Answer:
Andrew must run 1 5/12 Laps to complete the race.
Step-by-step explanation:
Total lap run by Michael, Nate, Ben, and Andrew = 4 laps
laps run by Michael = 1 1/3 laps =(3+1)/3 = 4/3 laps
laps run by Nate =1/2 laps
laps run by ben =3/4 laps
Let laps run by Andrew = x
therefore
laps run by Michael + laps run by Nate + laps run by ben + laps run by Andrew = 4 laps
4/3 + 1/2 + 3/4 + x = 4
LCM of 3,2 and 4 is 12
=> (4*4 + 6*1 + 3*3+12x)/12 = 4
=> (16+6+9+12x)/12 = 4
=> 31+12x = 12*4 = 48
=> 12x = 48-31= 17
=> x = 17/12 = 1 5/12
Thus, Andrew must run 1 5/12 Laps to complete the race.
<h3>
Answer: Commutative Property Of Addition</h3>
Explanation:
This property is where we have A+B = B+A. In other words, adding two numbers can be done in any order. This can be extended to more than two values also.
So, 4+6 = 6+4 since both sides result in 10.
Overall it leads to 2+(4+6) = 2+(6+4) being a true equation as well. Both sides simplify to 12.
<span>The charity needs at least 1001 entries to raise more than $25000.
Let n be the number of race entries, i be the income of the race, and c the cost of the race.
The money that the 5K brings in is given by
i = 25 * n + 5000
The cost is given by
c = n * 5
Now if the charity needs to raise more than 25000 then the income minus the cost must be greater than 25000.
25000 < i - c
subsituting for i and c
25000 < i - c = (25 * n + 5000 ) - (5 * n)
simplifying results in
25000 < 25*n - 5*n + 5000
25000 < 20*n + 5000
20000 < 20*n
1000 < n
The number of entries must be greater than 1000. Since entries must be an integer the charity needs at least 1001 entries.</span>
Use the formula to find the area of a circle: Area = pi times radius squared
Divide each diameter by 2 to get the radius
Ex) 50/2=25
A= pi times 25 squared
25^2=625
A= 625 times pi
A=2,587.5
Answer:
about 4.17 hours
Step-by-step explanation:
To solve the time for each class, you do 2.5 hours divided by 3. Then you take this value and multiply by 5 since there are 5 classes. After you do this, you should get 4.17 hours roughly.
