Answer:
![\huge\boxed{\sqrt[3]{c^4}=c^\frac{4}{3}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%5Csqrt%5B3%5D%7Bc%5E4%7D%3Dc%5E%5Cfrac%7B4%7D%7B3%7D%7D)
Step-by-step explanation:
![\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\text{therefore}\\\\\sqrt[3]{c^4}=c^\frac{4}{3}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%5Ctext%7Btherefore%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7Bc%5E4%7D%3Dc%5E%5Cfrac%7B4%7D%7B3%7D)
Answer:
x - y + z = 8
2x +3y -z= -2
3x - 2y -9z = 9
Step-by-step explanation:
[1 - 1 1 8]
[2 3 -1 -2]
[3 -2 -9 9]
You don't even need the picture to solve this one.
You said that h = 5 cot(Θ) , and you said that Θ is 30 degrees.
All you need now is to find the cotangent of Θ, plop that into the equation,
and the solution practically jumps off the paper into your lap.
To find the cotangent of 30 degrees, you can use a calculator, look it up
in a book, read it off of a slide rule if you have one, draw a picture of a
30-60-90 right triangle etc. You'll find that the cotangent of 30 degrees
is √3 . That's about 1.732 .
So your equation is h = 5 (1.732) = <em>8.66 </em>(rounded)
Apparently, somebody gave you the equation, and asked you to find 'h'.
Once you had the equation, you didn't even need to know that 'h' has
anything to do with a triangle.
To get the function y = <span>-2+5sin(pi/12(x-2)), the maximum value can be determined by differentiating the function and equating it to zero. The value of x will give the maximum value of the function.
dy/dx = 5 cos (pi/12 (x-2)) (pi/12)
dy/dx = 5 pi/12 cos(pi/12 (x-2))
Equate to zero</span>:
<span>5 pi/12 cos(pi/12 (x-2)) =0
pi/12 (x-2) = 3pi/2
x = 8
Substituting,
y= -2 + 5sin( pi/12 (8-2)
y = -1.86
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