Answer:
I'm so sorry.. I'm not that good in maths.. I'm only good in English..
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)
Answer:
Step-by-step explanation:
Method #1
We can draw a <em>right triangle</em> on the graph upon where the points are located and use the Pythagorean Theorem:
* Whenever we talk about distance, we ONLY want the NON-NEGATIVE root.
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Method #2
We can use the Distance Formula:
![\sqrt{[-x_1 + x_2]^{2} + [-y_1 + y_2]^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E%7B2%7D%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E%7B2%7D%7D%20%3D%20D)
<em>N</em>[−3, 2] <em>M</em>[−6, 0]
![\sqrt{[3 - 6]^{2} + [0 + 2]^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B3%20-%206%5D%5E%7B2%7D%20%2B%20%5B0%20%2B%202%5D%5E%7B2%7D%7D%20%3D%20D)
![\sqrt{[-3]^{2} + 2^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-3%5D%5E%7B2%7D%20%2B%202%5E%7B2%7D%7D%20%3D%20D)
* Whenever we talk about distance, we ONLY want the NON-NEGATIVE root.
** You see? It does not matter which method you choose, as long as you are doing the work correctly.
I am delighted to assist you anytime.
A semicircle's perimeter is half the perimeter of the complete circle.
The perimeter of the complete circle is its circumference which is found by the equation: πD.
Then the equation of the perimeter of the semicircle is πD/2
In this case D = 24 in, then the perimeteir is π (24in/2) = 12π in ≈ 37.7 in
Answer: The exact length is 12π inches and the approximate length is 37.7 inches
-64 1/3
disregard the negative sign for the moment...
to turn a mixed number into an improper fraction, take ur whole number (63) and multiply it by ur denominator (3) and u get (64 * 3) = 192....now add that to ur numerator (1) and u get (192 + 1) = 193...now put that over ur original denominator (3) and u get 193/3.....now put ur negative sign back....and u have - 193/3.....simplified form, does not reduce
