Answer:
Step-by-step explanation:
From the picture attached,
Graph (A)
Domain of the function : (-∞, ∞) or a set of real numbers
Range of the function : (-∞, ∞) or a set of real numbers
Graph (B),
Domain : (-∞, ∞) Or a set of real numbers
Range : (-∞, 6] Or x ≤ 6
Graph (C),
Domain : [-2, ∞) or x ≥ -2
Range : (-∞, ∞) Or a set of real numbers
Therefore, Graphs (A), (B) have the domain of all real numbers and Graph (C) has the range of all real numbers.
Since all the parabolas are not the functions,
Therefore, Graph (C) is not a function. Graphs (A) and (B) are the functions.
let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.
think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.
now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area. Well, what is the circumference of that shaded area?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D12%20%5Cend%7Bcases%7D%5Cimplies%20C%3D2%5Cpi%2012%5Cimplies%20C%3D24%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthree%20quarters%20of%20it%7D%7D%7B24%5Cpi%20%5Ccdot%20%5Ccfrac%7B3%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%206%5Cpi%20%5Ccdot%203%5Cimplies%2018%5Cpi)
well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20C%3D18%5Cpi%20%5Cend%7Bcases%7D%5Cimplies%2018%5Cpi%20%3D2%5Cpi%20r%5Cimplies%20%5Ccfrac%7B18%5Cpi%20%7D%7B2%5Cpi%20%7D%3Dr%5Cimplies%209%3Dr%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bdiameter%20is%20twice%20the%20radius%7D%7D%7Bd%3D18%7D~%5Chfill)
B. He received 29% discount on the car.<span />
Answer: 55.5 (A.)
Step-by-step explanation:
Since angle A = 29 and angle B = 41, angle C must be equal to 110
180 = m<A + m<B + m<C
180 = 29 +41 + m<C
180 = 70 + m<C
110 = m<C
Therefore, side c must be the longest, side b must be the second longest, and side a must be the shortest.
Since side length a, angle A, and angle B are known, one can use the law of sines to solve for side b.
Law of Sines: sinA/a = sinB/b = sinC/c
sinA/a = sinB/b
sin29/41 = sin41/b
b(sin29/41) = sin41
b = 41(sin41)/(sin29)
b = 55.48
b = 55.5
divide 35 by 3600
divide 38 by 60
add them together
35/3600 = 0.009722
38/60=0.6333
0.6333+0.009722 = 0.64305
so 13.64305 degrees. round answer as necssary
