You have to take $8,550 times 0.12 (12%) to find out the number it is being decreased by.
8,550 x 0.12 = 1,026
And then, you have to subtract 1,026 from 8,550 three times for three years and your answer becomes 5,472.
$5,472
Answer:
$132
Step-by-step explanation:
First we need to find the total number of woods required. This would be total area of fence divided by area of each wood. Remember to convert inch into feet.

Now we know the total number of planks, we need to find what is the cost of each plank. We know that one wood plank area is 2 sq. ft ( 6*0.33333). Therefore, the cost of each plank is = cost per area(sq.ft) multiply by total area (sq. ft).
cost per plank = 2 * $1.10 =$2.20.
We have total planks = 60 units and we have cost per plank = $2.20.
So, we can find total cost as 60*2.20= $132
<h3>
Answer: Choice B) 13</h3>
You chose the correct answer.
======================================================
Explanation:
Lines M and L are only parallel if and only if the corresponding angles are congruent.
The angles 127 and (9x+10) degrees are corresponding angles. They are both on the same side of the transversal, and they are to the right of each parallel line.
Set the two angle expressions equal to one another. Solve for x
9x+10 = 127
9x = 127-10
9x = 117
x = 117/9
x = 13
well, first off let's check those two points, we know it's centerd at (-26 , 120) and we also know it passes through (0 , 0), so the distance between those two points is its radius
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-26%7D~%2C~%5Cstackrel%7By_2%7D%7B120%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%28~~-26%20-%200~~%29%5E2%20%2B%20%28~~120%20-%200~~%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%28-26%29%5E2%20%2B%20%28120%20%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20r%3D%5Csqrt%7B%28%20-26%20%29%5E2%20%2B%20%28%20120%20%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%20676%20%2B%2014400%20%7D%20%5Cimplies%20r%3D%5Csqrt%7B%2015076%20%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill](https://tex.z-dn.net/?f=%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Chspace%7B5em%7D%5Cstackrel%7Bcenter%7D%7B%28%5Cunderset%7B-26%7D%7Bh%7D~~%2C~~%5Cunderset%7B120%7D%7Bk%7D%29%7D%5Cqquad%20%5Cstackrel%7Bradius%7D%7B%5Cunderset%7B%5Csqrt%7B15076%7D%7D%7Br%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%20~~%20x%20-%20%28-26%29%20~~%20%29%5E2%20~~%20%2B%20~~%20%28%20~~%20y-120%20~~%20%29%5E2~~%20%3D%20~~%28%5Csqrt%7B15076%7D%29%5E2%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%28x%2B26%29%5E2%2B%28y-120%29%5E2%20%3D%2015076~%5Chfill)