To find the area of the arena, you will need to find the areas of the rectangular spaces and the 2 semicircles. Because the formulas are given, I will just substitute in the values and show the work for finding the areas.
To find the perimeter, you will look at the distances of lines that take you around the space. Because two of these spaces are half circles, you will need to find the circumference of the full circle.
Also, the answers need to be given in meters, so all units given in centimeters will be divided by 100 to convert them to meters.
Perimeter:
C= 3.14 x 20 m
C = 62.8 meters
62.8 + 8 + 25 + 8 + 5 + 8 + 10 + 8 + 40= 174.8 meters for the Perimeter
Area:
A = 25 x 8
A = 200 square meters
A = 10 x 8
A = 80 square meters
A = 20 x 40
A = 800 square meters
A = 3.14 x 10^2
A = 314 square meters
Total Area: 314 + 800 + 80 + 200= 1394 square meters
Answers:
- a) x = 9
- b) arc JK = 68
- c) arc MJ = 112
- d) arc LMK = 248
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Explanation:
Arcs JK and KL form a semicircle, so they add to 180 degrees
(arcJK) + (arcKL) = 180
(5x+23) + (17x-41) = 180
22x-18 = 180
22x = 180+18
22x = 198
x = 198/22
x = 9
Then you'll use this x value to find arc JK and arc KL
arc JK = 5x+23 = 5*9+23 = 68
arc KL = 17x - 41 = 17*9-41 = 112
Since central angles MNJ and KNL are vertical angles, this means minor arcs MJ and KL are congruent arcs. So arc MJ is also 112 degrees
Arc LMK is basically nearly everything of the full circle, but we exclude out the portion from L to K (the shorter distance)
arc LMK = (full circle) - (measure of minor arc LK)
arc LMK = 360 - 112
arc LMK = 248
Answer: 12 friends.
Step-by-step explanation:
the data we have is:
Mei Su had 80 coins.
She gave the coins to her friends, in such a way that every friend got a different number of coins, then we have that:
The maximum number of friends that could have coins is when:
friend 1 got 1 coin
friend 2 got 2 coins
friend 3 got 3 coins
friend N got N coins
in such a way that:
(1 + 2 + 3 + ... + N) ≥ 79
I use 79 because "she gave most of the coins", not all.
We want to find the maximum possible N.
Then let's calculate:
1 + 2 + 3 + 4 + 5 = 15
15 + 6 + 7 + 8 + 9 + 10 = 55
now we are close, lets add by one number:
55 + 11 = 66
66 + 12 = 78
now, we can not add more because we will have a number larger than 80.
Then we have N = 12
This means that the maximum number of friends is 12.
If you're using a few larger intervals, then your histogram looks more stocky. If you imagine drawing one, it's because you're adding more values into the same category which can make the difference between two intervals much more noticeable. If you're using smaller intervals, however, you can much more accurately assess the difference between two different intervals. For that reason, the transition between one and another interval would look much more 'fluid'.
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