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sertanlavr [38]
3 years ago
5

Paul and his neighbors decided to build a fence along his farm. After three hours, they have 15 meters of fencing complete. They

decide to take a 2-hour break for lunch and then resume building the fence. After four more hours, they have a fence that is a total of 55 meters long. PLEASE HELP ME!!!! never been good with graphs,
the y-axis goes by 5's and goes all the way up to 55 starting from 5. the y-axis is the length of the fence in meters, and the x-axis goes to 10 by ones which is the number of hours
Mathematics
1 answer:
Goshia [24]3 years ago
8 0

Let the  be the time in hours, and  meters of fencing completed. 

We know that After three hours, they have 15 meters of fencing complete, our graph will go from the point (0,0) to the point (3,15). We also know that they decide to take a 2-hour break for lunch and then resume building the fence, so our graph will go from the point (3,15) to the the point (5,15). Finally, After four more hours, they have a fence that is a total of 55 meters long, so the final part of our graph will go from the point (5,15) to the point (9,55)


We can conclude that the graph of t vs y is:

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(45/360)2πr=9π/4
Lyrx [107]

Answer:

r=9

Step-by-step explanation:

45/360=1/8

1/8*2=2/8

2/8=1/4

1/4*pi*r=(9pi)/4

Divide both sides by 1/4 (the fours cancel each other out)

Pi*r=9*pi

Pi cancels each other out

R=9

Hope this helps

7 0
3 years ago
Paul opens a savings account with $350. He saves $150 per month. After how many months will Paul have $2150 in the account?​
Debora [2.8K]

Answer:

12 months

Step-by-step explanation:

2150-350=1800

1800÷150=12

3 0
2 years ago
Find 4 - (-11).<br> The difference is ?
Nana76 [90]

Answer:

15

Step-by-step explanation:

Step 1: distribute the negative: 4+11

Step 2: add up the numbers: 15

Hope that helps

7 0
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)

4 0
3 years ago
2x – y = 6 5x + 10y = –10
Studentka2010 [4]
2x-y=6;5x+10y=-10. -y=-2x+6. -y/-1=-2x+6/-1. y=2x-6. 5x+10y=-10. 5x+10(2x-6)=-10. 25x-60=-10. 25x=50. 25x/25=50/25. y=2x-6. y=(2)(2)-6. y=-2. Therefore, x=2 and y=-2
8 0
2 years ago
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