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zhannawk [14.2K]
2 years ago
11

The populations of two cities after t years can be modeled by -150t+50,000 and 50t+75,000 . What is the difference in the popula

tions of the cities when t=4 ?
Mathematics
2 answers:
Alex2 years ago
4 0
50t+34 t=4 is accelerating
creativ13 [48]2 years ago
3 0

Answer:

The difference in the populations of the cities when t=4 is 25,800.

Step-by-step explanation:

Given equations each of the city:

-150t+50,000

and

50t+75,000

To find the difference between the population when t=4, we first have to solve both equation when t=4.

For both equations, substitute 4 into t:

-150(4) + 50,000 = 49,400

50(4) + 75,000 = 75,200

Now subtract 49,400 from 75,200 and you get 25,800.

The difference in the populations of the cities when t=4 is 25,800.

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4 0
3 years ago
Let u,v,wu,v,w be three linearly independent vectors in R7R7. Determine a value of kk, k=k= , so that the set S={u−3v,v−2w,w−ku}
11Alexandr11 [23.1K]

Answer:

1/6

Step-by-step explanation:

For a set of vectors a, b and c to be linearly dependent, the linear combination of the set of vectors must be zero.

C1a + C2b + C3c = 0

From the question, we are given the set S={u−3v,v−2w,w−ku}, the corresponding vectors are a(0, -3, 1), b(-2,1,0) and c(1,0,-k)

The values in parenthesis are the ccoefficients of w, v and u respectively.

On writing this vector as a linear combination, we will have;

C1(0, -3, 1) + C2(-2,1,0) + C3(1,0,-k) = (0,0,0)

0-2C2+C3 = 0........ 1

-3C1+C2 = 0 ........... 2

C1-kC3 = 0 ….......... 3

From equation 2, 3C1 = C2

Substituting into 1, -2(3C1)+C3 = 0

-6C1+C3 = 0

-6C1 = -C3

6C1 = C3.…..4

Substitute 4 into 3 to have

C1-k(6C1) = 0

C1 = k6C1

6k = 1

k = 1/6

Hence the value of k for the set of vectors to be linearly dependent is 1/6

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