^{+}it declines from +∞ to 5 at x=∞.

1) In a pilot project,a rural township was was given recycling bins for separating and storing recyclable products.the cost in dollars for supply bins to p% of the population is giving

C=(50,000p)/(200-2p),0≤p<100

a)find the cost of giving bins to 45% of the population

b)find the cost of giving bins to 60% of the population.

C)find the cost of giving bins to 96% of the population

D) according to this model would it be possible to sumply bins to 100% of residents.

?

2)a)Determine the value of the function f as the magnitude of x increases.

b)is f(x) greater than or less than this functional value when x is positive and larger in magnitude

C) is f(x)greater than or less than this functional value when x is negative and larger in magnitude?

f(x)=5+(2)/(2x-6)

Tutors, sign in to answer this question.

Daisy,

For question 1a all you have to do is substitute p=45 in the cost equation and do the arithmetic as follows

C=50000*45/(200-2*450) =20,454 Do a simular calculation for p=60 and 96. For p= 100 the denominator becomes zero (200-2p)=0 when p=100 and the cost would be infinite i.e. very large so according to this model it would not be possible to supply bins.

2. This function starts at 5 at x=-∞ and decreases to -∞ at =3, (i.e. where2x-6=0) at x=3^{+}it declines from +∞ to 5 at x=∞.

Regards

Jim

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

C = ( 50,000 P) / ( 200 - 2 P)

a) 50000* ( 0.45) / ( 200 - 2( 0.45)) = 113.01

b) 50000 * ( 0.60) / ( 200 - 2( 0.0.60) ) = 150.91

c ) 50000* (0.96) / ( 200 -2( 0.96)) = 242.33

2)

f( X ) = 5 +( 2) / ( 2X - 6 )

2 / ( 2x - 6 )

2 X - 6 > 0

for : X > 3 2X - 6 > 0 the value of f( X ) = 5 +( 2) / (2X -6 ) Decreases.

for : X < 3 2x - 6 <0 the value of f(x ) decreases

b & c , it does not specify the function when referred to as this function

1)

C(p) = 50000 p/(200 - 2p), 0≤p<100

C(p) = 25000 p/(100 - p) = 25000/(100/p - 1)

C(45) = 25000/(100/45 - 1) ≈ $20,454.55

C(60) = 25000/(100/60 - 1) ≈ $37,500.00

C(p) = 50000 p/(200 - 2p), 0≤p<100

C(p) = 25000 p/(100 - p) = 25000/(100/p - 1)

C(45) = 25000/(100/45 - 1) ≈ $20,454.55

C(60) = 25000/(100/60 - 1) ≈ $37,500.00

C(96) = 25000/(100/96 - 1) ≈ $60,000.00

C(100) is undefined because there is division by zero.

2)

f(x) = 5 + 2/(2x-6) = 5 + 1/(x-3)

f(10^100) = 5 + 1/(10^100 - 3) ≈ 5 + 0

f(-10^100) = 5 + 1/(-10^100 - 3) ≈ 5 - 0

So as x —> ± inf, f(x) —> 5 ± 0

y = 5 is a horizontal asymptote approached by f(x):

from above on the right, and

from below on the left.

Thank you so much

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