since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)
Answer: 36 sec
Step-by-step explanation:
Let
x------> distance the kitten runs before being caught by the puppy
Remember that
The speed is equal to divide the distance by the time
v=d/t
solve for t
t=d/v
so
equate the time
x/20=(180+x)/25
25x=20(180+x)
25x=3,600+20x
25x-20x=3,600
x=3,600/5=720 m
t=d/v=720/20=36 sec
or
t=(720+180)/25=900/25=36 sec
Answer:
5
Step-by-step explanation:
-1+(1)= 0
0+(4)=4
1+4=5
Answer:
x = 4/7
Step-by-step explanation:
x^3 = 64/ 343 // - 64/ 343
x^3 - ( 64/ 343 ) = 0
x^3 - 64 / 343 = 0
1*x^3 = 64/ 343 // : 1
x^3 = 64/ 343
x^3 = 64/ 343 // ^ 1/3
x = 4/ 7
