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Correct! 

How does that happen? If one of the legs is 6, then the hypotenuse + the other leg is also equal to six, which is against the triangle inequality.

Answer: $$1.17$

Solution:

Case 1: Monica rolls a 1.

If Monica rolls a 1, then she loses 3 dollars.

Case 2: Monica rolls a 2.

If Monica rolls a 2, then she gains 2 dollars.

Case 3: Monica rolls a 3.

If Monica rolls a 3, then she gains 3 dollars.

Case 4: Monica rolls a 4.

If Monica rolls a 4, then she gains nothing.

Case 5: Monica rolls a 5.

If Monica rolls a 5, then she gains 5 dollars.

Case 6: Monica rolls a 6.

If Monica rolls a 6, then she gains nothing.

Adding all them up and then placing it over six (there are six possibillities) gives

$-3+2+3+0+5+0\over6$

Which is equal to $\frac76$, which rounded to the nearest cent, gives $$1.17$.

There are this many ways:

       a     b    c

1  = 180 60 9
2  = 126 70 9
3  = 120 72 9
4  = 90 90 9
5  = 180 36 10
6  = 130 39 10
7  = 120 40 10
8  = 105 42 10
9  = 90 45 10
10  = 80 48 10
11  = 75 50 10
12  = 66 55 10
13  = 60 60 10
14  = 110 30 11
15  = 120 24 12
16  = 100 25 12
17  = 70 28 12
18  = 60 30 12
19  = 45 36 12
20  = 40 40 12
21  = 156 20 13
22  = 84 20 14
23  = 70 21 14
24  = 35 30 14
25  = 90 18 15
26  = 60 20 15
27  = 40 24 15
28  = 30 30 15
29  = 120 16 16
30  = 48 20 16
31  = 45 18 18
32  = 36 20 18
33  = 30 20 20
34  = 28 21 20
35  = 24 24 20

How many real numbers are not in the domain of the function f(x)?

Hello Guest!

There are no whole numbers (integer) in the domain of the function f(x).

  !

9 + 14 = 23

∠H = 50º

∠K = 70º

∠L = 60

How about x = 1    ?   

   f(1) = 0

Your question isn't complete ;-;

given a, b such that 0

x = 32 / 4 

Hemisphere radius = circle diameter = square diagonal = square area = 2 

[ADE] / [ABCD] = 1/6 

The circumference of the cylinder's base is 6 units.

Radius = 1/2(6/pi)

Volume = r²pi*h

h = (27/2) / r²pi

sin(∠BAD) = h / 6

$\frac{c}{a} = x_1 x_2$

$2c = (-99 + \sqrt{8021})(-99-\sqrt{8021}) = 99^2 - 8021 = 9801 - 8021 = 1780$

$c = 890$

$\begin{align*} \frac{1}{1+\frac{1}{\sqrt{3}+2}} &= \frac{1}{\frac{3+\sqrt{3}}{\sqrt{3}+2}} \\ &= \frac{\sqrt{3}+2}{3+\sqrt{3}} \\ &= \boxed{\frac{3+\sqrt{3}}{6}} \end{align*}$

Solution:

$f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$

$\begin{align*} f^2(x) &= \frac{\frac{x-\sqrt{3}}{x\sqrt{3}+1}-\sqrt{3}}{\sqrt{3}\left(\frac{x-\sqrt{3}}{x\sqrt{3}+1}\right)+1} \\ &= \frac{\frac{x-\sqrt{3}}{\sqrt{3}x+1}-\sqrt{3}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}+1} \\ &= \frac{\frac{-2x-2\sqrt{3}}{\sqrt{3}x+1}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}+1} \\ &= \frac{-2x-2\sqrt{3}}{\left(x\sqrt{3}+1\right)\left(\frac{\left(x-\sqrt{3}\right)\sqrt{3}}{x\sqrt{3}+1}+1\right)} \\ &= \frac{-2x-2\sqrt{3}}{\frac{2\sqrt{3}x-2}{\sqrt{3}x+1}\left(\sqrt{3}x+1\right)} \\ &= \frac{-2x-2\sqrt{3}}{2\sqrt{3}x-2} \\ &= -\frac{2\left(x+\sqrt{3}\right)}{2\left(\sqrt{3}x-1\right)} \\ &= -\frac{x+\sqrt{3}}{\sqrt{3}x-1} \end{align*}$

$\begin{align*} f^3(x) &= f \left(-\frac{x+\sqrt{3}}{\sqrt{3}x-1} \right) \\ &= -\frac{\frac{x-\sqrt{3}}{x\sqrt{3}+1}+\sqrt{3}}{\sqrt{3}\left(\frac{x-\sqrt{3}}{x\sqrt{3}+1}\right)-1} \\ &= \frac{\frac{x-\sqrt{3}}{\sqrt{3}x+1}+\sqrt{3}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}-1} \\ &= -\frac{\frac{4x}{\sqrt{3}x+1}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}-1} \\ &= -\frac{4x}{\left(\sqrt{3}x+1\right)\left(\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}-1\right)} \\ &= -\frac{4x}{\left(-\frac{4}{\sqrt{3}x+1}\right)\left(\sqrt{3}x+1\right)} \\ &= \frac{4x}{\left(x\sqrt{3}+1\right)\frac{4}{x\sqrt{3}+1}} \\ &= \frac{4x}{4} \\ &= x \end{align*}$

$\begin{align*} f^4(x) &= f(f^3(x)) \\ &= f(x) \\ &= \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \end{align*}$

$f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$
$f^2(x) = -\frac{x+\sqrt{3}}{\sqrt{3}x-1}$
$f^3(x) = x$
$f^4(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$

$2021 \pmod 3 \equiv 2 \pmod 3$

$f^{2021}(x) = f^2(x) = \boxed{-\frac{x+\sqrt{3}}{\sqrt{3}x-1}}$

If you want to repost you need to put a link to the original question.

Request that people answer on the original. 

Then ask them to leave a little note on the repost saying that they have done so.

ok thank you  :)

I will have to think about it a bit.

To be honest,  I am not really sure what this question means ..but

$\frac{e^{ix}}{e^{iy}}\\ =\frac{cos(x)+isin(x)}{cos(y)+isin(y)}\\ =\frac{cos(x)+isin(x)}{cos(y)+isin(y)}*\frac{cos(y)-isin(y)}{cos(y)-isin(y)}\\~\\ =\frac{cos(x)cos(y)-icos(x)sin(y)+isin(x)cos(y) +sin(x)sin(y) }{cos^2(y)+sin^2(y)}\\~\\ =\frac{cos(x)cos(y) +sin(x)sin(y) +i[-cos(x)sin(y)+sin(x)cos(y) ] }{1}\\~\\ =cos(x-y)+isin(x-y)$

ALSO

$\frac{e^{ix}}{e^{iy}}\\ =e^{ (ix-iy) }\\ =e^{ i(x-y) }\\ =cos(x-y)+isin(x-y)$

I must be almost finished but i don't get what the question actually wants.   

LaTex

\frac{e^{ix}}{e^{iy}}\\
=\frac{cos(x)+isin(x)}{cos(y)+isin(y)}\\

=\frac{cos(x)+isin(x)}{cos(y)+isin(y)}*\frac{cos(y)-isin(y)}{cos(y)-isin(y)}\\~\\

=\frac{cos(x)cos(y)-icos(x)sin(y)+isin(x)cos(y)    +sin(x)sin(y) }{cos^2(y)+sin^2(y)}\\~\\

=\frac{cos(x)cos(y)  +sin(x)sin(y)  +i[-cos(x)sin(y)+sin(x)cos(y) ]    }{1}\\~\\

=cos(x-y)+sin(x-y)

\frac{e^{ix}}{e^{iy}}\\
=e^{ (ix-iy) }\\
=e^{ i(x-y)   }\\
=cos(x-y)+isin(x-y)

How did you get your answer? I don't understand.

Sorry, haven't learned this formula.

(12+15)/2*6 = 27/2*6 = 27*3 = 81.

There are this many of them:

1357 , 1375 , 1537 , 1573 , 1735 , 1753 , 3157 , 3175 , 3517 , 3571 , 3715 , 3751 , 5137 , 5173 , 5317 , 5371 , 5713 , 5731 , 7135 , 7153 , 7315 , 7351 , 7513 , 7531 , Total =  24 such numbers