Answer:
We need the following three rigid motions:
i) Reflection around y-axis, ii) Translation three units in the -y direction, iii) Translation four units in the -x direction.
Step-by-step explanation:
We need to perform three operations on pentagon ABCDE to create pentagon A'B'C'D'E':
i) Reflection around y-axis:
(Eq. 1)
ii) Translation three units in the -y direction:
(Eq. 2)
iii) Translation four units in the -x direction:
(Eq. 3)
We proceed to proof the effectiveness of operations defined above by testing point D:
1) 
Given.
2) 
By (Eq. 1)
3) 
By (Eq. 2)
4) 
By (Eq. 3)/Result
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.](https://tex.z-dn.net/?f=5m%5E2e%5E%7Bmt%7D%2B3me%5E%7Bmt%7D-2e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%285m%5E2%2B3y-2%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B3m-2%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B5m-2m-2%3D0%5C%5C%5C%5C%5CRightarrow%205m%28m%2B1%29-2%28m%2B1%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B1%29%285m-1%29%3D0%5C%5C%5C%5C%5CRightarrow%20m%2B1%3D0%2C~~~~~5m-1%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-1%2C~%5Cdfrac%7B1%7D%7B5%7D.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
![y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D2.8%5C%5C%5C%5C%5CRightarrow%20-A%2B%5Cdfrac%7BB%7D%7B5%7D%3D2.8%5C%5C%5C%5C%5CRightarrow%20-5A%2BB%3D14%5C%5C%5C%5C%5CRightarrow%20-5A-A%3D14~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7BUisng%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-6A%3D14%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B14%7D%7B6%7D%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B7%7D%7B3%7D.)
From equation (ii), we get
Thus, the required solution is
Answer:
Step-by-step explanation:
We have been given that at a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 11 m. The inner edge of the sidewalk is a circle with a radius of 9 m.
To find the area of the side walk we will subtract the area of inner edge of the side walk of lion pen from the area of the outer edge of the lion pen.
, where r represents radius of the circle.
Therefore, the exact area of the side walk is 
To find the approximate area of side walk let us substitute pi equals 3.14.
Therefore, the approximate area of the side walk is 
.
Answer:
58.25
Step-by-step explanation:
58 1/4 in decimal form = 58.25
