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alexdok [17]
1 year ago
13

Classify the solution set of each equation below as one of the following types of surfaces: A. parabolic cylinder, B. ellipsoid,

C. elliptic paraboloid, D. hyperbolic paraboloid, E. cone, and F. hyperboloid.

Mathematics
1 answer:
taurus [48]1 year ago
3 0

Answer:

the complete question is found in the attachment

1) D. hyperbolic paraboloid

2)C. elliptic paraboloid

3)E. cone

4)F. hyperboloid.

Step-by-step explanation:

The complete explanation is found in the attachment

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Students collected data about the capacities of their lungs by inflating balloons with a single breath. they measured the circum
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The given data is the following:

Student Trial 1  Trial 2 Trial 3 Average
-----------  --------  --------  --------  ------------
    1           66.0    66.5   68.5     67.0
    2          67.5     64.0   70.5     67.3
    3          60.3     60.5   60.5    61.0
    4          55.0     58.0   59.0    57.3

Let us check the reported averages.
Student 1:
Average = (66.0 + 66.5 + 68.5)/3 = 67.0   Correct
Student 2:
Average = (67.5 + 64.0 + 70.5)/3 = 67.3    Correct
Student 3:
Average = (60.3 + 60.5 + 60.5)/3 = 604    Incorrect
Student 4:
Average = (55.0 + 58.0 + 59.0)/3 = 57.3    Correct

Answer: Student 3
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1 year ago
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What’s 4x4x2-4 in a fraction over 2
JulijaS [17]

Answer:

it should be 12/2 or 6 i believe

Step-by-step explanation:

4x4 = 8

8x2 = 16

16-4 = 12

8 0
2 years ago
Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of
Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2  are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j

The unit vector in the direction of force will be

\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}

The magnitude of the force is 40lb, so the force will be

F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k

\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}

The magnitude of the force is 60lb, so the force will be

F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)

The resultant of the two forces is

F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k

The magnitude force will be

|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469

The angle formed by F and y axis

\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=

The angle formed by F and z axis

\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359

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$12 times X - 15

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2 years ago
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at an amusement park, Robin bought a t-shirt for $8 and 5 ticket for ride. she spent a total of $23. How much did each ticket co
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Each ticket cost $3

8+5x=23

5x=15

x=3

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2 years ago
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