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\(y = \dfrac{3}{2x}\\ v=10x+\dfrac 3 5 \cdot \dfrac{3}{2x} = 10x+\dfrac{9}{10x}\\ \dfrac{dv}{dx} = 10 -\dfrac{9}{10x^2}\\ \dfrac{dv}{dx}=0 \Rightarrow 10=\dfrac{9}{10x^2}\\ \dfrac{100}{9}=\dfrac{1}{x^2}\\ x^2 = \dfrac{9}{100}\\ x = \dfrac{3}{10}\)

\(\text{We need to confirm this critical point is a minimum}\\ \dfrac{d^2v}{dx^2} = \dfrac{9}{5x^3}\\ \text{$\left. \dfrac{9}{5x^3}\right |_{x=\frac{3}{10}} > 0$ so this is in fact a minimum}\)

\(x = \dfrac{3}{10} \Rightarrow \\ y = \dfrac{3}{2\cdot \frac{3}{10}} = 5\\ 10\left(\dfrac{3}{10}\right) + \dfrac{3(5)}{5} = 3+3= 6\)

\(\text{The minimum value of $10x+\dfrac{3y}{5},~x,y >0,~xy=\dfrac 3 2$ is 6}\)

2x^2  - 8x + 3y^2  + 6y  + 5  = 0

2x^2 - 8x + 3y^2  + 6y  =  - 5       compete the square on x , y

2 ( x^2 - 4x + 4)  + 3(y^2 + 2y + 1)   =   -5  + 8 + 3

2 ( x - 2)^2  +  3 ( y + 1)^2   =  6           divide both sides by 6

(x - 2)^2          (y + 1)^2

_______  +     ________    =       1

      3                    2

We have the form

(x - h)^2          ( y - k)^2

________  +  _________  =  1

     a^2                  b^2

The center of this ellipse  is  ( 2, - 1)

And the major  axis lies along  x

And the max value of x  =   ( 2 + a)  =  (2 + √3)  ≈  3.732

Here is a graph that shows this :  https://www.desmos.com/calculator/qu9bzxtafp

\(N-1 = 0 \pmod{6}\\ N=1 \pmod{6}\\ N=151, 157, \dots, 151+6k, \dots , 199\\~\\ \sum \limits_{k=0}^8 (151+6k) = 1575\)

105^4   = 121550625

We can solve this as follows:

Suppose that the first term is 1  and the last term is  15592

Then....the sum of these terms is    (15592) (15593) / 2  = 121563028

Then the difference in the sums is    121563028 - 121550625  =  12403

So....we need to find out how many of the first n positive integers sum to  12403

So  we have

n ( n + 1) / 2 =  12403

n^2 + n  = 24806

n^2 + n  - 24806  =   0

The square root of   24806  ≈  157....

So.... we might guess that the factoriztion of this is  ( n - 157) ( n + 158)  =0

Expanding this gives us  n^2  + n - 24806  = 0

Taking the positive value of n shows that n  = 157

So.....the first 157 positive integers sum to  12403

Which means that the sum of the integers from 158 to 15592   gives us 105^4  =  121550625

So....the first term is 158    the last is 15592   and the number of terms is   15592 - 158 + 1  = 15435

Sum it up from 1 to 15,592 =121,563,028
121,563,028 - 105^4 =12, 403
12,403 x 2 =24,806
Sqrt(24,806) =157.499 =~158 - the first term
15,592 - 158 + 1 =15,435 - number of terms.

....or is that   -89x * (100x^2) =  - 8900 x^3   ???

I see. Thanks CPhill!!

Ah, nice one. I would never have thought of that.

-89 *  (100x^2)  =

-89 * 100 * x^2  =

-8900x^2

Two circles of radius 1 are centered at (4,0) and (-4,0).

How many circles are tangent to both of the given circles and also pass through the point (0,5)?

A long time ago I wrote a program, see table, the calculation of the (Problem of Apollonius) I have derived by means of the vector calculation.

The third circle has the center (0, 5) and the radius 0.

1. Kreis: Mittelpunkt M1 und Radius r1: x1 = 4 y1 = 0 r1 = 1
2. Kreis: Mittelpunkt M2 und Radius r2: x2 = -4 y2 = 0 r2 = 1
3. Kreis: Mittelpunkt M3 und Radius r3: x3 = 0 y3 = 5 r3 = 0

\(\begin{array}{|c|l|c|c|c|} \hline & \text{Index } & x\text{ center} & y\text{ center}&r \text{ radius}\\ \hline 1&+++&0&1.66667&3.33333 \\ \hline 2 &&-8.88178e-16&1.66667&3.33333\\ \hline 3 &++-&0&1.66667&3.33333\\ \hline 4&&-8.88178e-16&1.66667&3.33333\\ \hline 5&+-+&-1.0328&1&4.13118\\ \hline 6&&-1.0328&1&4.13118\\ \hline 7&+--&-1.0328&1&4.13118\\ \hline 8&&-1.0328&1&4.13118\\ \hline 9&-++&1.0328&1&4.13118\\ \hline 10&&1.0328&1&4.13118\\ \hline 11&-+-&1.0328&1&4.13118\\ \hline 12&&1.0328&1&4.13118\\ \hline 13&--+&0&0&5\\ \hline 14&&0&0&5\\ \hline 15&---&0&0&5\\ \hline 16&&0&0&5\\ \hline \end{array}\)

A hyperbola centered at the origin has foci at \((\pm 7, 0)\), and passes through the point \((2, 12)\).
If the equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), where \(a\) and \(b\) are positive constants,
compute the ordered pair \((a, b)\).

\(\begin{array}{|lrcll|} \hline P(2,12): & \mathbf{\dfrac{2^2}{a^2} - \dfrac{12^2}{b^2}} &=& \mathbf{1} \\\\ & \dfrac{4}{a^2} - \dfrac{144}{b^2}&=& 1 \\\\ & 4b^2- 144a^2 &=& a^2b^2 \\ & 144a^2 &=& 4b^2-a^2b^2 \\ & 144a^2 &=& b^2(4-a^2) \quad | \quad a^2+b^2 = 7^2,\ \mathbf{b^2=49-a^2} \\ & 144a^2 &=& (49-a^2)(4-a^2) \\ & 144a^2 &=& 196-53a^2+a^4 \\ & 197a^2 &=& 196+a^4 \\ & a^4 - 197a^2+196 &=& 0 \\\\ & a^2 &=& \dfrac{197\pm\sqrt{197^2-4\cdot 196}} {2} \\ & a^2 &=& \dfrac{197\pm\sqrt{38025}} {2} \\ & a^2 &=& \dfrac{197\pm 195} {2} \\\\ & a^2 &=& \dfrac{197 + 195} {2} \\ & a^2 &=& 196, \ b^2 = 49-196 \quad \text{no solution $b$ is complex } \\\\ & a^2 &=& \dfrac{197 - 195} {2} \\ & a^2 &=& 1, \ b^2 = 49-1=48 \ \checkmark \\ & \mathbf{a} &=& \mathbf{1}, \ \mathbf{ b}= \sqrt{48} \mathbf{= 4\sqrt{3}} \\ \hline \end{array}\)

\(\mathbf{ (a,b) = (1,\ 4\sqrt{3}) }\)

\(2x^2 - 8x + 3y^2+6y+5=0\\ 2(x^2-4x) + 3(y^2+2y)+5=0\\ 2(x^2-4x+4-4)+3(y^2+2y+1-1) + 5 = 0\\ 2(x-2)^2-8+3(y+1)^2-3+5=0\\ 2(x-2)^2+3(y+1)^2=3\\ \left(\dfrac{x-2}{\sqrt{\dfrac 3 2}}\right)^2+(y+1)^2 = 1\)

\(\text{There are no $xy$ terms so the ellipse isn't rotated at all, thus the maximum $x\\$ coordinate will occur when $y=0$}\\ x-2 = \sqrt{\dfrac 3 2 }\\ x = 2 + \sqrt{\dfrac 3 2}\)

Travisio,

Your graph shows that the funtion is symmetrical about the y axis

this means that f(x)=f(-x)

and that means that the function is even

Do you understand now?

Yes I understood that.  :)

Next time you can send me a message and ask me to unlock it.

You can still write a new post but put an easy to use link and ask people to do the actual answer on the original post.

They can also put a comment on the new post saying that they have added an answer.

It is a bit of a rigmarole, sorry about that, but I do not see a better way.

Since |a-n|= |a - b| + |b - c| + |c - d| + .... + |m-n| and |n-a|=|m-n| + |n-o| + ... + |x - y| + |y - z| + |z - a|, and |a-n|=|n-a|, The maximum value of |a-n| is 10

|b - c| + |c - a|>=|a - b|,the maximum |a-b| can be is 10

If     f( -x )  =  f(x)     then the function is even.

If     f( -x )  =  -f(x)    then the function is odd.

\(f(x)\ =\ \frac{3}{2x^6-5}\)

                                           Plug in  -x  for  x

\(f(-x)\ =\ \frac{3}{2(-x)^6-5}\)

                                           And     (ab)^c  =  a^c b^c     so     (-x)⁶  =  (-1)⁶ x⁶

\(f(-x)\ =\ \frac{3}{2(-1)^6x^6\,-\,5}\)

                                           Because     -1 * -1  =  1,     -1  to an even power  =  1     and so     (-1)⁶  =  1

\(f(-x)\ =\ \frac{3}{2x^6\,-\,5}\)

                                           Now notice that  \(f(x)\ =\ \frac{3}{2x^6-5}\)  so we can substitute  f(x)  in for  \(\frac{3}{2x^6-5}\)

\(f(-x)\ =\ f(x)\)

Since  f(-x)  =  f(x) ,  the function is even.

\(y\ =\ \frac{x^3-x^2+x}{6x^2-9x}\)

mabey some one can help.

Thank you :)

\(R(p) = N(p) p\\ p = 450-5k\\ N(p) = 500+10k\\ R(p) = (500+10k)(450-5k) = \\ -50 k^2+2000 k+225000 = \\ -50(k^2-40k - 4500) = \\ -50((k-20)^2 - 400-4500) = \\ -50((k-20)^2-4900) = \\ -50(k-20)^2 + 245000\\~\\ \text{The maximum is at $k=20$, i.e. a price of \$350}\)

Jo's sum is easy   = (100)(101)/2  =  50 * 101   =  5050

Kate  would first start with 5  [ since 1 - 4 ]  are rounded to 0

Starting with 5 thru 14  she will have ten 10s

15 - 24   she will have  ten 20s

25 - 34 she will have  ten 30s

.

.

.

85 - 94 she will have ten 90s

95 - 100   she will have six 100s

So.....her sum is

10  [10  + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90]  + 6 [100] =

10  [ (10) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) ] + 6[100]  =

10 (10)  [  sum of the first nine positive integers ] + 6 [ 100]

100  [  (9)(10) / 2 ]  + 6[100]  =

100 [ 45]  + 6[100] =

100 [ 45 + 6 ]  =

100 [ 51 ] =

5100

So....the positive difference is    5100 - 5050  =  50