398/74 as a mixed number in simplest form
2 answers:
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Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.
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Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote
I would write it as
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For problem 3, we first need to convert the runner's speed from mph to feet per second.
Since the runner needs to travel 90-12 = 78 ft, this means
So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.
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The question is now: how much of a lead does the runner need in order to beat the throw?
Well the runner needs to get to second base in under 0.96 seconds.
Let's calculate the distance based on that, and based on the speed we calculated earlier above.
This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet
This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.
As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.
Answer:
a.No
b.No
c.No
Step-by-step explanation:
a.No,Such set does not exist .A set of natural numbers is N
Every point of this set is an isolated point but no accumulation point
Accumulation point:It is defined as that point a of set Swhich every neighborhood contains infinitely many distinct point of set
Isolated point : it is defined as that point a of set S which neighborhood does not contain any other point of set except itself
Interior point of set :Let .Then a is called interior point of set when its neighborhood is a subset of set S.
When a set is uncountable then interior point exist it is necessary for interior points existance .
Boundary points :Let .If every non empty neighborhood of a intersect S and complement of S.
Every member of a set is a boundary point
b.No, such set does not exist .A non empty set with isolated point then the set have no interior points .By definition of interior point and isolated point .For example.set of natural numbers
c.No, Such set does not exist ,for example set of natural every point is an isolated point and boundary point.By definition of boundary point and isolated point