the parabola y=(x-h)^2 + k has vertex at (h,k). That should help
I think you mean C'(x) = 2x+30 so, C(x) = x^2 + 30x + k since C(2) = 220, 220 = 4+60+k k = 156 so, C(x) = x^2 + 30x + 156
if the distance is x, x/40 = cos 32° evaluate that and solve for x
the hexagon consists of 6 triangle, each with base=6 and height=3√3
5t+8b = 1376 6t+7b = 1426 I don't get integer values for the prices (in cents). Is there a typo?
Math log equations/exponential form
Beats me. A log is a log, not an exponential form. You have 1/2 log2 + 2logw + 3logx - 5logy - 7logz similarly for (b) do you have an example of another problem which is solved, indicating what you wish to achieve?
sin2x = cos(x+30) when x=120, sin 240 = cos 150 There are other values, but you have to solve a 4th-degree equation to get them. 2sinx cosx = cos30 cosx - sin30 sinx square both sides, collect terms, convert sin^2 = 1-cos^2, and square both sides again. Gets messy
Algebra II Part 2
An excellent grapher can be found at http://rechneronline.de/function-graphs/ Just enter -3*cos(x) as your function
algebra II Part 2
for h=1, 1 = 7 cos(pi/3 t) cos(pi/3 t) = 1/7 pi/3 t = 1.427 t = 1.363 and so on for the others
visit http://rechneronline.de/function-graphs/ enter your function. Change the range to be -20 to 20 and you will see the graph.
For Further Reading