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Alex777 [14]
2 years ago
8

An airplane is at a location 800 miles due west of city X. Another airplane is at a distance of 1,200 miles southwest of city X.

The angle at city X created by the paths of the two planes moving away from city X measures 60°. What is the distance between the two airplanes to the nearest mile? Assume that the planes are flying at the same altitude.
Mathematics
1 answer:
zaharov [31]2 years ago
6 0

Answer:

The distance between the two airplanes (to the nearest mile) is 1058 miles.

Step-by-step explanation:

An airplane A is at a location 800 miles due west of city X. So AX = 800 miles.

Another airplane is at a distance of 1,200 miles southwest of city X. So BX = 1200 miles.

The angle at city X created by the paths of the two planes moving away from city X measures 60°. So angle ∠AXB = 60°.

In triangle ΔAXB, AX = 800 miles, BX = 1200 miles, ∠AXB = 60°.

Using law of cosines:-

AB² = AX² + BX² - 2 * AX * BX * cos(∠AXB).

AB² = 800² + 1200² - 2 * 800 * 1200 * cos(60°).

AB² = 640000 + 1440000 - 2 * 960000 * 1/2

AB² = 2080000 - 960000

AB² = 1120000

AB = √(1120000) = 1058.300524

Hence, the distance between the two airplanes (to the nearest mile) is 1058 miles.

Read more on Brainly.com - brainly.com/question/12103659#readmore

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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
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Answer

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Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

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Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

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Similarly solving for Josh we obtain

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Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

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t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

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<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
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