Please help. Basic. Brainliest will be given.

28.7 m at 46.9°

Explanation:

The x component of the displacement is:

x = 6 m cos 0° + 25 m cos 57°

x = 19.6 m

The y component of the displacement is:

y = 6 m sin 0° + 25 m sin 57°

y = 21.0 m

The total displacement is found with Pythagorean theorem:

d = √(x² + y²)

d = 28.7 m

And the direction is found with trig:

θ = tan⁻¹(y/x)

θ = 46.9°

Distance of displacement = 28.71 m

Angle of displacement = 46.91 degrees

Explanation:

convert all individual displacements to simple vectors.

6m east  -> <6,0>

25m at 57 -> <25cos57, 25sin57>

Total displacement = sum of two vectors

= <6+25cos57, 0+25sin57>

=<19.62, 20.97>

Magnitude of displacement

= sqrt(19.62^2+20.97^2)

= 28.71

Direction of displacement

= atan2(20.97,19.62)

= 46.91 degrees

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

neither accurate nor precise.