Answer:
It costs $154.88 to build the fences.
Step-by-step explanation:
Perimeter of square = 
We are given that A garden is in the shape of a square with a perimeter of 56 feet.
So,
Side =
So, Side of square park = 14 feet
One fence is around the perimeter of the garden, whereas the second fence is 2 feet from the first fence on the outside.
Outer length = 14+2+2=18 feet
So, Perimeter of outer fence = 
Cost of fencing 1 foot = $1.21
Cost of fencing 56 feet =
Cost of fencing 72 feet = 
Total cost = 67.76+87.12=154.88
Hence It costs $154.88 to build the fences.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20N%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B10%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B3%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%20NA%3D%5Csqrt%7B%286%2B3%29%5E2%2B%283-10%29%5E2%7D%5Cimplies%20NA%3D%5Csqrt%7B130%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20D%28%5Cstackrel%7Bx_1%7D%7B6%7D~%2C~%5Cstackrel%7By_1%7D%7B-1%7D%29%20%5C%5C%5C%5C%5C%5C%20AD%3D%5Csqrt%7B%286-6%29%5E2%2B%28-1-3%29%5E2%7D%5Cimplies%20AD%3D4%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
now that we know how long each one is, let's plug those in Heron's Area formula.
![\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D%5Csqrt%7B130%7D%5C%5C%20b%3D4%5C%5C%20c%3D%5Csqrt%7B202%7D%5C%5C%5B1em%5D%20s%3D%5Cfrac%7B%5Csqrt%7B130%7D%2B4%2B%5Csqrt%7B202%7D%7D%7B2%7D%5C%5C%5B1em%5D%20s%5Capprox%2014.81%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B14.81%2814.81-%5Csqrt%7B130%7D%29%2814.81-4%29%2814.81-%5Csqrt%7B202%7D%29%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B324%7D%5Cimplies%20A%3D18)
3 n
- = -
4 10
cross multiply to get 4n=(3)(10)
multiply 3 and 10. 4n/4 = 30/4
n = 7.5
Answer:
it depends on the rest of the equation
Step-by-step explanation:
sss is used for using 3 sides to determine if the triangles are congruent, asa uses an angle then a side then an angle, sas uses a side then an angle then a side, ssa uses a side, then another side, then an angle
Answer:
Step-by-step explanation:
to find the inverse first switch the x and y
y=log(x+3) inverse is x=log(y+3)
then change to exponential form to solve
turns to
a is 10
b is y+3
c is x
subtract 3 from both sides
