The circumference should be 37.68 because area= pi *r^2. So if area=pi*36, then the radius must be 6 which makes the diameter 12 and circumference=pi*diameter and 3.14*12= 37.68. So circumference of the circle is 37.68!
Answer:
(r o g)(2) = 4
(q o r)(2) = 14
Step-by-step explanation:
Given
![g(x) = x^2 + 5](https://tex.z-dn.net/?f=g%28x%29%20%3D%20x%5E2%20%2B%205)
![r(x) = \sqrt{x + 7}](https://tex.z-dn.net/?f=r%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%207%7D)
Solving (a): (r o q)(2)
In function:
(r o g)(x) = r(g(x))
So, first we calculate g(2)
![g(x) = x^2 + 5](https://tex.z-dn.net/?f=g%28x%29%20%3D%20x%5E2%20%2B%205)
![g(2) = 2^2 + 5](https://tex.z-dn.net/?f=g%282%29%20%3D%202%5E2%20%2B%205)
![g(2) = 4 + 5](https://tex.z-dn.net/?f=g%282%29%20%3D%204%20%2B%205)
![g(2) = 9](https://tex.z-dn.net/?f=g%282%29%20%3D%209)
Next, we calculate r(g(2))
Substitute 9 for g(2)in r(g(2))
r(q(2)) = r(9)
This gives:
![r(x) = \sqrt{x + 7}](https://tex.z-dn.net/?f=r%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%207%7D)
![r(9) = \sqrt{9 +7{](https://tex.z-dn.net/?f=r%289%29%20%3D%20%5Csqrt%7B9%20%2B7%7B)
{
![r(9) = 4](https://tex.z-dn.net/?f=r%289%29%20%3D%204)
Hence:
(r o g)(2) = 4
Solving (b): (q o r)(2)
So, first we calculate r(2)
![r(x) = \sqrt{x + 7}](https://tex.z-dn.net/?f=r%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%207%7D)
![r(2) = \sqrt{2 + 7}](https://tex.z-dn.net/?f=r%282%29%20%3D%20%5Csqrt%7B2%20%2B%207%7D)
![r(2) = \sqrt{9}](https://tex.z-dn.net/?f=r%282%29%20%3D%20%5Csqrt%7B9%7D)
![r(2) = 3](https://tex.z-dn.net/?f=r%282%29%20%3D%203)
Next, we calculate g(r(2))
Substitute 3 for r(2)in g(r(2))
g(r(2)) = g(3)
![g(x) = x^2 + 5](https://tex.z-dn.net/?f=g%28x%29%20%3D%20x%5E2%20%2B%205)
![g(3) = 3^2 + 5](https://tex.z-dn.net/?f=g%283%29%20%3D%203%5E2%20%2B%205)
![g(3) = 9 + 5](https://tex.z-dn.net/?f=g%283%29%20%3D%209%20%2B%205)
![g(3) = 14](https://tex.z-dn.net/?f=g%283%29%20%3D%2014)
Hence:
(q o r)(2) = 14
Answer:
Here you go, I think
Step-by-step explanation:
From the identity:
![sec(x)= \frac{1}{cos(x)}](https://tex.z-dn.net/?f=sec%28x%29%3D%20%5Cfrac%7B1%7D%7Bcos%28x%29%7D%20)
![f(x)=sec(x)= \frac{1}{cos(x)}](https://tex.z-dn.net/?f=f%28x%29%3Dsec%28x%29%3D%20%5Cfrac%7B1%7D%7Bcos%28x%29%7D%20)
the inverse of f is g such that f(g(x))=x,
we must find g(x), such that
![\frac{1}{cos[g(x)]}=x](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bcos%5Bg%28x%29%5D%7D%3Dx%20)
thus,
![cos[g(x)]= \frac{1}{x}](https://tex.z-dn.net/?f=cos%5Bg%28x%29%5D%3D%20%5Cfrac%7B1%7D%7Bx%7D%20)
![g(x)=cos^{-1} (\frac{1}{x})](https://tex.z-dn.net/?f=g%28x%29%3Dcos%5E%7B-1%7D%20%28%5Cfrac%7B1%7D%7Bx%7D%29%20)
Answer: b. g(x)=cos^-1(1/x)
The opposite angles are equal to are supplementary to each other or equal to each other.
<h3>What is a Quadrilateral Inscribed in a Circle?</h3>
In geometry, a quadrilateral inscribed in a circle, also known as a cyclic quadrilateral or chordal quadrilateral, is a quadrilateral with four vertices on the circumference of a circle. In a quadrilateral inscribed circle, the four sides of the quadrilateral are the chords of the circle.
The opposite angles in a cyclic quadrilateral are supplementary. i.e., the sum of the opposite angles is equal to 180˚.
If e, f, g, and h are the inscribed quadrilateral’s internal angles, then
e + f = 180˚ and g + h = 180˚
by theorem the central angle = 2 x inscribed angle.
∠COD = 2∠CBD
∠COD = 2b
∠COD = 2 ∠CAD
∠COD = 2a
now,
∠COD + reflex ∠COD = 360°
2e + 2f = 360°
2(e + f) =360°
e + f = 180°.
Learn more about this concept here:
brainly.com/question/16611641
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