Answer:
x = 3.25; y = 5.5; angle measures are (cw from bottom left): 96, 84, 96
Step-by-step explanation:
The two angles above the horizontal line form a linear pair, so they are supplementary. Their measures add to 180 deg. That let's you find the value of x.
28x - 7 + 24x + 18 = 180
52x + 11 = 180
52x = 169
x = 169/52
x = 3.25
Now use the lower left angle and the upper right angle. They are vertical angles, so they are congruent. That allows you to solve for y.
12y + 30 = 24x + 18
We know that x = 3.25.
12y + 30 = 24(3.25) + 18
12y + 30 = 78 + 18
12y = 66
y = 66/12
y = 5.5
I don't know what the problem is asking for since you don't show.
x = 3.25; y = 5.5; angle measures are (cw from bottom left): 96, 84, 96
Answer:
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Step-by-step explanation:
Check the picture below.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-9%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%20~%5Chfill%20AB%3D%5Csqrt%7B%5B%208-%201%5D%5E2%20%2B%20%5B%200-%20%28-9%29%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AB%3D%5Csqrt%7B7%5E2%2B%280%2B9%29%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B7%5E2%2B9%5E2%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B130%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}](https://tex.z-dn.net/?f=B%28%5Cstackrel%7Bx_1%7D%7B8%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B9%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%20~%5Chfill%20BC%3D%5Csqrt%7B%5B%209-%208%5D%5E2%20%2B%20%5B%20-8-%200%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20BC%3D%5Csqrt%7B1%5E2%2B%28-8%29%5E2%7D%5Cimplies%20%5Cboxed%7BBC%3D%5Csqrt%7B65%7D%7D)
now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.
We can write the sequence out more fully, as we can see each time it is divided by 6.
60, 60/6, 60/6^2, 60/6^3, and so on.
Therefore we know the sequence can be written as

You can think of this as a graph, i.e. y=60/6^(x-1)
As a result, as x tends to infinity, y tends to 0 (since it effectively becomes 60/infinity). Therefore the sequence
converges toward zero.