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nikdorinn [45]
2 years ago
9

Write a linear function that models each situation. a 24. The enrollment at a local community college was 5,000 in 1960. In the

year 2000, the enrollment was 15,000. Assuming that the enrollment grows linearly since 1960, write a linear equation that represents the enrollment of the college x years after 1960.​
Mathematics
1 answer:
tia_tia [17]2 years ago
8 0

9514 1404 393

Answer:

  y = 250x +5000

Step-by-step explanation:

The "initial" enrollment is 5000, when x=0 years after 1960.

The rate of change of enrollment is ...

  change of enrollment / change in time

  = (15000 -5000)/(2000 -1960)

  = 10000/40 = 250 . . . . students per year

Then the desired equation is ...

  y = 250x + 5000

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What is the solution to 2x-8 <12?
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x < 10

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a paintball Court charges an initial entrance fee plus a fixed price for ball pease represents the total price n dollars as func
Evgen [1.6K]

The complete question is;

A paintball court charges an initial entrance fee plus a fixed price per ball.

P represents the total price (in dollars) as a function of the number of balls used n.

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Cost of 10 balls = 8 dollars

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8 0
3 years ago
Is 2 a solution of the inequality 4 + a ≤ 6? *
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Answer:

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3 years ago
A home improvement contractor is painting the walls and ceiling of a rectangular room. The volume of the room is 1584 cubic feet
liq [111]

Answer:

The room dimensions that will minimize the cost of the paint are 12 ft x 12 ft x 11 ft.

Step-by-step explanation:

We can find first the volume equation using the formula of the volume of a box.

V= xyz

Thus we get the constraint function

1584 = xyz

Then since we are asked to minimize the cost, we can write the cost function which is the area of each one of the walls and ceiling multiplied by the painting cost.

C=0.11 xy+ 2(0.06)xz+2(0.06yz \\ C =0.11 xy+0.12xz+0.12yz

Lagrange Multipliers to find minimum cost.

We can continue finding the partial derivatives to build the system of equations required for Lagrange Multipliers method.

C_x=\lambda V_x \\ C_y = \lambda V_y \\ C_z = \lambda V_z

And the constraint function

xyz=1584

So we get

0.11y+0.12z=\lambda yz \\ 0.11x+0.12z=\lambda xz \\ 0.12x+0.12y=\lambda xy\\ xyz=1584

We can multiply each side of each equation by the dimension which is missing to get the full volume on the right side.

0.11xy+0.12xz=\lambda xyz \\ 0.11xy+0.12yz=\lambda xyz \\ 0.12xz+0.12yz=\lambda xyz

Then we can set each the equations equal to each other, so from the first one and the second equation we get

0.11xy+0.12xz= 0.11xy+0.12yz

We can subtract 0.11xy from both sides.

0.12xz=0.12yz

And we can divide both sides by 0.12z to get

x=y

We can repeat the process by setting the first and third equation equal to each other.

0.11xy+0.12xz= 0.12xz+0.12yz

We can subtract 0.12 xz from both sides

0.11xy=0.12yz

And we can solve by z

z= \cfrac{0.11x}{0.12}\\ z = \cfrac{11x}{12}

So if we replace that as well y = x on the constraint for the volume euqation we get

1584=x(x)\left(\cfrac{11}{12}x\right) \\ 1584=\cfrac{11}{12}x^3

We can then solve for x

x^3 = \cfrac{1584(12)}{11}

And taking the cube root

x = \sqrt[3]{\cfrac{1584(12)}{11}}

x = \sqrt[3]{1728}

x=12 ft

So then we can use the equations we have found for y and z in terms of x

y = x \\ y = 12 ft

And

z= \cfrac{11x}{12}\\z= \cfrac{11(12)}{12} \\ z=11ft

Then the dimensions of the room that will minimize the cost are 12ft x 12 ft x 11 ft. Since you have to enter using commas you can write 12, 12, 11, please check as well if you have to insert the units that are feet for each.

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