To determine the amount of fencing needed to enclose the pool area, you will need to add 15 feet to both sides of both dimensions to find the new dimensions that will need enclosed.
28 feet + 15 feet + 15 feet = 58 feet
23 feet + 15 feet + 15 feet = 53 feet
The fence will need to be 2 x 58 + 2 x 53 or 222 feet long for the 53 feet by 58 feet area.
Answer:
![\[\frac{1}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B1%7D%7B24%7D%5C%5D)
Step-by-step explanation:
Let the total number of batteries produced daily be represented by x.
7/8 of the batteries are packaged immediately.
So the total number of packaged batteries = ![\[\frac{7x}{8}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B7x%7D%7B8%7D%5C%5D)
Remaining batteries = ![\[x-\frac{7x}{8}\]](https://tex.z-dn.net/?f=%5C%5Bx-%5Cfrac%7B7x%7D%7B8%7D%5C%5D)
= ![\[\frac{x}{8}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B8%7D%5C%5D)
Two third of the remaining batteries are sent to charity.
So the total number of batteries sent to charity= ![\[\frac{2}{3} * \frac{x}{8}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B2%7D%7B3%7D%20%2A%20%5Cfrac%7Bx%7D%7B8%7D%5C%5D)
= ![\[\frac{x}{12}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B12%7D%5C%5D)
Remaining batteries are faulty.
So the total number of faulty batteries = ![\[\frac{x}{8} - \frac{x}{12}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B8%7D%20-%20%5Cfrac%7Bx%7D%7B12%7D%5C%5D)
= ![\[\frac{3x - 2x}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B3x%20-%202x%7D%7B24%7D%5C%5D)
= ![\[\frac{x}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B24%7D%5C%5D)
So the faction of batteries which are faulty =
= ![\[\frac{x}{24} \div x\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B24%7D%20%5Cdiv%20x%5C%5D)
= ![\[\frac{1}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B1%7D%7B24%7D%5C%5D)
Answer:
21.71% increase
Step-by-step explanation:
increase = Increase ÷ Original Number × 100
12756/58753*100
0.217112318*100
21.71% increase.
MARK ME BRAINLIEST
Answer:Number Example Natural number are those number which we can count.It denoted by N
Step-by-step explanation:
since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.
we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%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%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B-5-%28-7%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%28-5%2B7%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B2%5E2%2B2%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B2%5Ccdot%202%5E2%7D%5Cimplies%20d%3D2%5Csqrt%7B2%7D~%5Chfill%20%5Cstackrel%7B~%5Chfill%20radius%7D%7B%5Ccfrac%7B2%5Csqrt%7B2%7D%7D%7B2%7D%5Cimplies%5Cboxed%7B%20%5Csqrt%7B2%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%7D%29%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20%5Ccfrac%7B-2-4%7D%7B2%7D~~%2C~~%5Ccfrac%7B-5-7%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-6%7D%7B2%7D~%2C~%5Ccfrac%7B-12%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%5Cboxed%7B%28-3%2C-6%29%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Cqquad%20center~~%28%5Cstackrel%7B-3%7D%7B%20h%7D%2C%5Cstackrel%7B-6%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%5Csqrt%7B2%7D%7D%7B%20r%7D%20%5C%5C%5B2em%5D%20%5Bx-%28-3%29%5D%5E2%2B%5By-%28-6%29%5D%5E2%3D%28%5Csqrt%7B2%7D%29%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B6%29%5E2%3D2)