sqrt{3}$View answer 0 votes$\frac{x}{x^3-1}=\frac{x}{(x-1)(x^2+x+1)}\frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}x=A(x^2+x+1)+(Bx+C)(x-1)$Let$x=11=A(1^2+1+1)1=3AA=\frac{1}{3}$So,$x=\frac{...
$$f'(x)=\lim_{{h}\to{0}}\frac{(x+h)^{x+h}-x^x}{h}$$ $$=\lim_{{h}\to{0}}\frac{(x+h)^{x+h}-(x+h)^x+(x+h)^x-x^x}{h}$$ $$=\lim_{{h}\to{0}}\frac{(x+h)^x\left[(x+h)^h-1\right]}{h}+\lim_{{h}\to{0}}\frac{(x+... View answer 0 votes$$I = \int \ln(\ln \sqrt{x})^{\ln x} dx = \int \ln \left(\ln \left(x^{1/2}\right) \right)^{\ln x} dx = \int \ln \left(\frac{1}{2}\ln x \right)^{\ln x} dx= \int (\ln x) \cdot \ln \left(\...