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a +  1/b    =     (ba  + 1) / b

b + 1/a    =     (ba + 1)  / a

By Vieta

The  sum  of  the roots =  p   =      a(ba + 1)  + b (ba + 1)           (ab + 1) (a + b)

                                                       __________________  =     ____________

                                                                    ab                                    ab

The product of  the roots    = q  =    (ab + 1)^2  / ab

                       (ab + 1) ( a + b)             a + b

 p/q  =           _______________  =  ________     

                        (ab + 1)^2                   ab  +  1

14  *  5   =   $70

See  here  :    https://web2.0calc.com/questions/geometry-xi

Here is  a similar problem ....only PQ  is  a little  different    :

https://web2.0calc.com/questions/geometryyyyy-4

I wish I was as smart as you 🙁

Bisect CD  at  M

MD  =   8

Draw   OM  =  BA   =   12

We  have  right triangle  DOM   where  OD  =  the  radius of the circle   and  hypotenuse of DOM

Using  the  Pythagorean Theorem

OD   =sqrt  ( OM^2  + MD^2 )  =  sqrt  ( 12^2  +  8^2)  =  sqrt ( 144  + 64)  = sqrt  (208)   = 4sqrt (13)  = the  radius

Note  that  we  can  write  this  as  :

               1

x   =    _______                multiply  both  sides  by     3 - x

              3 -  x 

x ( 3 - x)   =  1               simplify

3x - x^2   =   1              rearrange  as

x^2  - 3x  +  1    =      0        complete  the  square on  x

x^2  - 3x  +  9/4    =    -1   + 9/4

( x - 3/2)^2   =    5/4          take  both roots

x  -3/2  =   ± sqrt (5)  / 2

x =  ±sqrt (5)/ 2   +  3/2       

x =  sqrt (3) /2   +  5/2        this  value of  x  makes  the  continued fraction  negative.....reject

x =  -sqrt (5)  / 2  +  3/2  =        [3  - sqrt 5]  /  2 

price per item would be lower

The ladder and the building and the ground form a right triangle will ladder being the hypotenuse....

pythagorean theorem      hyp² = 25² + d²

                                                       41^2 - 25^2 = d^2             d =~~ 32 ft

I get the following:

Thanks BWStar

You can use heron's formula but think about what I am saying anyway :)

The only way this can happen is if one triangle is acute and the other is obtuse.

there are many ways to tackle this problem, most are quite difficult but the answer can also be found using only basic logic and Pythagoras's theorem.

Draw the triangle 13,13, 10,    (Yes, you MUST draw it)

Cut it down the middle to form 2 congruent right-angled trianges.

For each of these right-angled triangles the hypotenuse is 13 and the base is 5, using pythag, the height is 12

I need another right-angled triangle with hypotenuse 13 and with the same area

SO

How about if the base is 12 and the height is 5.  That has to work    (draw it to see what I mean)

So the 3rd side of the second triangle (before it is cut in half) must be 12+12 = 24

You can leave a little note on this thread to say that you have answered on the original. (That is what I do) 

That way at least anybody who is perusing questions can see all of the answers that have been provided.

I have not looked at any of the answers in full but if your answer is the same as another earlier one, what was the purpose in providing it again?

I see your answer has a lot of detail and that is great and maybe that really did help the asker .... but maybe you could consider asking the poster what it is that he/she does not understand about the earlier answers.

Sometimes it can be something quite simple, that more detail still does not address.

If you can get the asker to interact with you you will be able to teach them MUCH more effectively.

Anyway, thanks for providing such a detailed answer.    

What is

\(\sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}}\)?

\(\begin{array}{|rcll|} \hline x &=& \sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}} \qquad \text{square both sides}\\ x^2 &=& 12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}} \\ x^2 &=& 12 + x \\ \mathbf{x^2-x-12} &=& \mathbf{0} \\ \hline x &=& \dfrac{ 1\pm \sqrt{1^2-4(-12)} } {2} \\\\ x &=& \dfrac{ 1\pm \sqrt{49} } {2} \\\\ x &=& \dfrac{ 1\pm 7 } {2} \\\\ \hline x_1 &=& \dfrac{ 1- 7 } {2} \\\\ x_1 &=& -3 \qquad \text{no solution $x$ must $> 0$ !}\\\\ x_2 &=& \dfrac{ 1+ 7 } {2} \\\\ \mathbf{x_2} &=& \mathbf{4} \\ \hline \end{array}\)

\(\sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}} = 4 \)

Here is a tip before posting in this forum for an answer: 

Graph it out! https://www.desmos.com/calculator is a good graphing calc

Look at the graph, and it's easy. try it out!

Your welcome! Have a nice night or day or afternoon! 

thanks for the clear, detailed explanation

well, I came back again to this site. 

you see the first and the last equation?

Subtract them such that the 3y is canceled out. 

$x+3y+2z-(-2x+3y+z)=1-7$
$3x+z = -6$

Neext, multiply the middle equation by $3$ to make $-9x+3y+15z=30$.

then, subtract the last equation from it to get: 
$-2x+3y+z-(-9x+3y+15z) = 7-30$
which is equal to $7x-14z=-23$

So here are the two equations we have: 

$3x+z = -6$

$7x-14z=-23$

Take the first equation and find z, which is $z=-6-3x$. Put the Z inside the second equation, $7x-14(-6-3x) = -23$, which is $7x+84+42x = -23$, or simply $49x=-107$

So, $x$ is equal to $-107/49$

Now, find z. Insert x into the equation, and get that $3(-107/49) + z = -6$, which means that $z=27/49$. 

Insert those two into the second ever equation to find y: 
$6 and 27/49 + y + 2 and 37/49= 10$, which in the end is equal to $y=9 and 15/49$

so the answer you seek is $(-107/49,27/49, 456/49)$

Where the denominator approaches zero there will be a vertical asymtope.....

Just tie those two together

so now you have 4 objects

they can be arranged in 4! ways. but of the 2 tied together, either could be first so there are 2 ways those 2 can be arranged.

Altogether there are 2*4! ways to arrange them

But if there were no restrictions then it would be  5! ways.

Can you finish it?

Start by squaring both sides. 

2sqrt(t-2) = sqrt(7-t)

sqrt(4t-8) = sqrt(7-t)

4t - 8 = 7 - t

5t = 15

t = 3

=^._.^=

f(x) = 5

if x >= -4:

x = 3

else:

x = 2

Since x >= -4, x = 3.

f(3) = 5

f(y) = 3

if y >= -4:

y = sqrt(7)

else:

y = 0

Since sqrt(7) >= -4, y = sqrt(7). 

I think there's only one value that works, sqrt(7). 

=^._.^=

We  have, in exponential  form  :

2^(1/2) (t  - 2)^(1/4)   =  (7 - t)^(1/4)       take  each  side to the  4th power

2^2 ( t - 2)  =  7- t

4  ( t - 2)  =  (7 - t)

4t  -  8  =  7  - t                   rearrange  as

5t  =  15

t  = 3

You can do this yourself

Treat the floor and ceiling functions like brackets, I mean, do them first.

The floor function, Like  L   brackets, means round down to the nearest integer (whole number, can be negative, 0 or positive)  

so

\(\lfloor 6.9 \rfloor = 6\)

The ceiling function does the opposite.   So the ceiling of 6.9= 7

Easy peazy.

so, to start you off

\(\lfloor -4-0.5\rfloor = \lfloor -4.5\rfloor = -5\)

(-infinity  , 3)  U  (3  , infinity )

Triangles PBC  and  PAD  are similar....so....

We  can  find   PC  as

PC  / 15   =  (PC + 10) /  27           cross-multiply

27 PC =  15  ( PC + 10)

27PC  = 15PC  +  150

27PC -  15PC =   150

12PC   =  150

PC   = 150  /12   =  12.5

And PB  = PC

So...perimeter  of   PAD   = 2 (12.5)  + 2 (10)  +  27   =    72