Answer:
8 minutes per mile
Step-by-step explanation:
Geri ran in a marathon race.
It took her 3 hours and 28 minutes to run 26 miles.
1 hour = 60 minutes
we will convert the time into minutes
3 hour 28 minutes = (3 × 60) + 28
= 180 + 28
= 208 minutes
∵ Geri ran 26 miles in 208 minutes
∴ She ran 1 miles =
= 8 minute
Geri ran 1 mile in 8 minute.
Given :
A bike path is 3 miles long. there are distance markers every path one fourth mile to the end of the path.
To Find :
Which number line correctly models this situation and the total number of distance markers.
Solution :
It is given that their are markers every 1/4 part of mile.
So, their are 4 markers per mile.
Numbers of markers in 3 miles long path is :
Therefore, the total number of distance markers in 3 mile path is 12.
Hence, this is the required solution.
Answer:
Step-by-step explanation:
The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.
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<h3>graphing</h3>The attachment shows a graphing calculator solution to the equation
-3x^2 + 6x + 6 = 0
The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.
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<h3>completing the square</h3>We can also solve the equation algebraically. One of the simplest methods is "completing the square."
-3x^2 +6x +6 = 0
x^2 -2x = 2 . . . . . . . . divide by -3 and add 2
x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square
(x -1)^2 = 3 . . . . . . . . written as a square
x -1 = ±√3 . . . . . . . take the square root
x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground
The numerical values of these solutions are approximately ...
x ≈ {-0.732, 2.732}
The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.
