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A  must walk  at a rate of   X mph

B must   walk at  at a rate of   Y mph

And when they switch destinations

Y / X    =  X  / Y  +    35/60

Y / X    =  X /Y   +  7/12

Y/X   - X/Y  =  7/12

Y^2   - X^2               7

_________   =       ____

       XY                    12

12 (Y^2 - X^2)  =  7XY

12Y^2  - 12X^2  =  7XY

12Y^2 - 7XY  - 12X^2   =  0

(4Y + 3X) (3Y - 4X)  =  0

Only the  second  factor gives us what we want

3Y   -  4x  =  0

4X  = 3Y

X  =  (3/4)Y

X/Y  = 3/4

So.....the ratio of A's speed to B's speed must be  X : Y    =         3  :  4

We will have a right triangle   with the string length the hypotenuse  

And we know the side opposite the angle of elevation

The sine relates the opposite side and the hypotenuse

We have

sin (40°)  =  opp  / hyp

sin (40°)  =  77/ hyp       rearrange as

hyp  =   77 / sin (40°)  =   119.79 ft   =   119.8 ft  =the string length 

We will have    4 lattice points

(x , y)  =  (24, 23)   ( 24, -23)  ( -24, 23)    (-24, -23)

If we multiplly  the first equation by   3/2  we get that

9x  + 6y  =   24

Since this is the same as the second equation......this system has infinite solutions   ......

2/3x+1/4y=3

-1/3x-5/4y=3

Multiply  both sides of each equation by 12  to  clear the fractions

8x  +  3y    =    36

-4x - 15y  =      36     

Multiply the second equation by 2

8x  +  3y    =  36

-8x  - 30y  =   72             add these equations

-27y  =  108         divide both sides by   -27

y  =  -4

And using  8x + 3y  =  36

8x + 3(-4)  = 36

8x  - 12  =  36   add 12 to both sides

8x  =  48      divide both sides by 8

x  =  6

Your answer is not correct. If you use the proper formula you will get it correct.

The prob of making exactly 2 shots in 6 tries is

\(\text{Prob (2 sucesses in 6 independent trials)}\\ = \;^6C_2\;*\;(\frac{211}{243})^2\;*\;(\frac{32}{243})^{(6-2)}\\~\\ = \;^6C_2\;*\;(\frac{211}{243})^2\;*\;(\frac{32}{243})^{4}\\~\\ =\quad 15\;* 211^2*\;32^4\;\div (243^6)\)

15*211^2*32^4/ 243^6 = 0.003401092481817478

That is a little lower than I expected But that is how probability goes.

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You can check if any mothod is correct by working out all the possible probabilities.

I mean

P(0 in) + P(1 in) + ......    +P(6 in)   and they have to add up to 1

x ( x^3 - 1) - 6(x^2 + x + 1)  =  0

Factor  (x^3 - 1)  as  a difference of cubes

x  ( x - 1) (x^2 + x + 1)  - 6( x^2 + x + 1)   =   0

(x^2  + x + 1)  [ x(x - 1) - 6  ]  = 0

(x^2 + x + 1) [  x^2 - x - 6 ] =  0

(x^2 + x + 1)  ( x - 3) ( x + 2)  =  0

The first factor doesn't  produce any real number roots  when set to 0 and solved

The  real roots   are  found  by setting the other two factors to 0 and solving for  x ⇒    x   =3   and  x  = -2

    If  2x + 7  is a factor, then  x  =  -7/2  is a zero....so.....

6(-7/2)^3 + 19(-7/2)^2  + c(-7/2)  +  35  =   0

-6 (343/ 8)  +  19 (49/4)   - (7/2)c  + 35  =  0

-2058 / 8  + 931/4  - (7/2)c +  35  = 0

[ -2058  + 1862  + 280 ] / 8   - (7/2)c  = 0

21/2  - (7/2)c   =  0

21 - 7c   =  0

21  =  7c

c  = 3

x*(x-1)(x^2+x+1)=6(x^2+x+1)

Notice that x^2+x+1 is on both sides so divide by it 

x*(x-1)=6

x^2-x-6=0 

(x-3)(x+2)=0

x= 3

x=-2 

For real numbers! 

Area  of square  =   [Side of Square ] ^2

Area  of hexagon  =  [Side of Hexagon]^2 * 3√3 / 2

So

[ Side of Square] ^2 =  [ Side of Hexagon] ^2   (3/2)√3

[ Side of Square  ]^2

___________________   =   (3/2)√3    =  (1.5)√3       take both roots

[ Side  of Hexagon] ^2   

Side  of Square

______________   =    √[ 1.5√3 ]

Side of Hexagon

So......the Side   of the square  =    √[1.5√3 ] * Side of the Hexagon

So......the perimeter  of the  Square  =    4√[ 1.5√3]  * Side of Hexagon

Let the Side of the Hexagon  =  A

So.....

The  perimeter of the  Square  =   4√[ 1.5√3] * A

And the perimeter of the Hexagon = 6A

So

Perimeter of Square                      4*√[ 1.5√3] * A                2 √[ 1.5√3]

_________________  =               _____________  =         __________

Perimeter of Hexagon                        6A                                    3

The correct interval is [-1,1].

I worked this out a different way and got a weird answer. Could you check my work to see if it is correct? 

In order to solve this problem we should first find the probability of making a ball into the bin on the first shot. It is not the probability of making one in 5 shots divided by 5 because you can make multiple shots out of 5. 

We can find the probability of making a ball into the bin on the first shot by making an equation. First, we need to find the chance we don't make a ball in the bin in 5 shots. That is \(1-\frac{211}{243}=\frac{32}{243}\) Next we can set p as P(making a ball into the bin on the first shot). This is the same probability of all the other shots because the probability is uniform across every shot. In terms of p, the chance of not making a ball in all of the shot is\( (1-p)^5\). Now we can make an equation. We get \((1-p)^5=\frac{32}{243}\). We can take the 5th root of both sides to get \((1-p)=\frac{2}{3} \)}. Solving for p we get \(p=1-\frac{2}{3}=\frac{1}{3}\).

Next, we can use complementary counting to find the probability of getting at least 2 shots in. `We can say P(0) is the probability of making no shots. P(1) is the probability of making one shot and so on. We can use this to make an equation. The probability of making at least 2 shots is 1-P(0)-P(1).

The probability of making 0 shots is \((\frac{2}{3})^6=\frac{64}{729}\)}. Next we have to find the probability of making 1 shot. The probability of that is \((\frac{2}{3})^5 \cdot \frac{1}{3}=\frac{32}{729}\). Now, we can find the probability of making at least 2 shots by plugging in the probabilities. We get \(1-\frac{64}{729}-\frac{32}{729}=\frac{633}{729}=\frac{211}{243}\). So, the probability of making two shots in 6 tries is \(\frac{211}{243}\)}.

Note that we can split this  as

2/1  +  2/9   +  2/81   +   .....+         and

1/3  +  1/27   +  1/243  + .... +

The  sum of the first series  =         2/  [ 1 - 1/9]  

The sum of the second series  =     (1/3)  / [ 1 - 1/9 ] 

So.....the sum  =    [ 2 +1/3]  / [ 1 - 1/9]   =   [ 7/3 ] / [ 8/ 9 ] =        63/ 24    =   21 / 8

WolframAlpha  shows several possibilities  :

https://www.wolframalpha.com/input/?i=x+%2B+yz+%3D+6%2C+++y+%2B+xz+%3D+6%2C++z+%2B+xy+%3D+6

Using sythetic division

5   [     1        0             30            - c   ]

                     5             25          275

       __________________________

           1        5            55             5

Which implies that

275 -  c  = 5

c  = 270

88888888888888

Sin x = 4/5          or     - 3/5

Cos x = -3/5        or     4/5                 

for quadrant II     sin is positive    so   4/5   

                          and cos is negative    so   -3/5

then tan = sin/cos =  4/5 / -3/5 =  - 4/3

∑ -1/2 3^(1 - n) ((-1)^n - 3), n, 1, ∞ =21 / 8

Since it it x < 2 it does not include '2' and there should be a hollow dot there

   since it says x = -3   for x > OR = 2   it DOES include '2' and there should be a soilid dot there...

Graph on the lower left.

Angle ACB  =  Angle XCB

Angle ABC  = Angle BXC

So.....by AA congruency......triangle CBA  is similar to triangle  CXB

Therefore

AB/ BC  =√3/ 1  = BX / XC ⇒  BX = √3 *  XC   ⇒   (BX)^2  = 3(CX)^2

And  by a property of  parallelograms,   AC  = 2CX

And

CX/ BC  =  BC / CA

BC^2  =  CA * CX

BC^2  =  2XC * CX

BC^2  = 2(CX)^2

BC  = √2 * CX

So .....using the Law of Cosines

BC^2  =  BX^2  + CX^2  -  2(BX)(CX) cos (CXB)

2(CX)^2  = 3(CX)^2 + (CX)^2  - 2(√3*CX * CX)  cos (CXB)

-2(CX)^2  =    -  2√3 * (CX)^2 cos (CXB)

1 / √3  =  cos (CXB)

So  the sin  of  CXB   =    √ [ 3 - 1]/ √3   =    √(2/3)

So....by the Law of Sines

sin XCB  / BX  =  sin (CXB) / BC

sin XCB / (√3 CX)  =  √(2/3) / (√2 CX)

sin XCB =   √3 * (√2/ √3)

                    _________     =      1

                           √2

sin (XCB)  =  1  = sin ACB   

arcsin (1)  =  ACB  =  90°

First, we can solve the inequalities as if they were equalities, so we solve the equation 3x - 5y = -6 and x + 2y = 3.  Eliminating y, we get x = 3/11.  Then y = 15/11, so (x,y) = (3/11, 15/11) is one corner of the region R.  The inequalities are symmetric around the origin, so the region R is also symmetric around the region.  So R is the rectangle with vertices (-3/11, 15/11), (3/11, 15/11), (-3/11, 15/11), and (-3/11, -15/11).

The maximum value of x + y is then given by the corners of the rectangle R, so the maximum is 3/11 + 15/11 = 18/11.

Solve for x:
(9 x^2 - 16)^3 + (16 x^2 - 9)^3 = (25 x^2 - 25)^3

Expand out terms of the left hand side:
4825 x^6 - 10800 x^4 + 10800 x^2 - 4825 = (25 x^2 - 25)^3

Expand out terms of the right hand side:
4825 x^6 - 10800 x^4 + 10800 x^2 - 4825 = 15625 x^6 - 46875 x^4 + 46875 x^2 - 15625

Subtract 15625 x^6 - 46875 x^4 + 46875 x^2 - 15625 from both sides:
-10800 x^6 + 36075 x^4 - 36075 x^2 + 10800 = 0

The left hand side factors into a product with seven terms:
-75 (x - 1) (x + 1) (3 x - 4) (3 x + 4) (4 x - 3) (4 x + 3) = 0

Divide both sides by -75:
(x - 1) (x + 1) (3 x - 4) (3 x + 4) (4 x - 3) (4 x + 3) = 0

Split into six equations:
x - 1 = 0 or x + 1 = 0 or 3 x - 4 = 0 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Add 1 to both sides:
x = 1 or x + 1 = 0 or 3 x - 4 = 0 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Subtract 1 from both sides:
x = 1 or x = -1 or 3 x - 4 = 0 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Add 4 to both sides:
x = 1 or x = -1 or 3 x = 4 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Divide both sides by 3:
x = 1 or x = -1 or x = 4/3 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Subtract 4 from both sides:
x = 1 or x = -1 or x = 4/3 or 3 x = -4 or 4 x - 3 = 0 or 4 x + 3 = 0

Divide both sides by 3:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or 4 x - 3 = 0 or 4 x + 3 = 0

Add 3 to both sides:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or 4 x = 3 or 4 x + 3 = 0

Divide both sides by 4:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or x = 3/4 or 4 x + 3 = 0

Subtract 3 from both sides:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or x = 3/4 or 4 x = -3

Divide both sides by 4:
 
x = 1   or    x = -1    or    x = 4/3    or    x = -4/3    or    x = 3/4    or    x = -3/4

a=1;b=1;p=0; cycle:d=a*10+b;if(d%(a*b)==0, goto loop, goto next); loop:printd,", ",;p=p+n; next:b++;if(b<10, goto cycle, 0);b=1;a++;if(a<10, goto cycle, 0);print"Total = ",p

11 , 12 , 15 , 24 , 36 , Total = 98

The probablility that I can get a crumpled up piece of paper into the recycling bin is 211/243.  So, if I shoot 6 pieces of paper, then the probablity I can get at least two into the recyccling bin is
\(\left( \frac{211}{243} \right)^6 + 6 \left( \frac{211}{243} \right)^5 \left( \frac{22}{243} \right) + 15 \left( \frac{211}{243} \right)^4 \left( \frac{22}{243} \right)^2 + 25 \left( \frac{211}{243} \right)^3 \left( \frac{22}{243} \right)^3 + 6 \left( \frac{211}{243} \right)^2 \left( \frac{22}{243} \right)^4,\)
because I get two into the recycling bin if I get two, three, four, five or six.  This works out to about 77.91%.

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Well, since we know the remainder is 5, We can see that x³+30x-c-5 when divided by x-5 leaves no remainder at all.

Given this, we can use knowledge to see that:

(x-5)(x²+5x+55) is equal to x³+30x-c-5.

We can multiply 55 and -5 to get -275, so c is 270.

-Random Person :)   

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BTW THIS WAS AN EDIT, THE ANSWER FOR C is 270 NOT -270!!!