Wednesday
June 19, 2013

Posts by Reiny


Total # Posts: 22980

geometry
5x + 8x + 11x = 60 , hit the wrong key

geometry
Since the sides must be in the same ratio... let the sides of the new triangle be 5x, 8x, and 11x 5x + 8x + 11x = 80 solve for x, then find 5x

Maths
a) for arithmetic: t3 = a+2d = 4/3 t6 = a+5d = 9 subtract them 3d = 23/3 d = 23/9 then a+2(23/9) = 4/3 a = -34/9 first 6 terms are: -34/9, -11/9 , 4/3 , 35/9 , 58/9 , 9 well, that checked out ok b) for geometric t3 = ar^2 = 4/3 t6 = ar^5 = 9 divide them r^3 = 27/4 = 54/8 , so ...

Math for Educators
like Joe said, the first condition limits you to odd numbers the second condition limits you to the numbers 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 (every odd multiple of 3 + 1) the third condition limit you to the numbers 5 9 13 1721 25 29 33 37 41 45 49 53 57 61 65 69...

math
yes, you are correct

Math
This is like finding the LCM of 36 and 48 36 = 2x2x3x3 48 = 2x2x2x2x3 LCM = 2x2x2x2x3x3 = 144 So it will happen 144 from Sept 17, 2012 BTW, Sept 17, 2012 was a Monday , not a Friday , so I will let you sort out the rest. http://www.searchforancestors.com/utility/dayofweek.html

Check on Test : Test
This looks like the reply I tried to send yesterday and wouldn't work. My square roots signs, which didn't come our correctly are done on my Mac by using " Option V " This has worked for me for years. Here is the original reply I tried posting √(5x-4) =...

maths
Ok, now you have my curiosity aroused. What is "multiplication using glazed paper" ??????

calculus
(1/3)x^(-2/3) + (1/2)y^(-1/2) dy/dx = 0 at (1,1) (1/3)(1) + (1/2)(1)dy/dx =0 dy/dx =-(1/2) / (1/3) = -3/2 so y-1 = (-3/2)(x-1) 2y - 2 = -3x + 3 2y = -3x + 5 y = (-3/2)x + 5/2 Damon, I had the same problem, now it seems to work

Precalc
let tan^-1 (-√3/3) = Ø then tanØ = -√3/3 I know tan 30° or tan π/6 = +√3/3 Ø must be in quadrants II or IV in degrees: Ø = 180-30=150° or Ø = 360-30 = 330° in radians: Ø = π - π/6 = 5π/...

Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>

For Further Reading

Search
Members
Community