<h3>Given</h3>
- a rectangle x units wide and y units high divided into unit squares
<h3>Find</h3>
- The total perimeter of the unit squares, counting each line segment once
<h3>Solution</h3>
For each of the y rows of squares, there are x segments at the top, plus another x segments at the bottom. The total number of horizontal segments is then
... horizontal segment count = (y +1)x
Likewise, for each of the x columns of squares, there are y segments to the left, plus another y segments to the right of the entire area. Then the total number of vertical segments is
... vertical segment count = (x+1)y
The total segment count is ...
... total segments = horizontal segments + vertical segments
.. = (y+1)x +(x+1)y
... total segments = 2xy +x +y
_____
<u>Check</u>
We know a square (1×1) has 4 segments surrounding it.
... count = 2·1·1 +1 +1 = 4 . . . . (correct)
We know the 3×3 window in the problem statement has 24 segments.
... count = 2·3·3 +3 +3 = 18 +3 + 3 = 24 . . . . (correct)
We know a 1×3 row of panes will have 10 frame elements.
... count = 2·1·3 +1 +3 = 6 +1 +3 = 10
It looks like our formula works well.
Is that the full
question, is their a picture with it
Answer:
$3, $11
Step-by-step explanation:
The profit function P(x) will be zero when either of its factors is zero. The left factor (x-3) is zero when x=3. The right factor (x-11) is zero when x=11.
Prices of $3 or $11 will result in $0 annual profit.
_____
<em>Additional comment</em>
A price halfway between these values, at (3+11)/2 = 7, will produce maximum profit, $32 million.
Answer:
P = 150 W * 60 sec/min * 60 min/hr = 540000 W-sec/hr
1 kW = 1000 W * 3600 sec = 3.6E6 W-sec
E (8 hrs) = 8 * .54E6 W-s = 4.32E6 W-sec
or E= 4.32/3.6 = 1.2 kW = .168 $/day
20/150 = .133 as much for a 20-watt bulb
For a LED bulb cost/day = .168 * .133 = $.022/da
Note 1 W = 1 Joule/sec
