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Agata [3.3K]
3 years ago
11

Which is the graph of the equation

Mathematics
2 answers:
bezimeni [28]3 years ago
8 0

Answer:

The correct answer is C

Step-by-step explanation:

Tbh I just guessed but I got a 100% on the test, so I hoped that helped

AysviL [449]3 years ago
8 0

Answer:

yeah its c

Step-by-step explanation:

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211 times 9=<—round the answer.
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211 × 9 is 1899 so rounding to nearest thousands would be 2000

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1.3( x + 2 )-( y - 6) <br> Help please
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Basically, we need to plug in the given values for the variables x and y into the given expression 1.3(x+2)-(y-6). First off, we can plug in all the given values into the expression, giving us 1.3(4+2)-(2-6). Now, perform the operations on the inside of the parentheses. Doing this, we get 1.3(6)-(-4). Now, we use the distributive property to simplify. This gives us 7.8+4. Finally, when we add the two numbers, we get \boxed{11.8}. Hope this helped!
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3 years ago
Next in sequence 9/16, -9/8, 9/4, -9/2, 9, ?
pogonyaev

Answer:

-18

Step-by-step explanation:

*2, *(-2), *2, *(-2) ...

6 0
2 years ago
Write down the first three terms of the sequence n*2 + 2n
rjkz [21]

Answer:

3,8,15

Step-by-step explanation:

5 0
3 years ago
4. A company manufactures tires to meet the annual demand of 125,000 production runs. One production run involves producing 100
lyudmila [28]

Answer:

a) Economic order or production quantity = 2,500 tires.

Number of production runs in a year = 50 runs

Hence, 2,500 tires should be produced in each of the 50 runs in a year to minimize total cost.

b) Minimum total inventory cost = Tsh 30,000

Step-by-step explanation:

The total cost for the tire production firm will be a sum of the total production cost and total inventory cost.

Total cost = Total Production cost + Total inventory cost

Total Production Cost = (Number of production runs in a year) × (Setup Cost of one production run)

Number of production runs in a year = (Annual demand)/(Number of units produced per production run)

Let the annual demand = D

Number of units produced per production run = Q

Setup Cost of one production run = S

Number of production runs in a year = (D/Q)

Total Production Cost = (DS/Q)

Total inventory Cost = (Average inventory level) × (Cost of holding 1 unit in inventory)

Average inventory level is usually assumed to be half of the number of units in a production run = (Q/2)

Cost of Holding a unit of product in inventory = H

Total inventory Cost = (QH/2)

Total cost = TC = (DS/Q) + (QH/2)

At minimum cost, (dTC/dQ) = 0

(dTC/dQ) = -(DS/Q²) + (H/2) = 0

(DS/Q²) = (H/2)

Q² = (2DS/H)

Hence,

Economic order/production quantity = Q = √(2DS/H)

For this question

D = Annual demand = 125,000 tires

S = Setup cost for one production run = Tsh 600

H = Holding cost for one unit in inventory = Tsh 24

Q = √(2×125000×600/24) = 2,500 units

Number of production runs in a year = (D/Q) = (125000/2500) = 50 production runs.

b) Total Inventory Cost = (QH/2)

At minimum total inventory cost, Q = 2,500

Minimum total inventory cost = (2500×24/2) = Tsh 30,000

Hope this Helps!!!

4 0
3 years ago
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