geno3141

Given a rhombus with diagonals  14  and  48  --  what is the perimeter?

Since the diagonals of a rhombus are perpendicular to each other, four congruent right triangles are formed.

Each of these triangles have bases of  7  and  24  so we can use the Pythagorean Theorem to find the side

of the rhombus.

c2  =  72 + 242     --->     c2  =  49 + 576     --->     c2  =  625     --->     c  =  25.

The perimeter will be  4 x 25  =  100

For your second question:  https://web2.0calc.com/questions/quadrilateral-problem_1

Given a rhombus with diagonals  14  and  48  --  what is the perimeter?

Since the diagonals of a rhombus are perpendicular to each other, four congruent right triangles are formed.

Each of these triangles have bases of  7  and  24  so we can use the Pythagorean Theorem to find the side

of the rhombus.

c²  =  7² + 24²     --->     c²  =  49 + 576     --->     c²  =  625     --->     c  =  25.

The perimeter will be  4 x 25  =  100

The perimete of a rectangle is 98 and has a diagonal of length 41.

If the perimeter is 98, then   2·length + 2·width  =  98   or   length + width  =  49.

Call the length  x  then the width will be  49 - x.

If you draw the rectangle with one diagonal, you will have drawn two congruent right triangles.

Consider either one of them:  one side will be  x  the other side will be  49 - x  and the hypotenuse will be  41.

x² + (49 - x)²  =  41²     --->    x² + 2401 - 98x + x²  =  1681     --->     2x² - 98x + 720  =  0

                                     --->     x² - 49x + 360  =  0     --->     (x - 9)(x - 40)  =  0     --->     x  =  9  or  x  =  40

--->     length  =  9   and   width  =  40     (or vice versa) 

--->     area  =  9 x 40  =  360 

Second problem:

1 + a² + a⁴ + ...  =  1/(1 - a²)

ab + a³b + a⁵b + ...  =  ab/(1 - a²)

b² + a²b² + a⁴b² + ...  =  b²/(1 - a²)

ab³ + a³b³ + a⁵b³ + ...  =  ab³/(1 - a²)

^...

Sum  =   1/(1 - a²)  +  ab/(1 - a²)  +  b²/(1 - a²)  +  ab³/(1 - a²)  +  ...

         =   1/(1 - a²) · [ 1 + ab + b² + ab³ + b⁴ + ab⁵ + b⁶ + ab⁷ + ...]

         =   1/(1 - a²)   [ (1 + b² + b⁴ + b⁶ + ...)  +  (ab + ab³ + ab⁵ + ab⁷ + ...) ]

         =   1/(1 - a²) · [ ( 1/(1 - b² ) + ( ab/(1 - b²) ) ]

         =   [ 1/(1 - a²) · 1/(1 - b² ) ] · [ 1 + ab ]

         =  ( 1 + ab ) / [ (1- a²)(1 - b²) ] 

First problem:

The first row is a geometric series with first term  1  and common ratio  a  whose sum is  1/(1 - a).

The second row is also geometric with first term  b  and common ratio  a  whose sum is  b/(1 - a).

The third row is also geometric with first term  b²  and common ratio  a  whose sum is  b²/(1 - a).

The sum of the rows is:  1/(1 - a)  +  b/(1 - a)  +  b²/(1 - a)  + ...  =   1/(1 - a) · [ 1 + b + b² + ... ]

The series  1 + b + b² + ...  is geometric with first term  1  and common ratio  b  whose sum is  1/(1 - b).

Therefore, the sum of the rows is:  1/(1 - a) · [ 1 + b + b² + ... ]  =  1/(1 - a) · 1/(1 - b)  =  1 / [ (1 - a)(1 - b) ]

If the side of an equilateral triangle is "a", the altitude of that triangle is:  a·sqrt(3)/2.

Since each of the equilateral triangles have a base of 1, the height of each of these triangles is: sqrt(3)/2.

So the distance from one vertex of the large square to its opposite vertex is  sqrt(3)/2 + 1 + sqrt(3)/2  or  1 + sqrt(3).

This is the length of each diagonal of a square.

If "d" is the length of a diagonal of a square, its area is:  A  =  ½·d².

So, the area of this square is:  ½·( 1 + sqrt(3) )²  or  2 + sqrt(3).

      2sin²(x) - 3sin(x) + 1  =  0

( 2sin(x) - 1 )( sin(x) - 1 )  =  0

Either:   2sin(x) - 1  =  0     --->     2sin(x)  =  1     --->     sin(x)  =  1/2     --->     x  =  sin^-1( 1/2 )   

           --->    x  =  pi/6  or  x  =  5pi/6

Or:     sin(x) - 1  =  0     --->     sin(x)  =  1     --->     x  =  sin^-1(1)      --->     x  =  pi/2

[ plus or minus all multiples of 2pi ]

4 of the cubes will have 3 sides painted  =  12 sides

20 of the cubes will have 2 sides painted  =  40 sides

28 of the cubes will have 1 side painted  =  28 sides

Total number of sides painted  =  80 sides

Total number of sides  =  6 x 64  =  384 sides

Pobability that the face showing is painted  =  80 / 384  =  4/19

You said that they can't be composite but you didn't say that they couldn't be negative:

[By definition, zero is neither prime nor composite.]

My suggestion:

     -2      7    -5

     -3      0     3

      5     -7     2

This has a sum of zero.

The sum could be either a 3, a 5, a 7, or an 11.

3  =  (1,2) or (2,1)

5  =  (1,4), (4,1), (2,3) or (3,2)

7  =  (1,6), (6,1), (2,5), (5,2), (3,4), or (4,3)

11  =  (5,6) or (6,5)

There are 14 winning possibilities out of a possible 36 possibilities:  14/36  =  7/18