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I have a crystal ball –compliments of Morgan Tud. I’ve not yet figured out the mystery to make it work. However, I do see very pretty things in the orb –usually my garden flowers or my own face. Sometimes it’s both at the same time –it’s really beautiful then.           

You probably need to include more information.   We do not have a crystal ball :)

Have you gone through the answer that our guest provided?

If there is an error perhaps you can spot it youself.  You can at least try to.

You can put it into the calculator on the home page like this:    npr(12,9)

But if you don't have a calculator that can do this.....

\(_nP_r=\frac{n!}{(n-r)!} \\~\\ _{12}P_9=\frac{12!}{(12-9)!} \\~\\ _{12}P_9=\frac{12!}{3!} \\~\\ _{12}P_9=\frac{12\,\cdot\,11\,\cdot\,10\,\cdot\,9\,\cdot\,8\,\cdot\,7\,\cdot\,6\,\cdot\,5\,\cdot\,4\,\cdot\,3!}{3!} \\~\\ _{12}P_9=12\,\cdot\,11\,\cdot\,10\,\cdot\,9\,\cdot\,8\,\cdot\,7\,\cdot\,6\,\cdot\,5\,\cdot\,4 \\~\\ _{12}P_9=79,833,600\)

Monday:         2     hrs

Tuesday:        1.5  hrs

Wednesday:   4     hrs

Thursday:       2     hrs

Friday:            2     hrs

Saturday:       1.5   hrs

Sunday:          0     hrs

Total:             13  hours in a week

If you swim 13 hours in a week for 43 weeks, that is

13 hours, done 43 times

13 * 43  =  559 hours

\(\frac{2x+1}{3}+7=\frac{3x-1}{4}\)                                      Multiply both sides of the equation by  4  .

\(4\,*\,(\frac{2x+1}{3}+7)=3x-1 \\~\\ \frac{4(2x+1)}{3}+28=3x-1\)                        Distribute the  4  .

\(\frac{8x+4}3+28=3x-1 \\~\\ 3\,*\,(\frac{8x+4}3+28)=3\,*\,(3x-1) \\~\\ 8x+4+84=9x-3\)          Multiply both sides of the equation by 3 and distribute.

\(8x+88=9x-3\)                                    Subtract  8x  from both sides of the equation.

\(88=x-3\)                                                Add  3  to both sides of the equation.

\(91=x\)

Notice that we could have multiplied both sides by  12  to begin with and this would get rid of both fractions in one step. 

Thank you for your help. But the answer I have is $15,895 per teacher. Can you please re-check your calculations?

Thanks for the problem.

Let's retrieve information from the question.

n(total) = 60 years

The interest rate is already effective; i=5%

An increase g=20% from years 50-60

Annual total = ?

Let's set up our notation:

60,000(P/A, 5%, 19) = X (Years 30 through 60 is the retirement stage - I'm finding the present worth (at years 30) of the annual income right before the 20% increase)

60,000((1-(1.2/1.05)^10)/(0.05-0.2))(P/F, 5%, 20) = Y (calculating the present worth of the 20% increase 10 year span then converting that value to year 30)

X + Y = Z (adding both values at year 30 together)

Z(P/F, 5%, 30)(A/P, 5%, 30) (Now converting the new value at year 30 to year 0 then finding the annual worth over tge 30 year contribution period of the monetary amount)

Let's evaluate:

60,000(12.0853) = $725,118.00

60,000((1-(1.2/1.05)^10)/(0.05-0.2))(0.3769) = $422,307.26

$725,118.00 + $422,307.26 = $1,147,425.26

$1,147,425.26(0.2314)(0.06505) = $17,271.69

Since the diameter of the circle forms the largest chord possible in the circle.....we can position the triangle's  base along the diameter, D = 12 cm

The triangle with the greatest area will have the radius - R -  as its height

So.....the largest possible area of the prescribed triangle  =  (1/2) D * R    =  (1/2) (12) (6) =  36 cm^2

This was answered a while back, but the answer ends up being 16

4 ways to place the X in the top row

2 ways to place the X  in the second row [ in the other 2 x 2  box not occupied by the first X ]

2 ways to place the X in the  third row  [ we can choose either of the two empty columns]

Only 1 way  to place the last X in the last empty column

So   ....4 x 2 x 2 x 1   =  16

Again, because we have a bisected angle in a triangle....we have that

AX / XB =  AC /BC

AX / XB  =  28 / 56

AX/ XB  =  1 / 2      which implies that

AX : XB  =  1 : 2  

Then  AX + XB   make up three equal parts of AB

And AX  is 1/3  of the three equal parts....so...

(1/3) 50  = 50/3  = AX

Because angle BAC is bisected  by AD, we have that

BD / DC  =  AB / AC       so

42 / DC = 105 /AC     which implies that

105 * DC  =  42AC

DC  =  (42/105)AC =  (2/5) AC  =  .4AC

So....using the Pythagorean Theorem we have that

AB^2  + BC^2  = AC^2

AB^2  + ( BD + DC)^2  = AC^2

105^2  + ( 42 + .4AC) = AC^2   simplify

105^2  + 42^2 + 33.6AC + .16AC^2  =  AC^2  

12789 + 33.6AC + .16AC^2  =  AC^2    subtract the left side from both sides

.84AC^2   - 33.6AC - 12789 =  0

Using the quadratic formula will give us the positive solution for AC  =  145 

Here's my approach.....however.....there may be better methods of attack

Draw  altitude AD (through O ) of triangle ABC to base BC

This altitude will bisect BC  so that DC  = 3

The triangle ADC is right  and since AC = 5  and DC = 3, then this is a "Pythagorean" Triple" 3 - 4 - 5  right triangle......so.....AD  = 4

And  the triangle ODC formed is also right.... and  we have  that

 OD = √[OC^2 - DC^2]  =  √[OC^2 - 3^2] = √[ OC^2 - 9  ]  =  height of  ODC 

And since OC = OA....we have that

OA + OD  =   AD =  4      ....so...substituting

OC + √[ OC^2 - 9  ]  = 4     subtract OC from each side

√[ OC^2 - 9  ] = 4 - OC      square both sides

OC^2 - 9 =  16 - 8OC  + OC^2     subtract OC^2  from each side

- 9 = 16 -  8OC     subtract 16 from both sides

-25 =  -8 OC    divide both sides by -8

25/8 = OC

So the height of triangle OBC = OD  = √[ OC^2 - 9  ] = √[ (25/8)^2 - 9  ] 

And its area  =  (1/2) *BC * OD  =    

(1/2) * 6 * √[ (25/8)^2 - 9  ]   = 

3 * √ (625/ 64 - 576/64 )  =

3 * √ (49/64)  =

3 * 7/8  =

21 / 8  units^2

Since apex angle BCA is bisected we have the following relationship :

AX / BX  =  AC / BC

AX / 24  =  21 / 28

AX / 24  =  3 / 4      cross-multiply

4 *AX  = 24 * 3

4 * AX  =  72       divide both sides by 4

AX  =  18

Thanks, Melody.....other than exposing the forum members  [ and guests ] to, perhaps, a new concept, I can't really take much credit for the answers. I just ran across that website one day. Before that, I had never heard of this, either !!!!

I am very impressed Chris,   I have never seen anything like that before  

Thanks!

[11x] / [3y^2] -  [ 7x] / [6y ]      the common denominator is 6y^2

[ 11x  * 2]  / [  3y^2 * 2 ]  -  [  7x * y ]/ [ 6y * y ]

[ 22x  -  7xy ] /  [ 6y^2 ]

Thank you

\(\frac{m-4}{m^2+8m+16}-\frac{m-4}{m+4}\)

We need to get a common denominator. To do that...multiply the second fraction by \( \frac{m+4}{m+4} \)  .

\(=\frac{m-4}{m^2+8m+16}-\frac{(m-4)(m+4)}{(m+4)(m+4)}\)

Multiply the numerator and denominator out.

\(=\frac{m-4}{m^2+8m+16}-\frac{m^2-16}{m^2+8m+16}\)

Now that we have a common denominator, we can combine the fractions.

\(=\frac{m-4-(m^2-16)}{m^2+8m+16} \\~\\ =\frac{m-4-m^2+16}{m^2+8m+16} \\~\\ =\frac{\mathbf{-1}m^2+m+\mathbf{12}}{m^2+\mathbf{8}m+\mathbf{16}}\)

This is (as JB calls it),”slop at the top!” Of course, Mr. BB mixed this slop with blarney; I would expect nothing less from the Blarney Master.   Mr. BB’s presentations are always s slop, but the contrast is quite sharp when you compare it to Heureka’s masterful presentation.

Let’s see:  you “don’t know what the formula is supposed to calculate!”  But you know it is about calculating the ‘escape velocity’” (it’s impressive that you know that).  “But you don’t need it here!”  That’s impressive too, because it’s a monumental metric for measuring Batshit Stupid!  

The asker is supposed to ignore the formula that gives the Schwarzschild radius, but know “…that if the Earth was squeezed into a black hole, the black hole would have a radius of 9 millimeters!”

How is he supposed to know that?  Oh, of course, ask your computer: it’s connected to the internet, and everything read on there is true; that’s a fact; you can read about on the internet!    

After he finds out (via magic incantation) that it is “9 millimeters!”  He’s supposed to know “The ‘photon sphere’ or the orbit of light would be: 1.5 x 9 =13.5 mm from the center of the black hole. Or, 13.5 - 9 =4.5mm from the ‘event horizon’.” He knows this the same way he knows the first part . . .  He asks his computer. No need for formulas or derivations. That’s just a waste of time!   

None of this matters because this is not the question.  The photon sphere is irrelevant here. Only the escape velocity is relevant. 

“That is the best I can do!.”

Oh surely not! Ask your computer this question:  What would the minimum orbital radius be if the partial were an electron?  

I can hardly wait!  I’ll be biding my time playing Monopoly with my dog and cat. It’s the cat’s turn to be the banker so it should be an interesting game!

-2*3.06+(7.6*6y^2)=240     simplify

-6.12 + 45.6 y^2  = 240     add 6.12 to both sides

45.6 y^2  =  246.12          divide both sides by  45.6

y^2  =  246.12 /  45.6  ≈   5.40

Thank you very much.