Answer:
it would be a 85 percent
Step-by-step explanation:
A(-7,-4) B(-2,0)
√[(x'-x)^2+(y'-y)^2]
√(-2-(-7)^2+(0-(-4)^2
√(5^2)+(4^2)
√25+16
√41
the distance is approximately 6.4 units
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:
10
Step-by-step explanation:
g(f(x)) = g(3x² - 4) = 2(3x² - 4) - 6
so
g(f(2) = 2(3*2² - 4) - 6 = 2*8 - 6 = 16 - 6 = 10