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3 years ago

Giải phương trình: Sin^4 x - cos^4 x =1/2

Mathematics

2 answers:

tamaranim1 [39]3 years ago

5 0

Hello,

$cos^4(x)=(cos^2(x))^2=(1-sin^2(x))^2=1+sin^4(x)-2*sin^2(x)\\\\sin^4(x)-(1+sin^4(x)-2sin^2(x))=\dfrac{1}{2} \\\\1-2sin^2(x)=\dfrac{1}{2} \\\\\\sin^2(x)=\dfrac{1}{4} \\\\sin(x)=\dfrac{1}{2} \ or\ sin(x)=-\dfrac{1}{2} \\\\\\modulo\ 2\pi: \\\\x=\dfrac{\pi}{6}\ or\ x=\dfrac{5*\pi}{6}\ or\ x=-\dfrac{\pi}{6}\ or\ x=-\dfrac{5*\pi}{6}$

Sliva [168]3 years ago

4 0

Step-by-step explanation:

...........MaRk As BrAiNeSt

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kvv77 [185]

Answer: He made $90 last week.

Assumption :

Let, Luis made $X last week.

He made $72 this week which is 80% of $X that he made last week

⇒ X * 80% = 72

⇒ X * 80/100 = 72

⇒ X = (72 * 100)/80

⇒ X = 90

Therefore, he made $90 last week.

To remember :

While solving this type of problems, be careful to understand which one of last time and current time is not given; just assume that one and input other conditions, the problem will be solved.

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3 years ago

The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

$V=\frac{1}{3}\pi r^2 h$

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

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Step-by-step explanation:

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