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Allisa [31]
2 years ago
12

Write an equation for a line perpendicular to y = 5x – 2 and passing through the point (-15,8) y=​

Mathematics
1 answer:
AleksandrR [38]2 years ago
8 0
The answer is y = -1/5x + 5

Perpendicular means the opposite reciprocal of the slope. The reciprocal of 5 would be 1/5. The opposite sign would be negative, so -1/5.

The equation that you use is y - y1 = m(x - x1)

The point is (-15,8).

x1 = -15
y1= 8
m= -1/5

y - 8 = -1/5(x - (-15)).

The 15 becomes positive.

y - 8 = -1/5(x+15)

Distribute

y - 8 = -1/5x - 3

Add 8 to both sides

y = -1/5x + 5

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Suppose that a company needs 1, 200,000 items during a year and that preparation for each production run costs $500. Suppose als
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Answer:

The number of items in each production run so that the total costs of production and storage are minimized is 8165 items/run

Step-by-step explanation:

We will use the following variables:

Q = Quantity being ordered

Q* = the optimal order Quantity: the result being sought

D = annual Demand for the item, over the year

P = unit Production cost

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run)

H = annual cost to Hold one unit

It is important to note which variables are annualized, which are per-order and which are per-unit.

Using the variables, here are the components of the first equation

Total Cost, TC = PC + SC + HC

PC = P x D :  Production Cost = unit Production cost times the annual Demand

SC = (D x S)/Q : Setting up Cost = annual Demand times cost per production setup, divided by the order Quantity (number of units)

HC = (H x Q)/2: Holding Cost = annual unit Holding cost times order Quantity (number of units), divided by 2 (because throughout the year, on average the warehouse is half full).

So TC = PC + SC + HC =  (P x D) + ((D x S)/Q) + ((H x Q)/2) = PD + (DS/Q) + HQ/2

To obtain the optimal order quantity, Q* that minimizes TC, at the minimum TC, dTC/dQ = 0

dTC/dQ = (H/2) – (D x S)/(Q²) = 0

(H/2) – (D x S)/(Q²) = 0

Solving for Q, which is Q* at this point.

(Q*)² = 2DS/H

Q* = √(2DS/H)

D = annual demand for the item = 200000

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run) = $500

H = annual cost to Hold one unit = $3

Q* = √(2×200000×500/3) = 8164.97 = 8165 items.

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