Do you remember how to find the area of a circle ?
Big Fat Hint: The area of a circle is (pi) x (its radius)² .
Can you find the area of the big circle ... the one with radius = 18 m ?
Do that. The whole big circle, just as if there was nothing inside it, and
no shaded region.
I'll wait.
Now, can you find the area of the small circle ... the one with radius = 12 m ?
Do that.
I'll wait.
The area of the shaded region is just
(the area of the big circle) minus (the area of the small circle) !
I got 180 pi or roughly 565.2 m². I could be wrong. You should check it.
The coordinates of the vertex for the quadratic function are given as follows: (100,10).
<h3>What is the vertex of a quadratic equation?</h3>
A quadratic equation is modeled by:
![y = ax^2 + bx + c](https://tex.z-dn.net/?f=y%20%3D%20ax%5E2%20%2B%20bx%20%2B%20c)
The vertex is given by:
![(x_v, y_v)](https://tex.z-dn.net/?f=%28x_v%2C%20y_v%29)
In which:
In this problem, the function is given as follows:
y = -x²/1000 + x/5
y = -0.001x² + 0.2x.
Hence the coefficients are a = -0.001, b = 0.2, c = 0 and the coordinates of the vertex are given as follows:
More can be learned about the vertex of a quadratic function at brainly.com/question/24737967
#SPJ1
Answer:
= 18 ( 3 + 2(\sqrt{3} + \sqrt{2} ))
Step-by-step explanation:
![6(\sqrt{36} +\sqrt{27}) + 6(\sqrt{27} +\sqrt{18}) + 6(\sqrt{18} + \sqrt{9} )\\= 6( {6} +\sqrt{9*3}) + 6( \sqrt{9*3} +\sqrt{9*2}) + 6(\sqrt{9*2} + 3 )\\= 6( {6} +3\sqrt{3}) + 6(3\sqrt{3} +3\sqrt{2}) + 6(3\sqrt{2} + 3 )\\=36 + 18\sqrt{3} + 18\sqrt{3} + 18\sqrt{2}+ 18\sqrt{2} + 18\\= 36+18 +18( \sqrt{3}+ \sqrt{3} +\sqrt{2} + \sqrt{2})\\= 54 + 18 (2\sqrt{3} +2\sqrt{2})\\=54 + 36\sqrt{3} + 36\sqrt{2}\\= 18 ( 3 + 2(\sqrt{3} + \sqrt{2} ))](https://tex.z-dn.net/?f=6%28%5Csqrt%7B36%7D%20%2B%5Csqrt%7B27%7D%29%20%2B%206%28%5Csqrt%7B27%7D%20%2B%5Csqrt%7B18%7D%29%20%2B%206%28%5Csqrt%7B18%7D%20%2B%20%5Csqrt%7B9%7D%20%29%5C%5C%3D%206%28%20%7B6%7D%20%2B%5Csqrt%7B9%2A3%7D%29%20%2B%206%28%20%5Csqrt%7B9%2A3%7D%20%20%2B%5Csqrt%7B9%2A2%7D%29%20%2B%206%28%5Csqrt%7B9%2A2%7D%20%2B%203%20%29%5C%5C%3D%206%28%20%7B6%7D%20%2B3%5Csqrt%7B3%7D%29%20%2B%206%283%5Csqrt%7B3%7D%20%2B3%5Csqrt%7B2%7D%29%20%2B%206%283%5Csqrt%7B2%7D%20%2B%203%20%29%5C%5C%3D36%20%2B%2018%5Csqrt%7B3%7D%20%2B%2018%5Csqrt%7B3%7D%20%2B%2018%5Csqrt%7B2%7D%2B%2018%5Csqrt%7B2%7D%20%2B%2018%5C%5C%3D%2036%2B18%20%20%2B18%28%20%5Csqrt%7B3%7D%2B%20%5Csqrt%7B3%7D%20%2B%5Csqrt%7B2%7D%20%2B%20%5Csqrt%7B2%7D%29%5C%5C%3D%2054%20%2B%2018%20%282%5Csqrt%7B3%7D%20%2B2%5Csqrt%7B2%7D%29%5C%5C%3D54%20%2B%2036%5Csqrt%7B3%7D%20%20%2B%2036%5Csqrt%7B2%7D%5C%5C%3D%2018%20%28%203%20%2B%202%28%5Csqrt%7B3%7D%20%20%2B%20%5Csqrt%7B2%7D%20%29%29)
Generally, you are told to approach these by "clearing fractions". That is, you generally multiply the equations by the least common denominator so all fractions and mixed numbers become integers.
Alternatively, you can simply do the arithmetic using the numbers given. You learned a long time ago how to add, subtract, multiply, and divide mixed numbers and fractions. Do these operations as necessary to solve the equations.
Answer:
![\sqrt{484} =22](https://tex.z-dn.net/?f=%5Csqrt%7B484%7D%20%3D22)
Step-by-step explanation: