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Never mind

This is very similar to all of your other postings over the last few days...are you catching on yet??????

This one is (assuming there is a parentheses around the numerator) ....

a^-2 / a^2b^-1     +   a^-1b / a^2b^-1    -   b^2 / a^2b^-1

b / a^4               +      b^2 / a^3             -   b^3 /a^2  

a^-2 b  (  a^-2    +      b a^-1                  -   b^2)

Here is a solution:

Thank you!!

Well, let's simplify.

When there dividing numbers with the same base but with different exponents, we subtract the second exponent from the first exponent. 

(m^2/5) / (m^(-8/5))

Subtract the exponents. 2/5 - (-8/5)

That would be 2/5 + 8/5, which equals to 10/5.

10/5 is 2. So the exponent would be 2.

Our final answer would be m^2

Thanks so much!

100,000x^15 - 25,000x^10 - 1/(32 x^10) + 2,500x^5 + 25/(8 x^5) - 125

=  5 x^(4-2)  y^(3-1)

    5 x^2 y^2

Here is a website for such things:

https://www.oddsshark.com/tools/odds-calculator

You have to bet $15 to win the $10 at 60%

a bet of $15 to win at 40% should win $22.50    (according to the calculator)

Perhaps   ?    60% bet  ==    15 x 40/60 = $10 win

                       40% bet  ==    15 x 60/40 = 22.50 win   ??

....or another way to look at it...

Total payout on $15 bet

bet on 60%   :    100/60  x $15  = $25 payout ($10 winnings)

bet on 40%   :    100/40  x   15  =   37.20  (22.50 winnings)

Expanding:

x^(1/3+4/3) - y^(-2/3-1/3)  - x^1/3 y^-1/3 + x^4/3 y^-2/3

x^5/3  - y^-1  - x^1/3 y^-1/3  +  x^4/3 y^-2/3

x^5/3 - y^-1   +  x ^1/3 y^-1/3 ( xy^-1/3  -1)

10)

Triangle ABC, inscribed in a circle, has AB = 15 and BC = 25. A tangent to the circle is drawn at B, and a line through A parallel to this tangent intersects BC at D. Find DC. 

Picture: https://latex.artofproblemsolving.com/e/f/6/ef688cf1577b8cd6d2c7da6f4b56dab5dee8e935.png

Let \(DC=x\)

Let \(BD = 25-x\)

Let O the centre of the circle.

Let r the radius of the circle.

\(\begin{array}{|rcll|} \hline \dfrac {\sin(\alpha)}{25-x} &=& \dfrac {\sin(\beta)}{15} \quad & | \quad \cos(90^{\circ}-\alpha) = \dfrac{15}{2r}\quad \cos(90^{\circ}-\beta) = \dfrac{25}{2r} \\ &&\quad & | \quad \dfrac{\sin(\alpha)}{15} = \dfrac{\sin(\beta)}{25} \\ &&\quad & | \quad \dfrac{\sin(\alpha)}{\sin(\beta)} = \dfrac{15}{25} \\ &&\quad & | \quad \dfrac{\sin(\alpha)}{\sin(\beta)} = \dfrac{3}{5} \\ \dfrac{\sin(\alpha)}{\sin(\beta)} &=& \dfrac{25-x} {15} \quad & = \quad\dfrac{3}{5} \\\\ \dfrac{25-x} {15} & = & \quad\dfrac{3}{5} \\\\ 5(25-x) & = & 45 \\ 25-x & = & 9 \\ x & = & 25-9 \\ \mathbf{ x } & \mathbf{=} & \mathbf{16} \\ \hline \end{array}\)

DC is 16

Thank you, CPhill

9)

Each side of quadrilateral ABCD is tangent to a circle. If AB = 84, BC = 71 and CD = 75, find DA.

Call  the two tangents that meet at A, x

Call the two tangents that meet at B, y

Call the two tangents that meet at C, z

Call the two tangents that meet at D, w

And we have the following system

x + y   =  84     ⇒    y   =  84  - x

y + z  =   71  ⇒   z = 71 - y ⇒    z  =  71  - (84- x)  ⇒  x - 13

z + w  = 75  ⇒   w  = 75 - z  ⇒  75 - (x - 13)  =  88 - x

w + x  =    (88 - x) + x   =  88   =   DA

That's very impressive,  heureka  !!!!

Here's the first part :

1^3   + 2^3  +  3^3   +  .....+  n^3   =  n^2 (n + 1)^2  / 4

Show that it's true for  n  = 1

1^3   =   (1)^2 (1 + 1)^2 / 4   =  4 / 4  =  1   ⇒   true

Assume that it;s true for n  = k

So  

1^3 + 2^3  + 3^3  +  .... +   k^3   =   k^2 (k + 1)^2 / 4

Show that it's true for k + 1

That is

1^3   +  2^3  + 3^3 +  .....+ k^3  + (k + 1)^3   =  [k+1]^2 [ (k +1) + 1) ]^2 / 4 

So we have

1^3   +  2^3  + 3^3 +  .....+ k^3  + (k + 1)^3   =  k^2 ( k + 1)^2 / 4  + (k + 1)^3

= [  k^2 (k + 1)^2  + 4(k + 1)^3] / 4

=  [ (k + 1)^2 ( k^2 + 4(k + 1) ]  / 4

= [ (k + 1)^2 ( k^2 + 4k + 4) ] /4

= [ (k + 1)^2 ( k + 2)^2 / 4 = 

= [ (k + 1)^2 [ (k + 1) + 1 ]^2 / 4

Which is what we wanted to show

For the second part....we can evaluate these sum-of-diferences  [ although there are probably faster ways, maybe  ??? ]

1         28         153          496          1225              2556

     27         125         343         729              1331

             98           218       386           602

                   120          168       216

                            48           48

We will generate a 4th power polynomial.....which is consistent with your second formula

Sorry... got to go to bed, now  !!!   ....LOL!!!

11)

Two circles, centered at A and B are externally tangent to each other, and tangent to a line L. A third circle, centered at C is externally tangent to the first two circles, and the line L. If the radii of circle A and circle B are 9 and 16, respectively, then what is the radius of circle C?

Picture: https://latex.artofproblemsolving.com/4/d/a/4da24a4edb9693476d2766d290f52dd917498a6f.png

\(\text{Pythagorean theorem:} \)

\(\begin{array}{|lrcll|} \hline (1) & (r_a+r_b)^2 &=& (r_b-r_a)^2 + s_1^2 \\ & r_a^2+2r_ar_b+r_b^2 &=& r_b^2-2r_br_a+r_a^2+s_1^2 \\ & 4r_ar_b & = & s_1^2 \\ & \mathbf{s_1} &\mathbf{=}& \mathbf{2\sqrt{r_ar_b}} \\\\ (2) & (r_b+r_c)^2 &=& (r_b-r_c)^2 + s_3^2 \\ & r_b^2+2r_br_c+r_c^2 &=& r_b^2-2r_br_c+r_c^2 + s_3^2 \\ & 4r_br_c &= & s_3^2 \\ & \mathbf{s_3} &\mathbf{=} & \mathbf{2\sqrt{r_br_c}} \\\\ (3) & (r_a+r_c)^2 &=& (r_a-r_c)^2 + s_2^2 \\ & r_a^2+2r_ar_c+r_c^2 &=& r_a^2-2r_ar_c+r_c^2 +s_2^2 \\ & 4r_ar_c & = & s_2^2 \\ & \mathbf{s_2} &\mathbf{=} & \mathbf{2\sqrt{r_ar_c}} \\\\ & \mathbf{s_1} &\mathbf{=}& \mathbf{s_2 + s_3} \\ & 2\sqrt{r_ar_b} &=& 2\sqrt{r_ar_c} + 2\sqrt{r_br_c} \quad & | \quad : 2 \\ & \sqrt{r_ar_b} &=& \sqrt{r_ar_c} + \sqrt{r_br_c} \\ & \sqrt{r_ar_b} &=& \sqrt{r_c}(\sqrt{r_a} + \sqrt{r_b}) \\ & \sqrt{r_c} &=& \dfrac{\sqrt{r_ar_b}} {\sqrt{r_a} + \sqrt{r_b}} \quad & | \quad \Rightarrow \frac{1}{\sqrt{r_c}} = \frac{1}{\sqrt{r_a}} + \frac{1}{\sqrt{r_b}} \\ & r_c &=& \dfrac{ r_ar_b } {(\sqrt{r_a} + \sqrt{r_b})^2} \quad & | \quad r_a = 9 \quad r_b=16 \\ & r_c &=& \dfrac{ 9\cdot 16 } {(\sqrt{9} + \sqrt{16})^2} \\ & r_c &=& \dfrac{ 144 } {(3+4)^2} \\ & r_c &=& \dfrac{ 144 } {49} \\ \hline \end{array}\)

The radius of circle C is \(\mathbf{\tfrac{144}{49}} \)

Here's my best effort.......

Out of any 5 marbles drawn, 2 need to be green.....and the probability of a green on any draw is 6/10

So we have :

C(5,2) * (.6)^2 * (.4)^3    =  144/625   =  23.04 %

This solution for the inverse of this function is g(x)=x-LambertW(e^x+1)+1. However there is a way around

y=e^x+ax+b

y-b=e^x+ax=Y

(Y^-1)'=Int(1/Y')dx

By computing the integral above and reverting to original variable you will get

g(x)=x/a-[ln(e^x+a)]/a+b

4. A triangle is formed by connecting the points (-5,0), (0,6), and (5,0). The resulting triangle is then rotated about the x-axis to form a solid. The original triangle is then rotated about the y-axis to form a different solid. What is the positive difference between the number of cubic units in the volumes of the two resulting solids? Express your answer in terms of pi.

The first object needs a full rotation about x to form a solid and it will form two cones....each with a radius  of 6  and a height of 5

So....their volume will be ;

(2/3)pi (6)^2 (5)  =  120 pi  unit's^3

The second object only needs a half rotation around y to form a solid and it will form a cone with a radius if 5 and a height of 6

So....its volume will be :

(1/3) pi (5^2) (6)  =  50 pi units^3

So...the diference between their volumes is    70 pi  units^3

3.The total number of degrees in the sum of the interior angles of two regular polygons is 1980. The sum of the numbers of diagonals in the two polygons is 34. What is the the positive difference between the numbers of sides of the two polygons?

Call the number of sides of the first polygon, a   and the number of sides of the second polygon, b

The sum of the interior angles is given by

[ (a - 2) + (b - 2)]  * 180  =  1980

[ a + b] - 4    =  11

[ a + b ]  =  15

b  =  15 - a

The number of diagonals in a polygon  =  n(n - 3) / 2    where n is the number of sides

So....the sum of the diagonals is

(a) (a - 3) / 2   + (b)(b - 3)/ 2  =  34

a^2 - 3a + b^2 - 3b  =  68

a^2 + b^2 - 3(a + b)  = 68

a^2 + b^2  - 3(15)  = 68

a^2 + b^2  =  113

a^2 + (15 - a)^2  = 113

a^2 + a^2  - 30a  + 225  = 113

2a^2 - 30a  + 112  = 0

a^2 - 15a + 56  =   0

(a - 7) (a - 8)  =  0

So...a  = 7 , b = 8  or a  = 8, b = 7

So....the positive difference between their number of sides  =1

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a 4x4 square array of dots like the one shown below?

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a 4x4 square array of dots like the one shown below?



Two triangles are congruent if they have the same traced outline, possibly up to rotating and flipping. This is equivalent to having the same set of 3 side lengths.

total 9

8)

In triangle ABC, AB = 6, BC = 8  and CA = 10. A circle centered at B is tangent to AC. What is the radius of this circle?

Here's one way to do this: 

Let  B = (0,0), A  = (0,6)  and C  = (8,0)

The side  AC  has a slope  =  -6/ 8  =  -3/4

And the equation of this line  is

y = (-3/4)x + 6

We need to  find a line perpendicular to this passing through (0,0)

The slope of this line is  4/3     and its equation is:

y  = (4/3)x

To find out where these two lines intersect we have

(-3/4)x + 6  =  (4/3)x

6  =  (4/3  + 3/4) x

6 = ( 25/12) x

(12 * 6) / 25  = x

72/25  = x

And   y  = (4/3) (72/25)   =  96/25

So  the radius of a circle centered at B and tangent to AC  is

sqrt  [ (72/25)^2  + (96/25^2 ]  = sqrt  [ 72^2  + 96^2 ] /25  =

sqrt [ 14400] / 25 =

120/25  =

24/5  

1)

Circle O is a unit circle. Segment AS has length 12/5 and is tangent to circle O at A. If P is the intersection of OS with circle O, find length PS.

AS  = 12/5      AO  = 1

So   ....  

SO  =  sqrt [ (12/5)^2  + 1 ]  =   sqrt [  144 + 25] / 5  =  13/5

So...PS   =   SO  - PO  =   13/5  -  1  =     8/5

2)

Angle A : Angle P : Angle ASP are in ratio 1 : 2 : 2. Find the degree measure of angle BSA.

Angle A  =  36°......Angle P, ASP  =  72°

Angle  ASP  =  (1/2)minor arc AS  =  angle BSA  =  72°

3)

If angle B = 39 degrees and arc PS = 116 degrees, find the degree measure of arc AS.

Angle B =  (1/2) ( arc AS - arc AP)

38  =  (1/2) (arc AS - arc AP)

76  =  arc AS - arc AP     (1)

And

arc AS + arc AP  + arc PS  = 360

arc AS + arc AP   +  116  =  360

arc AS  + arc AP  =  244   (2)

Add  (1)  and (2)

arc AS - arc AP   =  76

arc AS +  arc AP  = 244

2 arc AS  =  320      divide by 2

arc AS  =  160°

4)

Points A and B are on a circle centered at O, and point P is outside the circle such that PA and PB are tangent to the circle. If angle OPA = 32 degrees, then what is the measure of minor arc AB, in degrees?

Draw radii OA, OB...so  OAPB forms a quadrilaterlal....the sum of its interior angles = 360°

Angles OAP, OBP  =  90°  and OPA  =  32°, then angle BPA  =  64°

So...angle OAB  =  360 - 2(90) - 64  =  116°

And  OAB  is a central angle  intercepting minor arc AB, so its measure is also 116°

5)

Circle O and circle P, with radii 3 and 5, respectively, are both tangent to line L at H. Enter all possible lengths of OP separated by commas.

{Need a pic, here  }

6)

Given regular pentagon ABCDE, a circle can be drawn that is tangent to DC at D and to AB at A. What is the number of degrees in minor arc AD?

Call the center of the circle O, connect  OA  and OD

And OABCD  forms another pentagon whose interior angles sum to 540°

Angles  ODC and angle OAB   =  90°

Angles DCB and CBA  =  108°

So  angle DOA  =  540 - 2(90) - 2(108)  = 144°  ...this is a central angle in the circle intercepting minor arc  AD  ....so it also measures 144°

I do not know the answer yet, but to start off....why did you multiply the cost of the defective chairs x 60 instead of 10?    Each defective chair adds $10 to the cost  soooooo   500 x 70 + 10(#defective chairs) = total cost