Kevin is 333 years older than Daniel. Two years ago, Kevin was 444 times as old as Daniel.

Mathematics

2 answers:

tia_tia [17]3 years ago

5 0

Answer: the equations are

k = d + 33

k - 2 = 4(d-2)

Step-by-step explanation:

Let the present age of Kevin be represented by k

Let the present age of Daniel be represented by d

Kevin is 33 years older than Daniel. This means that the expression for their current age is

k = d + 33

Two years ago, Kevin was 4 times as old as Daniel. This means that 2 will be subtracted from their present ages to depict two years ago. Therefore, Two years ago for Daniel will be d-2

Two years ago for Kevin will be k - 2

Remember that Kevin was 4 times as old as Daniel two years ago, it becomes

k - 2 = 4(d-2)

babunello [35]3 years ago

5 0

Answer: $k= d+3$

$k-2 = 4(d-2)$

Step-by-step explanation:

Let k be Kevin's age and Let d be Daniel's age.

Given : Kevin is 3 years older than Daniel.

i.e. Age of Kelvin = Age of Daniel+ 3

i.e. In terms of variables , k= d+3 (1)

Two years ago,

Age of Kelvin = k-2

Age of Daniel = d-2

Since Two years ago ,Kevin was 4 times as old as Daniel.

i.e. In terms of variables : $k-2 = 4\times(d-2)$

$k-2 = 4(d-2)$ (2)

From (1) and (2) the system of equations represents this situation are

$k= d+3$

$k-2 = 4(d-2)$

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Answer:

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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.