Use the distance formula: x=sqrt((x2-x1)^2+(y2-y1)^2)
Plug in ordered pairs for the variables:
x=sqrt((2-2)^2+(8-3)^2)
x=sqrt(0+25)
x=sqrt(25)
x=5 or -5
Since distance is ALWAYS positive, it's 5 units
Hope this helped!
Answer:
Yes
Step-by-step explanation:
Answer:
C. 7 hours
Step-by-step explanation:
Let the time for the trip out be represented by t. Then the time for the return trip is t-1. The distance was the same for both trips, so we have ...
distance = speed × time
300t = 350(t -1)
300t = 350t -350 . . . . eliminate parentheses
350 = 50t . . . . . . . . . . add 350-300t
7 = t . . . . . . divide by 50
The trip out took 7 hours.
Answer:
7.7
Step-by-step explanation:
To find the intersection points of the line and the circle we have to set up a system with their equations and solve. The system would look like this:
To solve, substitute 1 for x in the second equation to get:
Solving, we get:
Therefore, the two points of intersection are 
and 
. The distance between these two points (the length of the chord in the circle) is 
which is 7.745966692414... which is 7.7 rounded to the nearest tenth.
Hope this helps :)
The probability of losing a data packet is 0.010.
Therefore the probability of successfully sending a data packet (not losing a data packet) is
p = 1 - 0.010 = 0.99
In 100 packet transmissions (independent events), the probability of success is
0.99¹⁰⁰ = 0.3660
The probability of losing a data packet is
1 - 0.366 = 0.6340
Answer:
The probability of resending a data packet is 0.6340
