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
Answer:
Step-by-step explanation:
1.) Apply the distributive law
5x + 5 * 2
2.) Multiply the numbers
5 * 2 = 10 is equivalent to 5x + 10
3.) Group like terms
5x - 3 + 10
4.) Add/subtract the numbers
-3 + 10 = 7 is equivalent to 5x + 7
There aren't really any instructions for this, so I'll just go ahead and find the total amount of money for both months. The first month Adriano had 200 subscribers. 5 times 200 is 1000, so Adriano received $1000 that month. The second month 40 members joined, but 10 cancelled their subscription. Taking this into account, we'll just say that 30 members joined. 30 times 5 is 150, and 150 plus 1000 is 1150. Therefore, Adriano had to have received $1150 the second month.
I'm not entirely sure what the question was to begin with, but I hope this answers it!
Answer:
96
Step-by-step explanation:
Strange question, as normally we would not calculate the "area of the tire." A tire has a cross-sectional area, true, but we don't know the outside radius of the tire when it's mounted on the wheel.
We could certainly calculate the area of a circle with radius 8 inches; it's
A = πr^2, or (here) A = π (8 in)^2 = 64π in^2.
The circumference of the wheel (of radius 8 in) is C = 2π*r, or 16π in.
The numerical difference between 64π and 16π is 48π; this makes no sense because we cannot compare area (in^2) to length (in).
If possible, discuss this situatio with your teacher.
