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Troyanec [42]
2 years ago
14

Zippy company manufactured 10,000 units in December with a total product cost of $25,472 They had zero finished goods inventory

at the start of December. In December Zippy sold 7,364 units at a unit price of $5.26. Period expenses were $3,722. What is the amount of Zippy's operating income? Round any intermediate calculations to four decimal places.
Mathematics
1 answer:
Mnenie [13.5K]2 years ago
6 0

The amount of Zippy's operating income is

  • $16255.0592

<h3>Cost per unit is given as </h3>

= \frac{total cost of goods manufactured}{number of units}\\\\&#10;= \frac{25472}{10000}\\\\&#10;= $2.5472 per unit&#10;

So, cost of goods sold

= Number of units * cost per unit\\\\&#10;= 7364 * 2.5472 \\\\&#10;= $18757.5808&#10;

Therefore, operating income

= sales - cost of goods sold - period expenses\\\\&#10;= 7364*5.26 - 18757.5808 - 3722\\\\&#10;= $16255.0592&#10;

For more information on operating income, visit

brainly.com/question/16232387

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What is the length of the line segment with endpoints (11, −4) and (−12, −4
RUDIKE [14]

Answer:

23 units

Step-by-step explanation:

Use the distance formula: d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}

Plug in the 2 points:

d = \sqrt{(-12 - 11)^2 + (-4 + 4)^2}

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3 years ago
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If 2y^2+2=x^2, then find d^2y/dx^2 at the point (-2, -1) in simplest form.​
horrorfan [7]

Answer:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}

Step-by-step explanation:

We have the equation:

2y^2+2=x^2

And we want to find d²y/dx² at the point (-2, -1).

So, let's take the derivative of both sides with respect to x:

\frac{d}{dx}[2y^2+2]=\frac{d}{dx}[x^2]

On the left, let's implicitly differentiate:

4y\frac{dy}{dx}=\frac{d}{dx}[x^2]

Differentiate normally on the left:

4y\frac{dy}{dx}=2x

Solve for the first derivative. Divide both sides by 4y:

\frac{dy}{dx}=\frac{x}{2y}

Now, let's take the derivative of both sides again:

\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{x}{2y}]

We will need to use the quotient rule:

\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}

So:

\frac{d^2y}{dx^2}=\frac{\frac{d}{dx}[(x)](2y)-x\frac{d}{dx}[(2y)]}{(2y)^2}

Differentiate:

\frac{d^2y}{dx^2}=\frac{(1)(2y)-x(2\frac{dy}{dx})}{4y^2}

Simplify:

\frac{d^2y}{dx^2}=\frac{2y-2x\frac{dy}{dx}}{4y^2}

Substitute x/2y for dy/dx. This yields:

\frac{d^2y}{dx^2}=\frac{2y-2x\frac{x}{2y}}{4y^2}

Simplify:

\frac{d^2y}{dx^2}=\frac{2y-\frac{2x^2}{2y}}{4y^2}

Simplify. Multiply both the numerator and denominator by 2y. So:

\frac{d^2y}{dx^2}=\frac{4y^2-2x^2}{8y^3}

Reduce. Therefore, our second derivative is:

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We want to find the second derivative at the point (-2, -1).

So, let's substitute -2 for x and -1 for y. This yields:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(-1)^2-(-2)^2}{4(-1)^3}

Evaluate:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(1)-(4)}{4(-1)}

Multiply:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2-4}{-4}

Subtract:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{-2}{-4}

Reduce. So, our answer is:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}

And we're done!

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OR
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4 0
4 years ago
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