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Tresset [83]
3 years ago
5

8 + 6x - 10x = 16 - 8x

Mathematics
1 answer:
Svetllana [295]3 years ago
4 0
The answer is x=2

Remember, first add/substract the like terms, in this case 6x-10x to get -4x. Then, add 8x to negative 8x to cancel it out and do the same thing to -4x. Do this with all your terms and you should get 4x=8. Then divide both sides by 4 to get x=2.
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Makes no sense I wish it did
34kurt

Answer:

<BCD=105 Degrees

Step-by-step explanation:

To solve this problem, you need to use linear pairs

Linear pairs are two angles on the same line divided by a single line between them. Linear pairs will always add up to 180 as they are supplementary.

Therefore, 180-75 will give you x, which is 105

8 0
3 years ago
If a family uses 3.5 gallons of milk per week, how much will his family need for four days?
Ad libitum [116K]

Answer:

2

Step-by-step explanation:

3.5 gallons = 7 days

3.5/7 = 0.5

0.5 x 4 = 2

3 0
3 years ago
Read 2 more answers
Aflati un numar natural care impartit la 17 da catul 23 si restul 11.
Anit [1.1K]
I cannot understand what you are saying 
5 0
3 years ago
Is -0.025 greater or less than -0.14?
AlekseyPX

Answer:

less than

Step-by-step explanation:

-0.025 is further away from -0.14 in the number line.

8 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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