In 2005, there were 14,100 students at college A, with a projected enrollment increase of 750 students per year. In the same year, there were 42,100 students at college B, with a projected enrollment decline of 1250 students per year. According to these projections, when will the colleges have the same enrollment? What will be the enrollment in each college at that time?

Set up two equations and set equal to each other. Let number of years = x:

College A = 14100+750x

College B = 42100-1250x

Set equal:

14100 + 750x = 42100 - 1250x

Subtract 750x from both sides:

14100 = 42100 - 2000x

Subtract 42100 from both sides:

-28000 = -2000x

Divide both sides by -2000:

x = -28000 / -2000

x = 14

It will take 14 years for the schools to have the same enrollment.

Enrollment will be:

14100 + 750(14) = 14100 + 10500 = 24,600

(a)2019 (14 years after)

(b)24,600

Step-by-step explanation:

Let the number of years =n

College A

Initial Population in 2005 = 14,100

Increase per year = 750

Therefore, the population after n years = 14,100+750n

College B

Initial Population in 2005 = 42,100

Decline per year = 1250

Therefore, the population after n years = 42,100-1250n

When the enrollments are the same

14,100+750n=42,100-1250n

1250n+750n=42100-14100

2000n=28000

n=14

Therefore, in 2019 (14 years after), the colleges will have the same enrollment.

Enrollment in 2019 =42,100-1250(14)

=24,600

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584

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