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harkovskaia [24]
2 years ago
13

A) 1.0 mile b) 5.0 mile c) 5.8 mile d) 0.8 mile ​

Mathematics
1 answer:
nordsb [41]2 years ago
3 0

Answer:

d) 0.8 mile

Step-by-step explanation:

Coordinate of object B = (6, 6)

Coordinate of object C = (1, 4)

Coordinate of the archaeologist = (6, 1)

✍️Distance between Object B (6, 6) and the archaeologist (6, 1):

Distance = |6 - 1| = 5 units = 5 miles

✍️Distance between Object C (1, 4) and the archaeologist (6, 1):

Using the distance formula, d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, we would have:

d = \sqrt{(6 - 1)^2 + (1 - 4)^2}

d = \sqrt{(5)^2 + (-3)^2}

d = \sqrt{25 + 9}

d = \sqrt{34} = 5.8 units = 5.8 miles (nearest tenth)

Comparing how far each object is from the archaeologist, object C is 0.8 mile (5.8 - 5 = 0.8) farther than object B.

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Airida [17]

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89.1° or -1.4°  

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

(3) \, t =\dfrac{2}{\cos \theta}

Substitute (3) into (2)

-50 = 600t \sin \theta - 4.9t^{2} =  600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )

Set up a quadratic equation

\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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3 years ago
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