SecretSquirrel3

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

Area of a circle = pi*r^2          (*)

they give you the diameter of the circle so using the relationship btwn diameter, d, and radius, r ---> r = d/2

thus the radius of this particular circle is 15/2.

Now plugging the values into equation (*), A = pi*(15/2)^2 -----> A = 225pi/4

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) 

First solve the quadratic by using the quadratic formula, x = (-(-11)+/- sqrt((-11)^2-4(5)(4)))/2(5) ---> x = (11+/- sqrt(41))/10. 

sps. a = (11+sqrt(41))/10 and b = (11-sqrt(41))/10

then a+b just equals 22/10 = 11/5 

the question asks for the reciprocal therefore the answer is 5/11.

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                -x^2

x^2-10x-10 = -4x-16                step 2: add 4x to both sides

      +4x          +4x

x^2-6x-10 = -16                     step 3: add 16 to both sides

          +16    +16

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).