CPhill

29.85 /  1.5  =   19.9 per pound

20.28 / 1.2   =  16.9 per pound

150 + 2W  = 195 - 1W     where W is the number of weeks it will take before they weigh the same

Add 1W to both sides......subtract 150 from both sides

3W   = 45    divide both sides by 3

W  = 15 weeks

8.6km x  1000m/1km x  100cm / 1m   =  860,000 cm

(Probability of rolling a 6 with 2 dice) = 5/36

sin t = y/r

r = √[ x^2 + y^2 ]  = √[ (2)^2 + (-4)^2 ]  =  √20  =   2√5

sin t  =  -4/ [2√5]  =  -2 / √5  ≈   -0.89

And to find t, we have

arcsin ( -2 / √5 )  = t  = -63.43°  =  296.57°  .....since (2,-4)  occurs in the 4th quadrant

Ah.......thanks, raindropsfurrever and heureka......I wasn't thinking in terms of "h" being the height

BTW........here's how heureka gets his formula......

Let a and  be legs......

Then the area  is  (1/2) ab

Now.......draw an altitude, h, to the hypotenuse, C

Then the area  is   (1/2)ch

Equating areas,  we have

(1/2)AB  = (1/2) ch    

ab  = ch      divide by  c

ab/c  = h

We can solve this with a proportion

6/15  = H/40     cross-multiply

6 * 40 = 15 * H      

240 = 15 * H       divide both sides by 15

240/15 = H  =  16 ft = the flagpole height

-20 – 17j = -14j + 7    add 17j to both sides.....subtract 7 from both sides

-27 =  3j       divide both sides by 3

-9 = j

12W - 12 = 13W   subtract 12W from both sides

-12 = W