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Thanks, Tiggsy  .....!!!!!

(My previous post is invisible. This one is better, anyway.)

Trained monkey! I don’t want to LMAO, but you are funny as heel! sit!, rollover!.

Do you think the monkey will learn a new behavior from your training?

Are you going to claim your copyright GA?

The basic technique for dealing with each of these problems is that of completing the square.

The first one,  $\displaystyle t^{2}-9t-36$.

Take a half of the coefficient of the linear term, (the 9t term), -9/2, square it, 36/4, add it to the expression and subtract it as well.

That gets you

$\displaystyle t^{2}-9t +\frac{36}{4}-36-\frac{36}{4}$, which is still equal to the original,

and which can be written as

$\displaystyle \left(t-\frac{9}{2}\right)^{2}-45$ .

The minimum value for that will be -45 occurring when t = 9/2, since that makes the expression inside the bracket zero. For any other value of t, the squared term will be greater than zero, making the whole expression greater than -45.

For the method to work, the coefficient of the squared term has to be 1. If the coefficient is not equal to 1, it has to be removed as a factor.

For example, if we had, (not one of yours),

$\displaystyle -2x^{2}-8x+17$,

we would proceed as follows.

$\displaystyle -2(x^{2}+4x)+17=-2(x^{2}+4x+4-4)+17$

$\displaystyle =-2\{(x+2)^{2}-4\}+17$ ,

$\displaystyle =-2(x+2)^{2}+8+17\\=25-2(x+2)^{2}$

Looking at that, its maximum value will be 25, occurring when x = -2, (since that makes the expression inside the bracket equal to zero).

The two variable example further down, $\displaystyle x^{2}+y^{2}-4x+2y$, can be dealt with in much the same way.

Write it as

$\displaystyle x^{2}-4x+y^{2}+2y$ ,

and complete the square on x and y separately. It's then easy to see what the minimum value is, and the values of x and y that give that value.

Tiggsy.

I wasn’t trying to be mean to Rosala, honest. I was just a little surprised by her post.  

Rosala’s social posts often awed me as a presentation of a quintessential, idealized young life on the other side of the world from where I live. Though she lives so far away, I could always clearly see the essence of my granddaughter in her.

Thank you! And thanks for your help!! 

Very nice, hectictar........!!!!

This is a little tricky.....since we have 4 identifiable turning points, the degree of the polynomial will be one greater  = 5

If we didn't have two intermediate turning points occuring between two roots, the degree would be 3 

Note that the function x^4 + 1   has no real roots, but a degree of 4...so....we cannot solely rely on the roots to give us the degree.......also....it only has one turning point.....so.....we can't always rely on that to tell us anything about the degree either!!!!

g(x)  =  $5-\sqrt{4-x}$

When  x = 3 , the slope of the curve  $=\,\lim\limits_{h\to0}\frac{g(3+h)-g(3)}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{(5-\sqrt{4-(3+h)})-(5-\sqrt{4-3})}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{5-\sqrt{4-3-h}-4}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{1-\sqrt{1-h}}{h} \\~\\ =\,\lim\limits_{h\to0}(\,\frac{1-\sqrt{1-h}}{h}\,)\,(\,\frac{1+\sqrt{1-h}}{1+\sqrt{1-h}}\,) \\~\\ =\,\lim\limits_{h\to0}\frac{1-(1-h)}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{h}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{1}{1+\sqrt{1-h}} \\~\\ =\,\frac{1}{1+\sqrt{1-0}} \\~\\ =\,\frac12$

Look at the graph and see that it does look like that line has a slope of $\frac12$ .

We want an equation of a line that has a slope of  $\frac12$  and passes through the point  ( 3, g(3) ) .

g(3)  =  $5-\sqrt{4-3}$  =  5 - 1  =  4

So the equation of our line in point - slope form is

y - 4  =  $\frac12$(x - 3)

And in slope - intercept form, this is...     y  =  $\frac12$x + $\frac52$

The area of the larger square is  9 square units

Note that we also have four other triangles bordering the smaller triangle each with an area of

(1/2) * 1 * 2  =   1 square unit....so 4 of them have an area of 4 square units

So....the area of the smaller triangle is   [ 9 - 4]  =  5 square units

So....the area of the smaller triangle is 5/9 that of the larger triangle

This has been asked before.......

Note that with

6/2 (1+ 2)   we perform the division, first  ......so we have

3 ( 1 + 2)  =

3 (3)  =    9

But......notice the difference with

6 / [ 2 ( 1 + 2) ]  =

6 / [ 2 *3]  = 

6 / 6  =    1

This points out the importance of parentheses/brackets.......

This is how it sounds broken up. Forgive me for how it sounds all broken up...

Edit: The word I put was starred out because that word is not allowed but here is what the stars mean:

H**l - low

  ^^

  el

So, big 'H' sound little 'ell' sound  and little 'low' sound

PEMDAS, parentheis first, you get 3

Then we have 6/2(3)

You use a rule where you flip them when encountering a fraction

So i did this

6/2 x 1/3 = 6/6 = 1

 Any constand has a value of (1) in the denomintaor, we just cant see it.

38%  =  .38  =   38 / 100  =   [ 19 * 2 ]  / [ 50 * 2 ]  =  19 / 50

(  [ (x + Δh) + 1 ] / [ (x + Δh) - 1 ]  -  [x + 1] / [ x - 1] ) / Δh  

{ get a common denominator }

(  [ (x + Δh) + 1 ]   [ x - 1]  -   [x + 1]  [ (x + Δh) - 1 ] ) /  ( [ Δh  [ (x + Δh) - 1 ]  [ x - 1] )  

{Simplify}

( x^2 + x Δh + x - x -  Δh - 1  -  x^2 -  xΔh + x - x -  Δh + 1 )  / ( [ Δh  [ (x + Δh) - 1 ]  [ x - 1] )    

( -2Δh)  / ( [ Δh  [ (x + Δh) - 1 ]  [ x - 1] )  

{ divide out   Δh  )

-2  / (   [ (x + Δh) - 1 ]  [ x - 1] )

Let  Δh → 0   and we have that the derivative is

-2  / ( x - 1)^2

WolframAlpha verifies this : 

https://www.wolframalpha.com/input/?i=derivative+%5B+x+%2B+1%5D+%2F+%5B+x+-+1%5D

I'm not great at Physics but I believe the equation we need is :

a_c  =   v^2  / r

Where  a_c  is the centripetal acceleration, v is the velocity and r is the radius

Note..... convert 2.75 km to meters  =   2750 m....this is the circumference....so to find the radius we have  r  =   2750 / [2 pi ]   =  [ 1375 / pi ]  m

So we have

a_c = [25m/s]^2 / [1375 / pi ] m  = 

[ 625 m^2 / s^2]  / [ 1375 / pi ] m   = 

 [ (5/11 ) pi ]   m / s^2 ≈ 

 [ 0.4545 pi ]   m / s^2 ≈

1.428 m / s^2

One last thing, Sekai

The  "1"    and  "2"   are known as "subscripts"

"Indexes"  usually refer to exponents

P.S.  ......if the "d"  is supposed to be a part of the second denominator, we have

0  =  I₁ / r  + I₂ /  (r  + d)      get a common denominator

0 =   [ I₁ ( r + d)  +  I₂ r ]  /  [ r (r + d) ]      multiply boh sides by   [ r (r + d) ]

0 *  [ r (r + d) ]  =   [ I₁ ( r + d)  +  I₂ r ] 

0  =   [ I₁ ( r + d)  +  I₂ r ]    simplify

0  =  I₁ r + I₁d +   I₂ r       subtract  I₁d  from both sides

- I₁d     =    I₁ r  +   I₂ r      factor the right side

- I₁d   =   r  [ I₁ + I₂ ]         divide both sides by  [ I₁ + I₂ ] 

- I₁d  /   [ I₁ + I₂ ]    =   r

Thank you !

(I also learned to make an index. I₁ <-- haha.)

Oh...OK....I understand now

We have that

0  =  I₁ / r  + I₂ / r  + d     subtract   d from both sides

-d  =   I₁ / r  + I₂ / r       note that the right side can be written as

-d =   [  I₁ + I₂ ]  / r      multiply both sides by r

-rd  =   I₁ + I₂         divide both sides by -d

r =     [  I₁ + I₂ ]  /  -d 

r  =   - [ I₁ + I₂ ] / d  

How often is the interest rate of 19% compounded?? Is it daily, weekly, monthly, quarterly, semi-annually or annually? Because, without knowing that you will get wrong answer.

The 1 at I1 and the 2 at I2 are just an index. In this exaple I1 is electricity number 1 and I2 is electricity number 2 :)!

Theoretical I could write : 0 = y/r - x/r+d 

and now I want to solve for r :3

If the triangle has a right angle then that means that two sides are identical. So if the short side is 6cm, and the longer side is 18cm then the middle side will be 6cm as well. If the long side is 18cm then the middle side will be:

SOLUTION: 6cm

Is the equation

0   =  I^(1/r) - I^(2/r) +  d     ????

Or is it

0  =  I^1  / r  +  I^(2) / r  + d      ????

The remaining side will be given by :

sqrt ( 18^2 - 6^2 )  = sqrt (288)  =  sqrt (144 * 2)  =  sqrt (144) * sqrt (2)  =  12sqrt(2)  cm  ≈ 

16.97 cm