To calculate the area of the black material on the flag, we need the shape and the dimensions of the black material itself.
<em>Since the question is incomplete, as the dimension of the flag and the dimension of the black material are not given, I will provide a general explanation</em>
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Assume the shape of the black material is a rectangle.
The area will be calculated as:
Take for instance;
The area is:
Assume the shape of the black material is a square.
The area will be calculated as:
Take for instance;
The area is:
Assume the shape of the black material is a triangle
The area will be calculated as:
Take for instance;
The area is:
So, in general.
You need to first get the shape of the black segment on the flag, then calculate the area using the appropriate formula.
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
arctan(x) is less than π/2 for x > 0. Therefore, if we say g(x) = 13 (π/2) / eˣ, then f(x) < g(x) for all values of x > 0.
g(x) = 13 (π/2) / eˣ
g(x) = 13π/2 · (1/e)ˣ
So g(x) is a geometric series where r = 1/e. Since |r| < 1, the series converges.
Since g(x) converges, the smaller f(x) also converges.
Answer what? This question? Oh, I'm answering it right now. Did you have another question out? Well, I can't see it.
Answer:
D. Pythagorean
Step-by-step explanation:
Given the identity
cos²x - sin²x = 2 cos²x - 1.
To show that the identity is true, we need to show that the left hand side is equal to right hand side or vice versa.
Starting from the left hand side
cos²x - sin²x ... 1
According to Pythagoras theorem, we know that x²+y² = r² in a right angled triangle. Coverting this to polar form, we have:
x = rcostheta
y = rsintheta
Substituting into the Pythagoras firnuka we have
(rcostheta)²+(rsintheta)² = r²
r²cos²theta+r²sin²theta = r²
r²(cos²theta+sin²theta) = r²
(cos²theta+sin²theta) = 1
sin²theta = 1 - cos²theta
sin²x = 1-cos²x ... 2
Substituting equation 2 into 1 we have;
= cos²x-(1-cos²x)
= cos²x-1+cos²x
= 2cos²x-1 (RHS)
This shows that cos²x -sin²x = 2cos²x-1 with the aid of PYTHAGORAS THEOREM
