CPhill

If it takes 15 laborers 40 days to dig 40 acres...then they are digging 1 acre per day. And each laborer must dig 1/15 of an acre per day.

So 20 laborers would dig 20(1/15) acres = (4/3) acres per day. So 60/(4/3) = 45 days to dig 60 acres.

We can also cut a circle with radius R in half by removing a circular area with a radius = R/√2 from within the circle...

To prove this, we have

pi*R^2 - pi*r^2 = (1/2)*pi*R^2   ......where r is the radius we're looking for........so.......

(1/2)pi*R^2 = pi*r^2   divide both sides by pi

(1/2)R^2 = r^2   take the positive root of both sides

r = (1/√2)R = R/√2

Note that......as long as the circular area removed lies totally witin the circle, it doesn't have to be concentric with the larger circle!!!

Note that (n+2)! is greater than n! by (n+2)(n+1)......so using 398 = n, we have

(398+2)*(398 + 1) = 400*399 = 159,600

I can just add 1 to yours...and I'd have you beat..of course, you could add 1 to mine and have me beat...and we could continue this indefinitely.....thus proving that there IS no "highest" number !!!

First let x = 0

So

3(0) - y = 5

-y = 5     so... y = -5      and there's one point.... (0, -5)

Next. let y = 0 and we have

3x- 0 = 5

3x = 5    divide both sides by 3  and x= 5/3

So the second point is (5/3, 0)

Now....just plot both points on a graph and draw a line through them !!!

x^-1 = 1/x

So we have

1/x > x

Note that there are NO positive integer values that make this true!!! These two functions intersect at x = 1 and that's the first positive integer. At all x's greater than this, 1/x < x  !!!!

We can convert the minutes to a decimal = 52/60 = .86666

So, sin(37.86666) = about .613

Because the function in the numerator is > 0 for all real numbers (thus, the graph never intersects the x axis), we only need to concern ourselves with the denominator.....!!!

This function will be undefined when the denominator = 0

So.....setting 2x^2 + x - 6 to 0 and factoring, we have

(2x -3)(x + 2) = 0       and setting each factor to 0, we have x = -2 and x = 3/2...so the graph is undefined at these two values (it has vertical asymptotes at these two values)....we just need to look at 3 intervals

When x < -2, the rational function > 0   (test -3 to see this is true)

And when x > -2 but < 3/2, the rational function < 0  (test 0)

And when x > 3/2 , the rational function > 0 (test 2)

So, the solution is

(-∞, -2) U (3/2, ∞)

First the number of possible sets = C(52,2) = 1326

a) Two spades.......there are 13 spades and we want to choose any set of 2 of them = C(13,2) = 78

So.... 78/1326 = 1/17 = about .059 = about 5.9%

b) Both are the same suite.....there are 13 cards in one suit.....so C(13,2) = 78. But there are 4 suites, so we have  ...

[4*C(13,2)] /1326 = 4/17 = about .235 or about 23.5%

c) Two kings .....   there are 4 kings and we want to choose 2 of them......so C(4,2) = 6

And..... 6/1326  = 1/221 = about .0045 = about .45%

c^3 + 4c > 5c^2      subtract 5c^2 from both sides

c^3 + 4c - 5c^2 > 0   rearrange

c^3 - 5c^2 + 4c > 0     set this = 0 and factor

c^3 - 5c^2 + 4c = 0

c(c^2 -5c + 4) = 0

c(c-4)(c-1) = 0

And setting each factor to 0 , we have that c =0, c = 4 and c =1

Now, adding back the inequality sign, we have

c(c-4)(c-1) > 0   

And if c < 0 ,   c(c-4)(c-1) < 0

And if c > 0 but < 1 ,  c(c-4)(c-1) is > 0

And if c > 1 but < 4,  c(c-4)(c-1) is < 0

And if c > 4, c(c-4)(c-1) > 0