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

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.

Use Law of cosines

d = √(800^2 + 1200^2 - 2·800·1200·COS(60°)) = 1058 miles

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2