Bet...but you really need to learn how to to this bc you use this in all types of math
All you have to do is distribute then combine like terms
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Step-by-step explanation:
let the unknown members be X
At first, the club had 20 + X members
After the first year,
2( 20+X)
The second year,
2( 20 +X/2) +40
The third year,
3[2(20+ X/2) +40]
Total number =
20+X + 2(20+X) + 2( 20 +X/2) +40 +3[2(20+ X/2) +40]
= 20+ X + 40+ 2X + 40 +2X/2 +40 +3[40+2X/2 +40]
= 20+40+X +2X +40+X+ 40 + 3[ 40+40+X]
= 60 +3X +80+X + 3[80+X]
= 140+4X + 240+3X
=380 + 7X
-380=7X
X = -54.3
We are asked in the problem to evaluate the integral of <span>(cosec^2 x-2005)÷cos^2005 x dx. The function is an example of a complex function with a degree that is greater than one and that uses special rules to integrate the function via the trigonometric functions. For example, we integrate
2005/cos^2005x dx which is equal to 2005 sec^2005 x since sec is the inverse of cos. The integral of this function when n >3 is equal to I=</span><span>∫<span>sec(n−2)</span>xdx+∫tanx<span>sec(n−3)</span>x(secxtanx)dx
Then,
</span><span>∫tanx<span>sec(<span>n−3)</span></span>x(secxtanx)dx=<span><span>tanx<span>sec(<span>n−2)</span></span>x/(</span><span>n−2)</span></span>−<span>1/(<span>n−2)I
we can then integrate the function by substituting n by 3.
On the first term csc^2 2005x / cos^2005 x we can use the trigonometric identity csc^2 x = 1 + cot^2 x to simplify the terms</span></span></span>
Answer:
86
Step-by-step explanation: