All Answers 358524 Answers

e^i*1.5708 ?

$$\boxed{
\mathrm{e}^{\mathrm{i}\,\varphi} = \cos\left(\varphi \right) + \mathrm{i}\,\sin\left( \varphi\right)
}$$

$$\varphi= \frac{\pi}{2}=1.57079632679$$

$$\mathrm{e}^{\mathrm{i}\,\cdot1.57079632679 }=
\mathrm{e}^{\mathrm{i}\,\cdot\frac{\pi}{2} } = \cos\left(\frac{\pi}{2}\right) + \mathrm{i}\,\cdot\sin\left( \frac{\pi}{2}\right)\\
= 0 + \mathrm{i}\,\cdot1\\
= \mathrm{i}\\$$

e^i*1.5708  = i

Find an angle x where sin x = cos x

$$\cos(x)-\sin(x)=0 \\\\
a\cdot \cos(x) + b\cdot \sin(x) = c \qquad | \qquad a=1 \qquad b=-1 \qquad c= 0\\
a\cdot \cos(x) + b\cdot \sin(x) = c \quad | \quad : a \\
\cos(x) + \frac{b}{a} \cdot \sin(x) = \frac{c}{a} \quad | \quad c = 0 \\\\
\cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \\
\small{\text{
$
\text{We set } \tan{(\varepsilon)} = \frac{b}{a} \quad \text{ we have } a = 1 \text{ and } b = -1 \text{ so } \varepsilon = \arctan{(-1)} = -\frac{\pi}{4}
$
}}\\\\
\cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \quad | \quad \tan{(\varepsilon)} = \frac{b}{a}\\\\
\cos(x) + \tan{(\varepsilon)}\cdot \sin(x) = 0 \\
\cos(x) + \frac{ \sin{(\varepsilon)} } { \cos{(\varepsilon)} } \cdot \sin(x) = 0 \quad | \quad \cdot \cos{(\varepsilon)} \\
\small{\text{
$
\cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x) = 0 \quad | \quad \cos{ (x-\varepsilon )} = \cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x)
$}}\\
\cos{ (x-\varepsilon )} =0 \quad | \quad \pm \arccos{}\\
x-\varepsilon = \pm \arccos{(0)} = \pm \frac{\pi}{2}\\
x-\varepsilon = \pm \frac{\pi}{2} \\
x= \varepsilon \pm \frac{\pi}{2} \quad | \quad \varepsilon = -\frac{\pi}{4}\\
x= -\frac{\pi}{4} \pm \frac{\pi}{2} \\\\
x_1= -\frac{\pi}{4} +\frac{\pi}{2} = \frac{\pi}{4} \\
x_1= 45\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$\\x_2= -\frac{\pi}{4} -\frac{\pi}{2}
= -\frac{3}{4} \cdot \pi
= -\frac{3}{4} \cdot \pi + 2\pi
= \frac{5}{4} \cdot\pi \\
x_2 = 225\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$k=0,1,2\cdots$$

Here is the graph.

I think Anonymous forgot to check his/her answer here

Taking it from this point, we have

x + ( x - 1)  = 7

2x - 1  = 7    add 1 to both sides

2x = 8     divide both sides by 2

x = 4

And using x + y = 7

4 + y = 7   ....  so.....y = 3

And the answer is  (x, y) = (4, 3)

The semi-annual answer given by Anonymous is correct

For the annual compounding, we have

1500(1.03)^4 = $ 1688.26

2+3=5

30/5 = 6  that is one share

2*6:3*6

12:18

This will occur at 45° ± n180°  where n is an integer

Compounded annually: $1680

Semi-annually $1689.74

A=P(1+rt)

Can you draw a triangle that is not a triangle?

$${\sqrt{{\mathtt{1\,235\,454\,686}}}} = {\mathtt{35\,149.035\: \!349\: \!494\: \!301\: \!591\: \!1}}$$

I would be interested in how you got 1.298

sinx = cos x @45 degrees

$${\sqrt{{\mathtt{50}}}} = {\mathtt{7.071\: \!067\: \!811\: \!865\: \!475\: \!2}}$$

$${\sqrt{{\mathtt{7}}}} = {\mathtt{2.645\: \!751\: \!311\: \!064\: \!590\: \!6}}$$

For example, when you flip a coin numerous times, it should land on heads 50% of the time. If it lands on heads all of the time, you know that it is not a fair coin.

However there is a very small possibility that it will land on heads for a large number of times. You cannot get this information from life because it is unpredictable.

$$\left({\frac{{\mathtt{1}}}{{\mathtt{11}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{11}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{121}}}} = {\mathtt{0.008\: \!264\: \!462\: \!809\: \!917\: \!4}}$$

Hum a few bars and I'll try and fake it.

1.52 Sin(x) = 1.00 Sin(60)

sin x=$$\frac{sin 60}{1.52}$$

sin x=0.569

x=arcsin 0.569

x=34.73

(-2x)²*(x²)

4x²*x²=4x⁴

A little like rattlesnake.

X=$${\mathtt{45}}{\mathtt{\,\times\,}}{\mathtt{96}} = {\mathtt{4\,320}}$$

The theoretical probability is that 50% of the time it will be heads.

Unless I am missing something in the picture-

Area A= Area B

$${\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}} = {\mathtt{6.708\: \!203\: \!932\: \!499\: \!369\: \!1}}$$