CPhill

https://web2.0calc.com/questions/simplify_39168

1/x  + 1/y =  1/15

( x + y)  / (xy)  = 1/15

(xy)    / ( x + y)  =  15

xy =  15 ( x + y)

xy  =15x + 15y

xy - 15y =  15x

y ( x - 15)  =  15x

y =   15x / ( x - 15)

x          y

16      240

18       90

20       60

24       40

30       30

40       24

Smallest  x + y =  40 + 24  =  64

The pyramid  EFGH  will  have the same base and the  sme height as pyramid DEFG....so....its volume also   =  18

-6x + 3y  = -18

3y = 6x -18      divide through by 3

y = 2x - 6

The slope of the line we're looking for = 2

x intercept  of  14x +7y - 14  =0

Let y = 0

14x - 14 =  0

14x  =14    divide through by 14

x =1

So.....we have the point (1,0)

Equation of line

y = 2 (x -1)

y = 2x  - 2

2x - y  - 2   =  0

                    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