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murzikaleks [220]
2 years ago
7

The components of vector are ax and ay (both positive), and the angle that it makes with respect to the positive x axis is θ. Fi

nd the angle θ if the components of the displacement vector are (a) ax = 12 m and ay = 12 m, (b) ax = 20 m and ay = 12 m, and (c) ax = 12 m and ay = 20 m.
Mathematics
1 answer:
Andru [333]2 years ago
6 0

we can use formula

\theta=tan^{-1}(\frac{a_y}{a_x})

(a)

we are given

a_x=12,a_y=12

now, we can plug values

\theta=tan^{-1}(\frac{12}{12})

\theta=tan^{-1}(1)

\theta=\frac{\pi}{4}rad

(b)

we are given

a_x=20,a_y=12

now, we can plug values

\theta=tan^{-1}(\frac{12}{20})

\theta=tan^{-1}(\frac{3}{5})

\theta=30.96^{\circ \:}

(c)

we are given

a_x=12,a_y=20

now, we can plug values

\theta=tan^{-1}(\frac{20}{12})

\theta=tan^{-1}(\frac{5}{3})

\theta=59.03^{\circ \:}


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C

Step-by-step explanation:

We can notice that the balance beam has in one side 7 balls and in the other one 5 balls and a square

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Let x be the square mass

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There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the
Dmitrij [34]

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

a) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

16.28.27.26.15 = 4,717,440

c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = <u>14.13.12!</u> = <u>14.13 </u>= 91

           12! 2!            2    

C₁₆,₂ = <u>16.15.14!</u> = <u>16.15 </u>= 120

           14! 2!            2    

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = <u>14.13.12.11.10!</u> = <u>14.13.12.11 </u>= 1001

                  10! 4!               4.3.2.1    

C₁₆,₄ = <u>16.15.14.13.12!</u> = <u>16.15.14.13 </u>= 1820

                  12! 4!               4.3.2.1    

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

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Triangle XYZ is obtuse which list could represent the angle measures of triangle XYZ?
asambeis [7]
An obtuse triangle must have an angle greater than 90degrees (cannot equal 90). Therefore, A. would be the correct answer. This is because A. is the only set of angles that includes an angle greater than 90. Also, all angle measures add up to equal 180degrees.
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