I can't read that language but I'll guess it says write a second degree equation then solve for n when d=10.



Answer: n² - 3n - 20 = 0
That doesn't factor so there is no integer n solution.
That means there are no polygons with 10 diagonals.

Some of the important "given" information is outside of the photo.
We need to know that the two triangles are similar.
And we need to know that the WHAT ? of angle M is 9/40.
Answer:

Step-by-step explanation:
Data
wide=20yd
length=40yd
Hedge area
=
If the area of rectangle is A=base * hight and our case L*W. then:

So 
From the given problem the same size is n = 70. Df = n – 1 =
70 – 1 = 69.
The population mean is u = 30, sample standard deviation is
s = 10 and the sample size is n = 70. Then,
t = x – u / s sqrt of n
1.995 = x – 30 / (10 sqrt of 70)
1.995 = x – 30 / 1.1952
X – 30 = 1.995 (1.1952)
X – 30 = 2.3844
X = 30 + 2.3844
X = 32.3844
The sample mean is 32.3844
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