Answer:
m = 6 7/8 or 55/8.
Step-by-step explanation:
6= m - 7/8
6 + 7/8 = m - 7/8 + 7/8
m = 6 7/8 or 55/8.
The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is
(x - x^2)^2 * sqrt(3)/4.
Integrating from x = 0 to x = 1, we have
[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4
= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144.
Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...
Price is reduced by 45%. Which means that the new price is 55% of the original.
Therefore, (55/100)*17 = $9.35 =new price
Hey
périmètre = 2 ( x + 2 x - 5) = 2 x + 4 x - 10 = 6 x - 10
Let's say that in the beginning he weighted x and at the end he weighted x-y, y being the number of kg he wanted to loose.
first month he lost
y/3
then he lost:
(y-y/3)/3
this is
(2/3y)/3=2/9y
explanation: ((y-y/3) is what he still needed to loose: y minus what he lost already
and then he lost
(y-2/9y-1/3y)/3+3 (the +3 is his additional 3 pounts)
(y-2/9y-1/3y)/3-3=(7/9y-3/9y)/3+3=4/27y+3
it's not just y/3 because each month he lost one third of what the needed to loose at the current time, not in totatl
and the weight at the end of the 3 months was still x-y+3, 3 pounds over his goal weight!
so: x -y/3-2/9y-4/27y-3=x-y+3
we can subtract x from both sides:
-y/3-2/9y-4/27y-3=-y+3
add everything up:
-19/27y=-y+6
which means
-19/27y=-y+6
y-6=19/27y
8/27y=6
4/27y=3
y=20.25
so... that's how much he wanted to loose, but he lost 3 less than that, so 23.25
ps. i hope I didn't make a mistake in counting, let me know if i did. In any case you know HOW to solve it now, try to do the calculations yourself to see if they're correct!
