Find the circumference of the circle

Mathematics

1 answer:

vaieri [72.5K]3 years ago

4 0

Answer:

72.22

Step-by-step explanation:

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Determine which consecutive integers do not have a real zero of between them. a. (–6, –5) c. (–2, –1) b. (–5, –4) d. (1, 2) Plea

svetoff [14.1K]

We have by the intermediate value theorem that if a continuous function takes values both above and below zero at 2 points, there is a zero of the function in-between. We have that polynomials are continues. Let's calculate f(-6) and f(-5). f(-6)=-36 while f(-5)=-1. Thus, we cannot conclude that there is a root between them.
F(-2)=8, f(-1)=-1, so there is a flip; a zero must exist between them.
F(1)=-1, f(2)=20, so again there is a change of signs.
f(-5)=-1, f(-4)=14 so there is a root still.
We have that the only choice that does not have a root between the integers is choice a.

7 0

3 years ago

I need help and the time is running please help

Alex Ar [27]

Answer:

I think its B

Step-by-step explanation:

I'm sorry if I'm wrong. I'm busy

7 0

3 years ago

Read 2 more answers

If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei

MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is ${11 \choose 3} = 165 .$ As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is ${7 \choose 3} = 35$ . Thus, there are only 35 ways to distribute the blackboards in this case.