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Ksivusya [100]
3 years ago
12

Solve the equations in Parts A and B using inverse operations. Check your solutions. In your final answer, include all of your w

ork.
Part A: 5 + x^2 = 2x^2 + 13
Part B: 5 + x^3 = 2x^3 + 13
Mathematics
2 answers:
12345 [234]3 years ago
4 0

Answer:

Part A:5=2x^2-x^2+13

5=x^2+13

0=x^2+13-5

0=x^2+8

Part B:5=2x^3-x^3+13

5=x^3+13

0=x^3+13-5

0=x^3+8

Step-by-step explanation:

Ludmilka [50]3 years ago
3 0
Part A:5=2x^2-x^2+13
5=x^2+13
0=x^2+13-5
0=x^2+8

Part B:5=2x^3-x^3+13
5=x^3+13
0=x^3+13-5
0=x^3+8
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Answer:

a) At break even point, Q = 380 units, Sales dollar = $5247.8

b) Margin of safety at Q = 400 units is 0.05 × 100 = 5

c) Number of cakes that need to be sold before a profit of $2200 is made = 595 cakes

Step-by-step explanation:

Let the number of units be Q

Cost equation = (fixed cost) + (Variable cost)

Fixed Cost = $3898.8

Variable Cost = Variable cost per unit × number of units produced = (2.32 + 1.11 + 0.12) × Q = 3.55Q

Cost equation = 3898.8 + 3.55Q

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Total cost = Revenue

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In sales dollars, R = 13.81Q = 13.81 × 380 = $5247.8

b) Margin of safety = 100 × (Current sales - Breakeven point sales)/(current sales)

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At Q = 400 units, Current sales = 13.81 × 400 = $5524

Margin of safety = 100 × (5524 - 5247.8)/5524 = 5

c) number of cakes that Cove must sell to generate $2,200 in profit.

Profit = Revenue - Total Cost

2200 = 13.81Q - (3898.8 + 3.55Q)

2200 = 10.26Q - 3898.8

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Q = 594.4 = 595 units

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