Answer:
152.7 yd
Step-by-step explanation:
I presume that there is a fence around the perimeter and separating the various plots.
AG = BF = CE = 18 yd
GH = DE = 12 yd
EO² = DE² + DO²
1 5² = 12² + DO²
225 = 144 + DO²
81 = DO²
DO = 9
BC = DO = EF = 9 yd
AC = DH = EG = 17 yd
AO² = AB² + BO² = 8² + 6² = 64 + 36 =100
AO = √100 = 10 yd
CO² = BC² + BO² = 9² + 6² = 81 + 36 = 127
CO = √127 yd
GO² = FG² + FO² = 8² + 12² =64 + 144 = 208
GO = √208 = 4√13 yd
Fencing needed
= (AC + DH + EG) + (AG + BF + CE) + AO + CO + EO + GO
= (3 × 17) + 3 × 18) + 10 + √127 + 15 + 4√13
= 51 + 54 + 25 + 11.27 + 14.42
= 152.7 yd
Answer:
the answer is C
Step-by-step explanation:
Use the law of cosines.
a2+b2−2abcosC=c2
Find the measure of angle C. It is the opposite side of c.
c2−a2−b2−2ab=cosC
cosC=13.62−22.52−182−2(22.5)(18)≈0.797
C=cos−10.797=0.649 rad=37.19∘
angle B:
a2+c2−2accosB=b2
cosB=b2−a2−c2−2ac
B=cos−1b2−a2−c2−2ac=cos−1182−22.52−13.62−2(22.5)(13.6)≈0.927 rad=53.13∘
angle A:
b2+c2−2bccosA=a2
A=89.68∘
Refer to this previous solution set
brainly.com/question/26114608
===========================================================
Problem 4
Like the three earlier problems, we'll place the kicker at the origin and have her kick to the right. The two roots in this case are x = 0 and x = 20 to represent when the ball is on the ground.
This leads to the factors x and x-20 and the equation 
We'll plug in (x,y) = (10,28) which is the vertex point. The 10 is the midpoint of 0 and 20 mentioned earlier.
Let's solve for 'a'.
This then leads us to:
The equation is in the form 
with 
The graph is below in blue.
===========================================================
Problem 5
The same set up applies as before.
This time we have the roots x = 0 and x = 100 to lead to the factors x and x-100. We have the equation 
We'll use the vertex point (50,12) to find 'a'.
Then we can find the standard form
The graph is below in red.
The matrix R represents the reflection matrix for the provided vertices, and graph A represents the pre-image and the image on the same coordinate grid.
<h3>What is the matrix?</h3>
It is defined as the group of numerical data, functions, and complex numbers in a specific way such that the representation array looks like a square, rectangle shape.
We have vertices shown in the picture.
Form a matrix using the vertices:
![= \left[\begin{array}{ccc}-3&5&6\\7&3&-5\\\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%266%5C%5C7%263%26-5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To reflect over the x-axis, multiply by the reflection matrix:
![= \left[\begin{array}{ccc}1&0\\0&-1\\\end{array}\right]\left[\begin{array}{ccc}-3&5&6\\7&3&-5\\\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%266%5C%5C7%263%26-5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\rm R= \left[\begin{array}{ccc}-3&5&6\\-7&-3&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Crm%20R%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%266%5C%5C-7%26-3%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The above matrix represent the reflection matrix.
Thus, the matrices R represents the reflection matrix for the provided vertices, and graph A represents the pre-image and the image on the same coordinate grid.
Learn more about the matrix here:
brainly.com/question/9967572
#SPJ1
