CPhill

Here's another way to see this.......2 men are just 2/3rds the labor of 3 men.....therefore....it should take them 3/2 times as long to do the job.....so....

3/2  * 20    =  30 minutes

y=9.65x+43    y=8.40x+58

Let's just set the equations equal

9.65x + 43  = 8.40x + 58      subtract 8.40x, 43 from both sides

1.25x  = 15      divide both sides by 1.25

x = 12

And we can use the first equation to find y

y = 9.65(12) + 43  =   

      115.8  +  43   =

       158.8

So....the solution is  {x , y}   = { 12, 158.8}

Here's my approach to this one

Let him start at(0,0)......first, he flies to the point given by:  (29cos 60, 29sin60)  = (14.5, 25.1)

Then ....he flies 20 km directly south  and ends up at (14.5, 5.1)

Then......he flies to the point given by (14.5 + 10 cos150, 5.1 + 10 sin150) = (5.83975, 10.1)

The distance from this point to the origin [ where he started]  =

sqrt [ 5.83975^2 + 10.1^2]  ≈  11.67 km ≈ 11.7 km

[ Just as Melody said.....!!!!  ]

sin(3Q)=cos(Q-6)

Since the sine and the cosine are co-function angles......we must have that

90 - 3Q   =  Q - 6        add 3Q and 6 to both sides

96  = 4Q    divide both sides by 4

24  = Q

Check

sin (3 * 24)  = cos ( 24 - 6)

sin (72)  = cos (18)    ........and this is true

Magnitude  = sqrt [ 5.5^2 + ( -6.5)^2 ]  = about 8.5  units

The answer is "C"

I don't understand the question......

2² + 5²   =

4 + 25  =

29

I)     5/ √2   multiply top/bottom by √2  = 5√2 / [√2 * √2]  = [5√2] / 2  

II)  10/ ∛4   multiply top/bottom by  4^2/3  = 10 * 4^2/3 / 4  = [ 5 * 4^2/3 ] / 2  

III)  [ 5 + √2] / [ 5 - √2]     multiply top/bottom by [5 + √2]

     ([5 + √2] [ 5 + √2]) / [ 25 - 2]   =  [25 + 10√2 + 2] / 23  = [ 27 + 10√2] / 23

IV)  √3 / arctan (√3)  =   √3 / [pi/ 3] =   [3 √3] / pi

V)   3 / [ 6^1/5 / 6^1/7 ]   = [3 * 6 ^1/7 ] / 6^1/5      multilply top/bottom by 6^4/5 =

     [ 3 * 6^1/7 * 6^4/5] / [6^1/5 * 6^4/5]  = [ 3 * 6^33/35]  / [ 6]  =     6^33/35 / 2

NO!!

15.0   = 15

150  = 150