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zimovet [89]
3 years ago
9

A ship at position (1, 0) on a nautical chart (with north in the positive y direction) sights a rock at position (7, 5). What an

gle θ does this vector make with due north? (This is called the bearing of the rock from the ship. Round your answer to two decimal places.) answer should be in radians east of north
Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:

0.69473828 radians east of north.

Step-by-step explanation:

Interpreting the information given on a Cartesian plane, the position of the ship will be (1,0) and that of the rock will be (7,5). Connecting the point (1,0) and (7,5) with a line and dropping a perpendicular line from point (7,5) to the x-axis to intersect the x-axis at (7,0), will form a right-triangle with base length 6 units and a height of 5 units.

<u>Finding the base length from points (1,0) and (7,0)</u>

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(7-1)²

d= √6²

d=6 units

<u>Finding the height from points (7,5) and (7,0)</u>

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(7-7)² + (0-5)²

d=√-5²

d=√25

d=5 units

So you now have a right triangle with base length 6 units and height 5 units.

To get the bearing of rock from ship position you apply the formula for tangent of an angle

Tan Ф = length of opposite side/length of adjustment side

TanФ=5/6

Tan Ф =0.83333333333

Tan⁻¹ (0.83333333333) =0.69473828 radians east of north.

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<h3>Answer:  53%</h3>

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Explanation:

The question asks "What percent of females participate in extracurricular activities?" This means we only focus on the second column. There are 36 women, and of this total, 19 are in extracurricular activities.

Dividing the two values leads to approximately 19/36 = 0.52777 which rounds to 0.53

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Roughly 53% of the female students participate in extracurricular activities.

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3 years ago
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yarga [219]

Answer:

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Step-by-step explanation:

For ABD ≅ ACD;

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PQ≅PR by given

QS≅RS by given

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8 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
yan [13]

Answer:

1. ║u-z║= 5.66

2.║v-w║=6.0

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4.║v-u║=7.81

Step-by-step explanation:

The vectors are given as;

u = <-1, -3>, v = <5,-8>, w = <5, -2>, and z = <3, 1>.

To find the magnitude of the vectors;

1. ║u-z║

<-1 - 3> = <-4  and < -3 - 1> = <-4

║<-4,-4> ║= √{ -4²+-4²} = √32 = 5.66

2.║v-w║

   <5,-8> - <5,-2>

   <5-5> , <-8--2>

   <0,-6>

    ║<0,-6>║= √{0²+ -6²} = √36 = 6

3. ║w-u║

    <5,-2> - <-1,-3>

    <5--1> , <-2--3>

    <6,1>

    ║6,1║= √{6²+1²} = √36+1 = √37 = 6.08

4.║v-u║

   <5,-8> - <-1,-3>

    <5--1> , <-8--3>

    < 6 , -5 >

    ║6,-5║= √{6²+-5²} = √36+25 =√61 = 7.81

   

   

   

   

   

5 0
3 years ago
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Answer:

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For this case the reliability of the sytem would be given by:

R= \prod_{i=1}^n R_i

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

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Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

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For this case the reliability of the sytem would be given by:

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R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

R = 0.95*0.95*0.5*0.79*0.6= 0.21389

8 0
3 years ago
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