If both of those times were on the same date and in
the same time-zone, then exactly 1 hour passed.
Answer:
Milli
Step-by-step explanation:
X, x+1, x+2, x+3 - four consecutive integers
![x+x+1+x+2+x+3=682 \\ 4x+6=682 \ \ \ |-6 \\ 4x=676 \ \ \ |\div 4 \\ x=169 \\ \\ x+1=169+1=170 \\ x+2=169+2=171 \\ x+3=169+3=172](https://tex.z-dn.net/?f=x%2Bx%2B1%2Bx%2B2%2Bx%2B3%3D682%20%5C%5C%0A4x%2B6%3D682%20%5C%20%5C%20%5C%20%7C-6%20%5C%5C%0A4x%3D676%20%5C%20%5C%20%5C%20%7C%5Cdiv%204%20%5C%5C%0Ax%3D169%20%5C%5C%20%5C%5C%0Ax%2B1%3D169%2B1%3D170%20%5C%5C%0Ax%2B2%3D169%2B2%3D171%20%5C%5C%0Ax%2B3%3D169%2B3%3D172)
The integers are 169, 170, 171 and 172.
3.5 / 9
(35 / 10) /9
35 / 10 ÷ 9
35 / 10 × 1/9
35 / 90
5 into 35 is 7, and 5 into 90 is 18
7 / 18
So it is 35 / 90 or 7 / 18.
Answer:
![(x-2)^2+(y+5)^2=36](https://tex.z-dn.net/?f=%28x-2%29%5E2%2B%28y%2B5%29%5E2%3D36)
Step-by-step explanation:
Recall that the equation of a circle of radius R centered at
is given by:
![(x-x_0)^2+(y-y_0)^2=R^2](https://tex.z-dn.net/?f=%28x-x_0%29%5E2%2B%28y-y_0%29%5E2%3DR%5E2)
For your particular case: R = 6,
= 2, and
= -5, therefore the equation becomes:
![(x-x_0)^2+(y-y_0)^2=R^2\\(x-2)^2+(y-(-5))^2=6^2\\(x-2)^2+(y+5)^2=36](https://tex.z-dn.net/?f=%28x-x_0%29%5E2%2B%28y-y_0%29%5E2%3DR%5E2%5C%5C%28x-2%29%5E2%2B%28y-%28-5%29%29%5E2%3D6%5E2%5C%5C%28x-2%29%5E2%2B%28y%2B5%29%5E2%3D36)