Number Two or B would be correct
The total area of the complete lawn is (100-ft x 200-ft) = 20,000 ft².
One half of the lawn is 10,000 ft². That's the limit that the first man
must be careful not to exceed, lest he blindly mow a couple of blades
more than his partner does, and become the laughing stock of the whole
company when the word gets around. 10,000 ft² ... no mas !
When you think about it ... massage it and roll it around in your
mind's eye, and then soon give up and make yourself a sketch ...
you realize that if he starts along the length of the field, then with
a 2-ft cut, the lengths of the strips he cuts will line up like this:
First lap:
(200 - 0) = 200
(100 - 2) = 98
(200 - 2) = 198
(100 - 4) = 96
Second lap:
(200 - 4) = 196
(100 - 6) = 94
(200 - 6) = 194
(100 - 8) = 92
Third lap:
(200 - 8) = 192
(100 - 10) = 90
(200 - 10) = 190
(100 - 12) = 88
These are the lengths of each strip. They're 2-ft wide, so the area
of each one is (2 x the length).
I expected to be able to see a pattern developing, but my brain cells
are too fatigued and I don't see it. So I'll just keep going for another
lap, then add up all the areas and see how close he is:
Fourth lap:
(200 - 12) = 188
(100 - 14) = 86
(200 - 14) = 186
(100 - 16) = 84
So far, after four laps around the yard, the 16 lengths add up to
2,272-ft, for a total area of 4,544-ft². If I kept this up, I'd need to do
at least four more laps ... probably more, because they're getting smaller
all the time, so each lap contributes less area than the last one did.
Hey ! Maybe that's the key to the approximate pattern !
Each lap around the yard mows a 2-ft strip along the length ... twice ...
and a 2-ft strip along the width ... twice. (Approximately.) So the area
that gets mowed around each lap is (2-ft) x (the perimeter of the rectangle),
(approximately), and then the NEXT lap is a rectangle with 4-ft less length
and 4-ft less width.
So now we have rectangles measuring
(200 x 100), (196 x 96), (192 x 92), (188 x 88), (184 x 84) ... etc.
and the areas of their rectangular strips are
1200-ft², 1168-ft², 1136-ft², 1104-ft², 1072-ft² ... etc.
==> I see that the areas are decreasing by 32-ft² each lap.
So the next few laps are
1040-ft², 1008-ft², 976-ft², 944-ft², 912-ft² ... etc.
How much area do we have now:
After 9 laps, Area = 9,648-ft²
After 10 laps, Area = 10,560-ft².
And there you are ... Somewhere during the 10th lap, he'll need to
stop and call the company surveyor, to come out, measure up, walk
in front of the mower, and put down a yellow chalk-line exactly where
the total becomes 10,000-ft².
There must still be an easier way to do it. For now, however, I'll leave it
there, and go with my answer of: During the 10th lap.
To obtain the total surface we have to calculate the surface of the 4 triangles and add up the areas (remember that the area of a triangle is (b*h)/2 , b is the base, h is the height ).
We will caculate first the area of the base triangle for that we considerer the fact that it is an equilateral triangle with sides of lenght 6 cm, now we calculate the height, I am going to draw please wait a moment
using the pythagorean theorem we have that
![\begin{gathered} h^2=6^2cm^2-3^2\operatorname{cm}=27cm^2 \\ h=\text{ }\sqrt[]{27\text{ }}cm \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2cm%5E2-3%5E2%5Coperatorname%7Bcm%7D%3D27cm%5E2%20%5C%5C%20h%3D%5Ctext%7B%20%7D%5Csqrt%5B%5D%7B27%5Ctext%7B%20%7D%7Dcm%20%5Cend%7Bgathered%7D)
Then, the area of the triangle is 6*h/2 = 3h = 15.59 cm^2.
Now we calculate the area of the other 3 triangles, notice that those triangles have the same base and height so we will calculate for one of them and multiply by 3. From the image we know that the height is 15cm and the base is 6 cm so the area is 45cm^2, and 45*3 cm^2 = 135cm^2.
Finally we add up all the areas:
Answer:
a) 
, b) 
, c) 
, d) 
Step-by-step explanation:
a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:
b) The general solution is found after separating variables and integrating each sides:
Where 
is the time constant and 
c) The time constant is:
The particular solution of the differential equation is:
d) The amount of radium after 300 years is:
