Answer the question for brainliest. (Simplify Please)

Mathematics

1 answer:

mr_godi [17]3 years ago

5 0

$1 \frac{2}{5} m - \frac{3}{5} ( \frac{2}{3} m + 1)$

$\frac{7}{5} m - \frac{6}{15} m - \frac{3}{5}$

$\frac{21}{15} m - \frac{6}{15} m - \frac{3}{5}$

$( \frac{(21 - 6)m}{15} ) - \frac{3}{5}$

$\frac{15}{15} m - \frac{3}{5}$

$1m - \frac{3}{5}$

$m - \frac{3}{5}$

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

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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.