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torisob [31]
2 years ago
7

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics
1 answer:
Elanso [62]2 years ago
7 0

Answer: Im pretty sure its 143 if it was degrees but yeah

Step-by-step explanation:

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Which of the following graphs represents the equation below?: -9x – 3y = -8
Mamont248 [21]
It’s C! that’s the answer, hope it helps
7 0
2 years ago
Help me pleaseee (10 points )
Gemiola [76]

Answer:

m<T = , m<M =  and m<Z =

Step-by-step explanation:

From the given ∆TMZ, let the measure angle T be represented by T.

So that,

m<M = 2T + 6°

m<Z = 5T - 50°

Sum of angles in a triangle =

T + (2T + 6°) + (5T - 50°) =

8T - =

8T =  +

   =

T =

  =

Therefore,

i. m<T =

ii. m<M = 2T + 6°

       = 2 x  + 6°

       =

m<M =

iii. m<Z = 5T - 50°

           = 5 x  - 50°

           =  - 50°

           =

m<Z =

5 0
3 years ago
You are on a low fat diet.
Alexxx [7]

Answer:

Probably just egg sandwich

Step-by-step explanation:

protein  

7 0
3 years ago
Read 2 more answers
PLEASE HELP WITH ALGEBRA WILL GIVE BRAINLIEST
nataly862011 [7]

Answer:

x^5/4

Step-by-step explanation:

The first one

8 0
3 years ago
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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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