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jek_recluse [69]
2 years ago
15

The bearing of P from Q is 046º.

Mathematics
1 answer:
a_sh-v [17]2 years ago
7 0

Answer:

  175°

Step-by-step explanation:

Bearing angles are usually measured clockwise from North. Reverse bearing angles differ from forward bearing angles by 180°. These relations and the usual angle sum relation for a triangle can be used to solve this problem.

Angle PQR will be the difference in the bearings from Q to P and Q to R:

  ∠PQR = 124° -46° = 78°

Triangle PQR is isosceles, so the base angle at P will be ...

  ∠QPR = (180° -78°)/2 = 51°

__

The bearing from P to R will be 51° less than the bearing from P to Q. The bearing from P to Q is 180° more than the bearing from Q to P.

  PR bearing = PQ bearing - ∠QPR

  = PQ bearing - 51°

  = (46° +180°) -51° = 175°

The bearing of R from P is 175°.

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a. the cost of a call for six minutes is 0.70.

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<h3>How to find the cost of each call ?</h3>

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rate schedule for an m-minute call from any of its pay phones 0.70.

c(m) = 0.70 when m<=6

c(m) = 0.70+ 0.24(m-6) when m>6 & m is an integer

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a. to find the cost of a call for six minutes.

already given that c(m) =0.70 when m=6.

Therefore, the cost of a call for six minutes is 0.70.

b. to find the cost of a 14-minute call.

according to given information

c(14) = 0.70 + 0.24(m - 6) \\ c(14) = 0.70 + 0.24(14 \: - 6)  \\c(14) = 0.70 + (0.24 \times 14) - (0.24 \times 6) \\c(14) = 0.70 + (3.36 - 1.44) \\ c(14) = 0.70 + 1.92 \\ c(14) = 2.62

Therefore,the cost of a 14-minute call is 2.62.

c.to find the cost of a 9 ½2-minute call

according to given information

c(9  \frac{1}{2} 2) = 0.70 + 0.24((9 \frac{1}{2} 2 - 6) + 1) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.24((9 - 6) + 1) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.24(3+ 1) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.24(4) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.96 \\ c(9  \frac{1}{2} 2) = 1.66

Therefore,the cost of a 9 ½2-minute call is 1.66.

Learn more about problems on cost of call, refer:

brainly.com/question/16584226

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