Answer:
<h2>35 different ways</h2>
Step-by-step explanation:
Since there are 7 students in a classroom to fill a front row containing 3 seats, we will apply the combination rule since we are to select 3 students from the total number of 7 students in the class.
In combination,<em> if r objects are to be selected from a pool of n objects, this can be done in nCr number of ways.</em>
<em>nCr = n!/(n-r!)r!</em>
Selecting 3 students from 7 students to fill the seats can therefore be done in 7C3 number of ways.
7C3 = 7!/(7-3)!3!
7C3 = 7!/(4)!3!
7C3 = 7*6*5*4!/4!*3*2
7C3 = 7*6*5/6
7C3 = 7*5
7C3 = 35
<em>Hence there are 35 different ways that the student can sit in the front assuming there are no empty seats.</em>
Answer:
FJ= 24
DM= 16
EM= 18
Step-by-step explanation:
The centroid is twice as close along any median to the side that the median intersects as it is to the vertex it emanates from.
So in the case of FJ, the median intersects the side at point J and extends from vertex F. This means that FM is twice as long as MJ. It tells us that MJ is 8, so MF would be 16. Now because FJ is just FM+MJ, <u>FJ is 24.</u>
It also tells us that DK is 24, which means that DM+MK = 24. If DM is twice as long as MK, that would make MK equal to 8 and <u>DM equal to 16</u>
<u />
Similarily, EM would be twice as long as ML. It says that ML=9 so that would make <u>EM=18</u>
QUE :
75 - [5 + 3 of (25 - 2 × 10)]
= 75 - [5 + 3 of ( 25 - 20)]
= 75 - [5 + 3 of 5]
= 75 - [5 + 15]
= 75 - 20
= 55
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Answer:
7r+63
Step-by-step explanation:
7(r+9)
7(r)+7(9)
7r+63
Hope this helps .-.
