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3 years ago

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Which equation will solve the following word problem? Vanessa spent $87.00 on entertainment for the week. Her rent costs twelve

dollars less than double that amount. How much does Vanessa spend on rent?
2 * $87 - $12 = R

2R = $87 + $12

2R - $12 = $87

2($87) - 2($12) = R

1 answer:

Lunna [17]3 years ago

5 0

Answer:

2 * $87 - $12 = R

Step-by-step explanation:

Let R represent the cost of Vanessa's rent.

Vanessa spent $87.00 on entertainment for the week.

Her rent costs twelve dollars less than double that amount. Breaking it down, the expression "double that amount" is expressed as (2 × 87). Subtracting 12 from double that amount, the final expression would become

R = 2(87) - 12

R =

R = 174 - 12

The correct option is

2 * $87 - $12 = R

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Answer:

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Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

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$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

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Another solution is (9/8,0).

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$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

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The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

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S2: (3/8,-3/8)

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S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

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S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

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Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

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3 years ago