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Crank
3 years ago
7

What is the value of k?​

Mathematics
2 answers:
iragen [17]3 years ago
8 0

Answer:

(4k+5)+ (6k+10) = 115

Exterior angle of a triangle is equal to the sum of two opposite interior angles

10k=15=115

10k= 100

k = 100/10

k = 10

Step-by-step explanation:

REY [17]3 years ago
8 0

Answer:

k = 10

Step-by-step explanation:

The 3rd angle that's not defined in the equation is supplementary to 115, so the angle is 180-115 = 65.

We can solve for the equation 65+(4k+5)+(6k+10)=180 -> 65+4k+5+6k+10=180 -> 10k+80 = 180 -> 10k = 100 -> k = 10.

To double-check(optional), we can plug k back into the equation, so 65 + 4*10 + 5 + 6*10 + 10 = 180 -> 100 + 65 + 5 + 10 = 180 -> 180 = 180, so we can confirm that k = 10.

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In how many ways can 100 identical chairs be distributed to five different classrooms if the two largest rooms together receive
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Answer:

There are 67626 ways of distributing the chairs.

Step-by-step explanation:

This is a combinatorial problem of balls and sticks. In order to represent a way of distributing n identical chairs to k classrooms we can align n balls and k-1 sticks. The first classroom will receive as many chairs as the amount of balls before the first stick. The second one will receive as many chairs as the amount of balls between the first and the second stick, the third classroom will receive the amount between the second and third stick and so on (if 2 sticks are one next to the other, then the respective classroom receives 0 chairs).

The total amount of ways to distribute n chairs to k classrooms as a result, is the total amount of ways to put k-1 sticks and n balls in a line. This can be represented by picking k-1 places for the sticks from n+k-1 places available; thus the cardinality will be the combinatorial number of n+k-1 with k-1, {n+k-1 \choose k-1} .

For the 2 largest classrooms we distribute n = 50 chairs. Here k = 2, thus the total amount of ways to distribute them is {50+2-1 \choose 2-1} = 51 .

For the 3 remaining classrooms (k=3) we need to distribute the remaining 50 chairs, here we have {50+3-1 \choose 3-1} = {52 \choose 2} = 1326 ways of making the distribution.

As a result, the total amount of possibilities for the chairs to be distributed is 51*1326 = 67626.

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