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2 years ago

5

If Morgan has 3 3/4 pounds of roast beef and Eric has 2 2/4 pounds of roast beef. how many more pounds of roast beef does Morgan

have than Eric?

1 answer:

ivann1987 [24]2 years ago

7 0

Answer:

1 1/4 pounds

Step-by-step explanation:

convert the fractions into improper fractions for easier calculations

15/4 - 10/4 = 5/4

5/4 pounds

covert back into mixed numbers: 1 1/4

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Step-by-step explanation:

Counterclockwise is going left, so if you start at the top, 270 degrees left of a clock will result in the same as 90 degrees clockwise.

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A) Goats that weigh exactly 50 or 60 pounds fall into two classes.

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3 years ago

Suppose the Willard gold nugget was turned into 4 equal sized gold bricks. If one of the bricks was sold how many ounces of the

Zielflug [23.3K]

788/4 = 197

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3 years ago

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Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

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

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:

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

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

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

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

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:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

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

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

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

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:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

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)

$H = -9$ (Saddle Point)

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

$H = 3$ (Local maximum or minimum)

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)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

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

S4: (0, - 9/8)

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

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

Solve for all values of x in simplest form. 9-4|3+5x|=5

Zinaida [17]

Answer:

$x=-2/5\text{ or } x=-4/5$

Step-by-step explanation:

So we have the equation:

$9-4|3+5x|=5$

First, subtract 9 from both sides:

$-4|3+5x|=-4$

Divide both sides by -4:

$|3+5x|=1$

Definition of Absolute Value:

$3+5x=1\text{ or } 3+5x=-1$

Subtract 3 from both sides:

$5x=-2\text{ or } 5x=-4$

Divide both sides by 5:

$x=-2/5\text{ or } x=-4/5$

And we're done!

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3 years ago

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