(x+15) (3x-35) If the angles are congruent, what is x?
1 answer:
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Answer:
90 km, N 46° E
Step-by-step explanation:
<em>A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.</em>
Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).
This will be the jet's distance from its starting point.
In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.
The <u>Law of Cosines</u> will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.
In this case, the known angle is 110° (angle B) so we will use the <u>Law of Cosines</u> respective to B.
- b² = a² + c² - 2ac cosB
Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).
- b² = (60)² + (50)² - 2(60)(50) cos(110°)
- b² = 6100 -(-2052.12086)
- b² = 8152.12086
- b = 90.28909602
- b ≈ 90 km
The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.
The <u>Law of Cosines</u> with respect to A:
- a² = b² + c² - 2bc cosA
Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).
- (60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
- 3600 = 8152.12086 - 6528.909602 cosA
- -4552.12086 = -6528.909602 cosA
- 0.6972252853 = cosA
- A = cos⁻¹(0.6972252853)
- A = 45.79519
- A ≈ 46°
The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.
9514 1404 393
Answer:
4mn/(3m+12)
Step-by-step explanation:
It is often helpful to factor expressions so that common factors can cancel.
Answer:
Step-by-step explanation:
Box plot is enclosed.
To find out 5 number summary
3.0, 3.5, 4.5, 5.5, 6.5, 6.5, 7.0, 7.0, 7.0, 7.0, 7.0, 7.5, 8.0, 9.0, 9.0, 9.0, 9.5, 9.5, 9.5, 9.5}
Minimum: 3.0
Quartile Q1: 6.5
Median: 7
Quartile Q3: 9
Maximum: 9.5
Besides we find that
Average (mean): μ=7.25
Absolute deviation: 30.5
Mean deviation: 1.525
Minimum: 3.0
Maximum: 9.5
Variance: 3.7125
Standard deviation σ=1.926
