e and r = 2e and j=1 <10
e + 2e + 1 <10
3e<9
e <3 soooo r < 6
a ~ b/ c2 k k is proprtionality constant calculate k by subbing in the initial values
10 = 8/2^2 k
k =5
a = 5b/c2
a = 5(20)/52 = .........
4x = 27 log both sides
x log 4 = log27
x = log 27 / log4 similarly 9y = 16 yields y = log 16 /log 9
xy becomes (log 27 log16) / (log4 log9) = 3
From an online summation calculator:
n/2 ( 2a + (n-1)d) n = 670 a = 1 d = 3
670/2 (2+ 669(3)) = 673015
Yah....that is why I was asking how kanisan got only '4'.....I wanted to see k's work so far....
I got 4 and -2 also.... ~EP
How did you get '4' Did you get another value for 'k' besides 4 ??? (Like when you factored a quadratic for k )
Whenx = -2 f(x) = y = 0
0 = (-2)2 + k(-2) +8
0 = 4 -2k + 8 I think you should be able to solve for 'k' now......
3/8 at 8% 5/8ths at 10%
3/8 x * .08 + 5/8 x *.10 = 92000 solve for x
(3x+1)(x+5) = 0 means x = -1/3 , -5
(x-3)(x-5) =0 means x = 3,5 just sayin'
