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Ber [7]
3 years ago
9

A camera manufacturer spends $2,000 each day for overhead expenses plus $9 per camera for labor and materials. The cameras sell

for $17 each.
a. How many cameras must the company sell in one day to equal its daily costs?
b. If the manufacturer can increase production by 50 cameras per day, what would their daily profit be?

250; $400
170; $240
118; $656
250; $850
Mathematics
1 answer:
hjlf3 years ago
6 0
A) let the number of cameras sold per day for breakeven be x

Total daily cost = 2000 + 9x

Total daily revenue = 17x

therefore for just covering expenses both cost and revenue must be equal

2000 + 9x = 17x

2000 = 17x - 9x = 8x

x = 2000/8 = 250 cameras


b) increasing production by 50 cameras per day will give a daily profit of;

50 * (17 - 9) = 50 * 8 = $400 (seeing that the fixed daily cost of $2000 remains unchanged)


It's a
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Please help mee pleaseee
nalin [4]

Answer:

The exact answer in terms of radicals is x = 5*\sqrt[3]{25}

The approximate answer is x \approx 14.62009 (accurate to 5 decimal places)

===============================================

Work Shown:

Let y = \sqrt[5]{x^3}

So the equation reduces to  -7  = 8-3y

Let's solve for y

-7 = 8-3y

8-3y = -7

-3y = -7-8 ... subtract 8 from both sides

-3y = -15

y = -15/(-3) ... divide both sides by -3

y = 5

-----------

Since y = \sqrt[5]{x^3} and y = 5, this means we can equate the two expressions and solve for x

y = 5

\sqrt[5]{x^3} = 5

x^3 = 5^5 Raise both sides to the 5th power

x^3 = 3125

x = \sqrt[3]{3125} Apply cube root to both sides

x = \sqrt[3]{125*25}

x = \sqrt[3]{125}*\sqrt[3]{25}

x = \sqrt[3]{5^3}*\sqrt[3]{25}

x = 5*\sqrt[3]{25}

x \approx 14.62009

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Step-by-step explanation:

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Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

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