9514 1404 393
Answer:
97 cm
Step-by-step explanation:
The perimeter is the sum of the side lengths. Starting from the upper left corner and working clockwise, the side lengths are ...
10 +3 +33 +15 +33 +3 = 25+2(36) = 97 . . . cm
The perimeter of the figure is 97 cm.
Answer:
(a) Growth

Step-by-step explanation:
Given
-- current
--- rate
Solving (a): Growth or decay
The question says his followers increases each day by 24%.
Increment means growth.
<em />
<em>Hence, it is a growth problem</em>
Solving (b): Equation to represent the scenario.
Since the rate represents growth, the equation is:

Substitute:
and 




Answer:
90 m
Step-by-step explanation:
The piece on the right side fits perfectly in the left side so you can move the triangle to the left in order to make a square. After that, you can do 10 times 9 because 5+4=9 to get 90m as the area.
<span>y=-4/x+1 is ambiguous, since it's not immediately clear whether you meant
-4
y = -4/x + 1 or y = ---------
x+1
I'm going to assume that the latter is what you meant.
1. Interchange x and y, obtaining:
-4
x = --------
y+1
2. Solve this for y, obtaining y+1 = -4/x, or xy + x = -4, or
-x - 4
xy = -x-4, or y = ---------
x
-1 -1 4
3. Replace y with f (x): f (x) = -1 - -----
x
This last result has the correct form.</span>
Check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

![\bf \textit{area of a segment of a circle}\\\\ A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right] \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=6\\ \theta =120 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA_y%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~-~sin%28%5Ctheta%20%29%20%20%5Cright%5D%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D6%5C%5C%0A%5Ctheta%20%3D120%0A%5Cend%7Bcases%7D)