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Lena [83]
3 years ago
14

A van will drive ten miles north, 15 miles south, and then five miles north again. The van gets 33 miles per gallon, and there i

s one gallon of gas left in the tank. Compute the relative distance and the total distance. Then decide if there is enough fuel for the trip.
Mathematics
2 answers:
gladu [14]3 years ago
7 0
The total distance is 30 miles 33 is the relative distance yes there is enough fule for the trip
Finger [1]3 years ago
4 0

Answer:

30 miles.

Step-by-step explanation:

A van will drive 10 miles north, 15 miles south, and then 5 miles north again.

Van will drive total distance = 10 + 15 + 5  = 30 miles.

Relative distance is 0 miles since the van is back where it started.

The total distance to travel is 30 miles and van gets 33 miles per gallon. There is one gallon of gas left in the tank so there is enough fuel for the trip.

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Step-by-step explanation:

\frac{Meg}{Tanner} = \frac{4}{3}

total dist. driven = dist. Meg drove + dist. Tanner drove

Meg = 112

Tanner = x    or how far did Tanner drive?

\frac{112}{x} = \frac{4}{3}

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What is the value of x? x + (-10) = 2 Enter your answer in the box
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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
MAVERICK [17]

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.

3 0
3 years ago
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