Problem 1
x = measure of angle N
2x = measure of angle M, twice as large as N
3(2x) = 6x = measure of angle O, three times as large as M
The three angles add to 180 which is true of any triangle.
M+N+O = 180
x+2x+6x = 180
9x = 180
x = 180/9
x = 20 is the measure of angle N
Use this x value to find that 2x = 2*20 = 40 and 6x = 6*20 = 120 to represent the measures of angles M and O in that order.
<h3>Answers:</h3>
- Angle M = 40 degrees
- Angle N = 20 degrees
- Angle O = 120 degrees
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Problem 2
n = number of sides
S = sum of the interior angles of a polygon with n sides
S = 180(n-2)
2700 = 180(n-2)
n-2 = 2700/180
n-2 = 15
n = 15+2
n = 17
<h3>Answer: 17 sides</h3>
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Problem 3
x = smaller acute angle
3x = larger acute angle, three times as large
For any right triangle, the two acute angles always add to 90.
x+3x = 90
4x = 90
x = 90/4
x = 22.5
This leads to 3x = 3*22.5 = 67.5
<h3>Answers:</h3>
- Smaller acute angle = 22.5 degrees
- Larger acute angle = 67.5 degrees
Your answer would be 4.23 as the answer will be a long one, estimated.
Answer:
Total number of possible combinations are 6
Length width
23 dm 2m
21 dm 4 dm
19 dm 6 dm
17 dm 8 dm
15 dm 10 dm
13 dm 12 dm
Step-by-step explanation:
We are given that
Perimeter of rectangular garden=50 dm
Width is even number.
Length is always longer than or equal to width.
Let length of rectangular garden=x
Width of rectangular garden=y
We have to find the possible number of combinations .
Perimeter of rectangular garden=



If y=2 dm
x=25-2=23 dm
If y=4 dm
x=25-4=21 dm
If y=6 dm
x=25-6=19 dm
If y=8 dm
x=25-8=17 dm
If y=10 dm
x=25-10=15 dm
If y=12 dm
x=25-12=13 dm
If y=14 dm
x=25-14=11 dm
x<y
It is not possible
Then, possible combinations are 6
Length width
23 dm 2m
21 dm 4 dm
19 dm 6 dm
17 dm 8 dm
15 dm 10 dm
13 dm 12 dm