Bro Use Math way it really works I’m telling you
Integrate indefinite integral:
Solution:
1. use substitution
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2. decompose into partial fractions
where A=1/2, B=1/2, C=-1
3. Substitute partial fractions and continue
4. back-substitute u=e^x
Note: log(x) stands for natural log, and NOT log10(x)
Prove we are to prove 4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work.
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2
= e^(3x) /2 + e^(-3x) /2
= cosh(3x)
= LHS
<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3,
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
The answer is 12 I believe
Answer:
B
Step-by-step explanation:
The score above the median is the upper quartile
The score below the median is the lower quartile
The upper and lower quartiles are the ends of the box plot
