The right triangle has a base = 19 - 13 = 6
By the Pythagorean Theorem, the height = sqrt (10^2 - 6^2) = sqrt (100 - 36) = sqrt (64) = 8
This is also the height of the rectangle on the left
So....the perimeter = 13 + 8 + 19 + 10 = 50
F
E D
tan D = 4 sin D
EF / DE = 4 EF / FD
1/ DE = 4 / FD
FD = 4 DE
sin F = DE / FD
sin F = DE / ( 4 DE)
sin F = 1 / 4
See here : https://web2.0calc.com/questions/3d-geometry_9
The number of possible outcomes = C(6,2) = 15
The only way a "2" can be achieved is by painting face "1" and face "2" both blue
So......14 ways
By the triangle inequality
21 + x > 29 or 21 + 29 > x
x > 8 x < 50
So the possible lengths are
49 - 9 + 1 = 41
f(x) + g(x) lead to the same result when added together.....so...
f (x) + g(x) = -2 - x and
f(x) + g(x) = x^3 + 2x + 2
Is not possible because adding the same two polynomials can't produce different outcomes
(p (x))^2 =
( x^4 - 3x + 2) ( x^4 - 3x + 2)
As you can see.....there will not be any multiplications of any two terms that will lead to an x^3 term !!!
Midpoint formula :
( add x coordinates / 2 , add y coordinates / 2 ) .....so.....
[ ( 9 -1) / 2 , ( 8 - 2) / 2 ] =
[ 8 / 2 , 6 / 2 ] =
( 4 , 3 ) = meeting point
j = m / k where m is the proportionality constant
42 = m /56
56* 42 = m = 2352
So
j = 2352 / 32 = 73.5
See my answer here : https://web2.0calc.com/questions/help_77088
Or see Melody's method here :https://web2.0calc.com/questions/let-xyz-be-an-equilateral-triangle-centered-at-o
