961. lcm(3,4,5) = 60 so k will be in the form ----> k = 60n + 1 for some integer n
the problem asks for largest k less than 1000 so: k < 1000 ----> 60n+1 < 1000
solving for n the inequality yields n < 17 and since we are looking for greatest int less than 1000 then n will equal 16 in this case to maximize k.
plugging in n = 16
k = 60(16) + 1 = 961
6. from A there are only two paths you can take (A -> C and A -> D) from either C or D there are three paths each and each subsequent choice leads to a node for which there are only possible path remaining. (i.e. A to C to F which means that from F you can only move to D then to E then to B; one path remaining at each of the remaining 3 nodes)
Set the two equations equal and solve.
2x^2-10x-10 = x^2-4x-16 step 1: subtract x^2 from the both sides
x^2-10x-10 = -4x-16 step 2: add 4x to both sides
x^2-6x-10 = -16 step 3: add 16 to both sides
x^2-6x+6 = 0 step 4: use quadratic formula x = (-b +/- sqrt(b^2-4ac))/2a
x = (-(-6) +/- sqrt((-6)^2-4(1)(6)))/2(1) step 5: plug in
x = (6 +/- 2*sqrt(3))/2 step 6: simplify
x = 3 - sqrt(3); 3 + sqrt(3) answers
The formula for compounded interest annually is P(1+r)^y where P = the initial investment, r = interest rate (decimal), and y = number of years. So in this case, plugging in the values given -> 800(1+0.08)^10 = 1727.14. This is the amount of money Ms. Montgomery has after 10 years (her initial investment is included).