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                    P

 6       Z                     Y            6     

Q                  X                      R

                 6sqrt 2

Let the right angle be at P

PZ = s  = the side of the  square

QR = 6sqrt 2  ....so...QX =  3sqrt 2 =  sqrt (18)

PX  =  diagonal of the  square  =   s* sqrt (2) = the altitude of the isosceles triangle

Using the Pythagorean Theorem

PQ^2  = QX^2 + PX^2

6^2  = (sqrt 18)^2  + (s * sqrt 2)^2

36  = 18 + 2s^2

36 - 18  = 2s^2

18 = 2s^2

9 = s^2

s = 3

Here's one way to  do this (maybe not what you're looking  for )

Rewrite the equation of the  line as

4y  = 3x + 2

3x - 4y + 2  =  0

Recognize  that  the center of the circle  is (0,0)

Using  the formula for the distance  between  a point and  a line we have

abs  ( 3(0) - 4(0) + 2)  /  sqrt [ 3^2 + 4^2 ] =   2 / 5   =  .4

This is the distance from  the center of the  circle to the line

The radius of the circle = sqrt 5

Using the Pythagorean Theorem, we can  find  1/2  the length of AB as follows

(1/2)AB   = sqrt [ (sqrt 5)^2  - (.4)^2 ] = sqrt [ 5  - .16 ] =  sqrt [ 5 - 4/25] =

sqrt [  (125 - 4) / 25 ] =   sqrt [ 121 / 25] = 11/5

Then  AB =   2 (11/5)  = 22/5  =  4.4 

https://web2.0calc.com/questions/geometry-problem_97858

Here : https://web2.0calc.com/questions/geometry-problem-due-today

Nothing shown....

Here :  https://web2.0calc.com/questions/hexagon_7

n = -12 , -6 , -4 , -3 , -2 , -1, 1, 2, 3, 4, 6 ,12

The sum =   0

                                 A

                              1

                           F                

                        2

                      D                      E

                 B                                    C

Because EF  parallel to CD

AF / AE = AD / AC

1 / AE = 3 / AC

AC / AE = 3 /1 = 3

And because DE parallel to BC

AB / AD =  AC / AE

AB/ AD  =  3

AB / 3  = 3

AB = 3 * 3  =  9

So

BD  =  AB - AD  =  9  -  3  =   6

Thank youu

Just  a difference of  squares

Let  (x + y + z)  =  a

Let  (m -3)  = b

So....we have

a^2  - b^2  =  ( a + b) ( a  - b)

Sub back in for  a, b

(a + b) ( a - b)   =

( x + y + z + m - 3)  ( x + y + z - (m - 3) )   =

( x + y + z + m -3 ) (x + y + z - m + 3)

And that's it  !!!!

(a ° b) (3) =  a ( b(3) )

b(3)  = 4 -(3)^2 + 4(3)^4  =  -5 + 324  = 319

a (319)  = 2(319) + 5 - 3(319)^2 + (319)^3   =  32,157,119

(b ° a)(3)  =  b (a (3))

a(3)  =  2(3) + 5 - 3(3)^2 + 3^3   = 11

b (11) =  4 - (11)^2 + 4 (11)^4  =  58447

(a ° b)(3)   -   (b ° a)(3)   =

32,157,119   -  58447    =

32,098,672

In general  (without detail)

Let B =(0,0)    C = (18,0)

Construct  a circle  with a radius of 10 with a center at B   and another with a radius of 14 at C

Let them intersect at A

Then construct  a circle with a radius of 18 at B  and another with a radius of 6 at C

Construct a line through AB  to intersect this first circle at D   and a line through AC  to intersect this second circle at E

Find the coordinates of D , E    and use the distance  formula to find DE  =  36

Using  the  Law of Cosines  we  have  that

BC^2  =  AB^2  + AC^2  - 2(AB* AC) cos  ( BAC)

9^2  = 5^2 + 8^2  - 2(5 * 8)  cos (BAC)

81 = 25  + 64  - 80 cos (BAC)

cos ( BAC)  =  [ 81  - 25 - 64 ]  / -80    =    -8  / -80    =   1/10

And applying it once  more we  have  that

DE^2  = AE^2  + AD^2  - 2(AE * AD) cos (BAC)

DE^2  = 10^2  + 16^2  - 2 ( 10 * 16)  (1/10)

DE^2 = 100 + 256  - 32

DE^2  =  356  -  32

DE^2 = 324

DE  = 18

When  Dirk finishes, Edith has run 14km  and Foley has run 9 km

So   Folley  runs  9/14 of Edith's distance at any time

So....when  Edith finishes, Foley has run   24  (9/14)  = 108/7  km

So....he is  24 - 108/7  = 60/7 ≈ 8.57 km behind

\( x \mathbin{\spadesuit} y = \frac{x^2}{y} \)

Find a such that   \(a \mathbin{\spadesuit} (a + 1) = 9\)

a^ 2  / ( a + 1)   = 9

a^2  = 9 (a + 1)

a^2 =  9a + 9

a^2 - 9a  =  9            complete the square on  a

a^2  - 9a + 81/4  = 9 + 81/4

(a - 9/2)^2  =  117/4    take both roots

a - 9/2  = sqrt (117) /2                  a  - 9/2  =  -sqrt (117) /  2

a =   (9 + 3sqrt (13)) / 2                a = (9 - 3sqrt (13))  / 2

Rewrite as

2x^2 -2x -10 =  0 

We have the form   mx^2 + nx + c

The sum of the roots a + b =  -n / m  = (- - 2) / 2 = 2/2 =   1

a + b = 1

(a + 3) + (b + 3)   =

a + b + 6  =

1 + 6  =  

7

x^2 -2x + 1 =  (x - 1) (x -1)  = (x -1)^2    has a repeated root of x =  1

Another way is to find the discriminant of each.....the discriminant that  = 0  is the one with a repeated root

Each additional worker hired cuts the job time  by 4 days

So.....to cut the job by 20 days.....5 additional workers should be hired

x^2 - 4x + y^2 - 6y =  15            complete the square on  x,y

x^2 -4x + 4 + y^2 -6y + 9 =  15 + 4 + 9

(x -2)^2   + (y - 3)^2  =  28

This is a circle centered at  (2, 3)  with a radius of sqrt (28)  = 2sqrt 7

Largest value for x =  2 + 2sqrt(7)

Thank you! also, 3 am? how do u stay up that late :_)

See my corrected answer....