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Harlamova29_29 [7]
2 years ago
3

at a certain middle school, the 6th-grade class of 390 increased by 10%, the 7th grade class of 350 increased by 22%, and the 8t

h grade class of 420 increased by 20%. Overall, what was the percentage increase of students for this middle school? Express your answer to the nearest tenth.
Mathematics
1 answer:
jonny [76]2 years ago
6 0

The overall percentage of increase of students for this middle school is about 17.2%

To do this you have to know the number of students in each grade and add them together and do this:

390 x 0.1 = 39 Students

350 x 0.22 = 77 students

420 x 0.2 = 84 students

Add all the number of students and do this:

(200 students over 1,160) x 100%

(Increased students over original students) x 100%

0.172 x 100% = 17.2%

The percentage increase of the students of the middle school is about 17.2%

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Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of
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Answer:

Step-by-step explanation:

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

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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

8 0
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