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Just count the options

2 in, 2 out leaves only 2 books.  they are the only ones that need to be thought about at all.  Where can they be?

I actually figured it out. My phrasing is probably a bit off but it makes sense (to me at least) 

We end up with 2pi(x)T + pi(T)^2 - cos(x+T) = -cos(x) after some simplifying, and distributing/expanding.  

If we let x = 0 

Then we end up with pi(T)^2 - cos(T) = -1. This makes our proof impossible. It's a contradiction because when we let x = 0, the function is not periodic. 

UNLESS, if we let T = 0, then our proof is valid. We end up with -cos(x) = -cos(x) which then makes our function periodic. However, this isn't what we are looking for, as we aren't trying to find what makes the function periodic. 

The roots of \(x^2 + 5x + 3 = 0\) are p and q.
Find \(\dfrac1p + \dfrac1q\).

\(\text{Set $x=\dfrac{1}{x}$ in $x^2 + 5x + 3$ }\)

\(\begin{array}{|rcll|} \hline x^2 + 5x + 3 &=& 0 \\\\ \Rightarrow \left(\dfrac{1}{x}\right)^2 + 5*\dfrac{1}{x} + 3 &=& 0 \\\\ \dfrac{1}{x^2} + \dfrac{5}{x} + 3 &=& 0 \quad | \quad * x^2 \\\\ 1 + 5x + 3x^2 &=& 0 \\\\ 3x^2+5x+1 &=& 0 \quad | \quad : 3 \\\\ x^2+\dfrac{5}{3}x +\dfrac13 &=& 0 \\ \hline \text{By Vieta:}~\dfrac1p + \dfrac1q &=& -\dfrac{5}{3} \\\\ \text{By Vieta:}~\dfrac1p * \dfrac1q &=& \dfrac{1}{3} \\ \hline \end{array}\)

I'd be interested in seeing a formal solution for this.   

https://web2.0calc.com/questions/let-find-the-function-k-x-such-that-f-is-its-own-inverse_1

This will help you, it is extremely similar. The only difference is one of the terms in the second equation. other than that, you are set.

Thank you so much!

value = 23150 * ( 1 - .013)ⁿ

15000 = 23150(.87)ⁿ

15000/23150   = .87ⁿ

log ( 15000/23150) = n log (.87)        I think you can take it from here, no ? 

If your amount (4) is greater than 0, then you would use the equation -2x - 5.

So, for an official way to show your work: 
 

f(4) = -2x - 5

f(4) = -2(4) - 5

f(4) = -8 - 5

f(4) = -13 

4 doesn't equal -13, neither is -13 greater than 0, therefore this equation is inaccurate.

I believe that your answer has no solution due to it being a false equation. Therefore, the amount of solutions that your equation has is none. Feel free to correct me if I'm wrong though, since it has been a while since I have reviewed this..!

From a3    to   a4   is     ONE   r       a3 * r = a4

                                                         100 * r = -500

                                                               r = -5

a1 * r * r = a3

a1 * (-5)(-5) = 100

a1 = 4

answer c

Selling 2/3 means he has 1/3 left      1/3 * 60

Selling 1/2 means he is still left with 1/2       1/2 *    1/3 * 60

     = 1/6 * 60 = 10 left          or   he sold 50 toys .

Let's fix up this question first.

~~~~~~~~~~~~~~~~~~~~~~~~~

Arturo had some marbles. 1/4 of the marbles were blue, 3/12 of the marbles were red, and the rest were yellow. There were 14 more yellow marbles than red marbles. How many marbles did Arturo have altogether?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now we can solve this problem.

       1       :        1         :        2

1/4 blue      3/12 red        yellow

     x      +       x       =        x + 14 

x = 14

Arturo had 56 marbles. 

Thank you so much and thank you for explaining it! I could not even remember learning that. It seems so simple now. I'll remember that.

To find geometric mean, the first thing you need to do is multiply your original numbers together. 

108 x 3 = 324. Then what you do next is find the square root of 328, or the number gathered from what you multiplied. The square root of 324 is 18, therefore your answer and the geometric mean between 108 and 3 is A. Hope this helps. :)

Edit: ​For a more official way of showing your work:

x = √ab

x = √(108 x 3)

x = √324

x = 18

I think you might be correct but I'm not too sure. If we wanted to make the proof impossible, we need an expression such as a positive equals a negative and/or non matching sides. So I do believe that you can't go further for your answer besides simplying both sides if applicable. 

The answer key given for this particular question is a dead end for me (can't see the equations), so I have to rely on others so I can understand this. There can be different solutions to this, so any general proof would work otherwise. 

Thanks for the help though! I will wait until some other users answer this so I can get more feedback and compare. 

Your wrong

Answered!

BRUHHHH

Using veita formula, the sum of the roots is 7/3 and the product is -5/3

7/3 / -5/3 = 7/3 * - 3/5 = - 7/5

I got it right -_-

 6r^2 + 7r + 8 = 6 + 3r + 9r^2       rearrange  as

3r^2  -  4r   -  2   =  0       

Using  the Quadratic Formula.....the largest  value  of  r  =

4  +  sqrt ( 4^2  - 4 (3)(-2)  ]                    4  +  sqrt  [ 40 ]          4  + 2sqrt [10]              

_______________________  =            _____________  =   _____________  =       

                2 * 3                                                 6                                6   

2  +  sqrt [10]

_____________

          3                           

Note  that  27   =  9 *  3

10!  =   10 *  9  *    ...... * 3  *  2  * 1

So.....the  remainder =   0

(x + 4)(x + 5) =   n        simplify  as

x^2  +  9x  + 20 - n   =   0

This will  have   real solutions  if  the disxriminant  is  ≥  0.....so.......

9^2  -  4 (20 - n) ≥    0

81  - 80  +  4n  ≥  0

1  +  4n   ≥     0

4n  ≥  -1

n ≥ -1/4

In other words.......any   positive   n   will give real solutions....so....the probability  of non-real solutions =  0%

I'm going to try my best, but since I'm not too familiar with these questions, it might be wrong. 

I'm going to follow the hint, and assume that f(x) is a periodic function of period T and try to prove that this equation is impossible. 

f(x + T) = f(x)

pi*(x + T)^2 - cos(x + T) = pi*(x)^2 - cos(x)

pi*x^2 + 2*pi*xT + pi*T^2 - cos(x + T) = pi*(x)^2 - cos(x)

2*pi*xT + pi*T^2 - cos(x + T) = - cos(x)

2*pi*T(x + T) + cos(x) = cos(x + T)

2*pi*T(x + T) + cos(x) = cos(x) * cos(T) - sine(x) * sine(T)

And... now I'm stuck. :((

Maybe someone else can give help?

I'm really sorry that I wasn't able to help more. 

=^._.^=

If  the  center  lies  on the  x axis,  then  let the  center =  (x, 0)

And using the  square  of the distance formula.....the distance  rom the  center to both points is the  same.....so....

(x + 3)^2   + (0 - 2)^2   =   ( x +2)^2  +  (0 + 2)^2

x^2  + 6x  + 9  +  4   =  x^2   + 4x  + 4  +  4           rearrange  as

2x  +  5  =  0

2x  =  -5

x = -5/2

The  center is   (-5/2 , 0)

And the  radius =  sqrt   [ (-5/2 + 3)^2  +  4  ]   =  sqrt  (17/4)  =  sqrt (17)  / 2

Here's  a graph  :   https://www.desmos.com/calculator/sd3oslkkql