Starting from 105 feet away a person on a bicycle rides towards a checkpoint and then passes it. the rider is traveling at a constant rate of 35 feet per second. THe distance between the bicycle and the checkpoint is given by the equation d= [105-35t] at what times is the bike 60 feet away from the checkpoint?
1.3 sec and 2.6 sec
1.3 sec and 4.7 sec****
1.1 sec and 2.6 sec
4.7 sec and 9.4 sec
Suppose P = {0,1,2,3,4,5,6,7,8,9} and V = {2,4,6,8,10,12,14,16,18,20} What is P O V
{0,1,2,3,4,5,6,7,8,9,10,12,14,16,18,20}
{0,1,3,5,7,9}
{10,12,14,16,18,20}****
{2,4,6,8}
Of 350 male athletes at a high school, some play only basketball, and some do both. if 250 of the males play basketball and 120 play both sports, how many of the males play baseball
a. 100
b. 220 **
c. 130
d. 120
In #2, it's not:
{10, 12, 14, 16, 18, 20}
It is:
{2, 4, 6, 8}
Good thing I checked that answer, or I would have had it wrong.
Looks good to me, if #2 means P∩V
Yes thanks
good nobody having a full on argument in this one
To find the times at which the bike is 60 feet away from the checkpoint, we can substitute the given distance value of 60 feet into the equation d = 105 - 35t and solve for t.
Here's how to do it step by step:
1. Start with the equation d = 105 - 35t, where d represents the distance between the bicycle and the checkpoint at time t.
2. Substitute the distance value of 60 feet into the equation: 60 = 105 - 35t.
3. Simplify the equation: 35t = 105 - 60.
4. Further simplify: 35t = 45.
5. Divide both sides of the equation by 35: t = 45 / 35.
6. Calculate the value: t ≈ 1.3 seconds.
Now we need to check if this value of t satisfies the given condition, i.e., if the bike is 60 feet away from the checkpoint at 1.3 seconds. We can substitute t = 1.3 into the original equation and check if the resulting distance is indeed 60 feet:
d = 105 - 35t
d = 105 - 35(1.3)
d ≈ 105 - 45.5
d ≈ 59.5 feet
Since the resulting distance is approximately 59.5 feet (close enough to 60 feet), the time t = 1.3 seconds is correct.
To find the second time, we can follow the same steps:
1. Start with the equation d = 105 - 35t.
2. Substitute the distance value of 60 feet into the equation: 60 = 105 - 35t.
3. Simplify the equation: 35t = 105 - 60.
4. Further simplify: 35t = 45.
5. Divide both sides of the equation by 35: t = 45 / 35.
6. Calculate the value: t ≈ 2.6 seconds.
Again, we need to check if t = 2.6 satisfies the given condition:
d = 105 - 35t
d = 105 - 35(2.6)
d ≈ 105 - 91
d ≈ 14 feet
Since the resulting distance is approximately 14 feet (not 60 feet), the time t = 2.6 seconds is incorrect.
Therefore, the correct times at which the bike is 60 feet away from the checkpoint are approximately 1.3 seconds and 4.7 seconds (not 2.6 seconds).