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3 years ago

The linear combination method is applied to a system of equations as shown.

Mathematics

1 answer:

Kay [80]3 years ago

4 0

Answer:

Step-by-step explanation:

Option 4th is correct

The solution of the system of equations is, (9, 3)

Step-by-step explanation:

Given the system of equations:

⇒ .....[1]

⇒ .....[2]

Add equation [1] and [2] we have;

Divide both sides by 2 we have;

Substitute x = 9 in [1] we have;

Subtract 9 from both sides we have;

2y = 6

Divide both sides by 2, we have;

y = 3

Therefore, the solution of the system of equations is, (9, 3)

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1. A buoy floats 19 yards from the eastern most point of a boat and 15 yards from the western most point of a second boat. The a

Black_prince [1.1K]

The laws of cosines and law of sines can be used given that two sides

and an included angle, or two angles a side are known.

Response:

1. The other angles in the triangle formed by the buoy are approximately;

31.1° and 40.9°

2. Distance of the helicopter from the first island is approximately;

14.5 miles

<h3>How is the Law of Sines and Cosines used?</h3>

Given parameters are;

Distance of the buoy from the easternmost point of a boat = 19 yards

Distance of the buoy from the westernmost point of the other boat = 15 yards

Angle formed from the buoy to the two boats = 108°

Distance between the two boats, d, is given by the law of cosines, as follows;

d² = 19² + 15² - 2 × 19 × 15 × cos(108°) = 586 - 570·cos(108°)

d = √(586 - 570·cos(108°))

By the law of Sines, we have;

$\dfrac{d}{sin(108^{\circ})} = \mathbf{\dfrac{15}{sin(Angle \ formed \ from \ the \ boat \ on \ the \ West, \ \theta_1)}}$

Which gives;

$sin(\theta_1) = \mathbf{ \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} }}$

The o

$\theta_1 = arcsin \left( \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} } \right) \approx \mathbf{31.1^{\circ}}$

The other angles formed in the triangle containing the buoy are;

θ₁ ≈ 31.1

θ₂ ≈ 180° - 108° - 31.1° ≈ 40.9°

2. Distance between the two islands = 20 miles

Angle of elevation with one island = 15°

Angle of elevation with the second island = 35°

Required:

The mileage (distance travelled) of the helicopter.

Solution:

Let A represent the island that has an angle of elevation to the helicopter

of 15°, and let B represent the other island.

Angle formed by the helicopter and the two island, θ, is found as follows;

θ = 180° - (15° + 35°) = 130°

By the Law of Sines, we have;

$\dfrac{20}{sin(130^{\circ})} = \mathbf{ \dfrac{Distance \ from \ island \ A }{sin(35^{\circ})}}$

Which gives;

$Distance \ of \ helicopter \ from \ island \ A = \mathbf{ \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ})}$

$Mileage \ from \ island \ A = \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ}) \times cos(15^{\circ}) \approx 14.5$

The mileage of the helicopter from the first island is approximately 14.5 miles

Learn more about the Law of Sines and Cosines here:

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