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3 years ago

10

;-;could someone help a girl out? I've got at least 8 pages of this left and this is the part I'm stuck on! I'll give you brainl

ist if you help and not be mean and take my points!
1) 89.610 -26.632
2) 29.00 +27.69
3) 71.0 -12.3
4) 26.000 +13.824
5) 51.00 -38.75
6) 54.70 + 9.39
7) 63.030 -59.688
8) 83.000 +77.841
9) 93.0 -32.2
10) 66.00 + 8.84
11) 33.970 - 8.851
12) 48.000 +44.63
13) 98.0 -69.9
14) 10.0 + 7.1
15) 90.0 -83.0

1 answer:

Zepler [3.9K]3 years ago

7 0

Answer:

1. 62.978 2. 56.69 3. 58.7 4.39.824 5. 12.25 6 64.09. 7. 3.342 8. 160.841 9. 60.8 10. 74.84 11 25.119 12. 92.63

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A farmer is going to divide her 30 acre farm between two crops. Seed for corn costs $15 per acre. Seed for wheat costs $30 per a

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Answer:

$3200

Step-by-step explanation:

The total area of land is 30 acre. Let x represent the area in acre that corn is planted and y represent the area in acre that wheat is planted. Therefore:

x + y = 30 (1)

Also, the farmer can spend at most $750 on seeds. Corn cost $15 per acre, maize cost $30 per acre. Hence:

15x + 30y ≤ 750 (2)

Also, x and y > 0

The profit is given by the function:

Profit = 100x + 110y

Graphing equation 1 and equation 2 and solving, we get the points:

(10, 20)

Hence the maximum profit is at:

Max profit = 100(10) + 110(20) = $3200

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2 years ago

A businessman was comparing his company's profits for two consecutive months. In November, n, were 0.04 thousand dollars more th

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N and d are both numbers of thousands of dollars.
Thus, if n = 1, that means $1000.

Here n = d + 0.4.
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Similarly, in D, d = 0.04 = n is completely wrong.
The "combined profit amount" for Nov. and Dec. is n + d = 3.15.

Only A matches this info. Your answer is A.

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2 years ago

Jenny is making a quilt from the pattern blocks shown below.

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Answer:

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2 years ago

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

Use the point-slope form, y - y1 = m(x - x1), to find the linear equation of a line that passes through the points

Andreyy89

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2 years ago

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