Answer:
2³*3²
8*9=72
2³*3²²
8*81=648
gcf of 72 annd 648
72=2*2*2*3*3
648=2*2*2*3*3*3
2*2*2*3*3=72
<h2>the gcf is 72</h2>
Step-by-step explanation:
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.
Answer:
square root of 16
Step-by-step explanation:
Answer:
Step-by-step explanation:
The quadrilateral has 4 sides and only two of them are equal.
A) to find PR, we will consider the triangle, PRQ.
Using cosine rule
a^2 = b^2 + c^2 - 2abcos A
We are looking for PR
PR^2 = 8^2 + 7^2 - 2 ×8 × 7Cos70
PR^2 = 64 + 49 - 112 × 0.3420
PR^2 = 113 - 38.304 = 74.696
PR = √74.696 = 8.64
B) to find the perimeter of PQRS, we will consider the triangle, RSP. It is an isosceles triangle. Therefore, two sides and two base angles are equal. To determine the length of SP,
We will use the sine rule because only one side,PR is known
For sine rule,
a/sinA = b/sinB
SP/ sin 35 = 8.64/sin110
Cross multiplying
SPsin110 = 8.64sin35
SP = 8.64sin35/sin110
SP = (8.64 × 0.5736)/0.9397
SP = 5.27
SR = SP = 5.27
The perimeter of the quadrilateral PQRS is the sum of the sides. The perimeter = 8 + 7 + 5.27 + 5.27 = 25.54 cm
