At that rate, how many minutes will it

Mathematics

1 answer:

Tcecarenko [31]3 years ago

5 0

Answer:

It would require 12 minutes for the plane's attitude to change by -5,400 meters.

Step-by-step explanation:

As the altitude (or height) of a plane landing at an airport-changes at a rate of -450 meters per minute.

∴ unit rate = -450 meters/per minute

We have to determine how many will it take for the plane's altitude to change by -5,400 meters.

Altitude change per minute = -450 meters

i.e.

-450 × 1 = -450 meters (altitude change in 1 minute)

-450 × 12 = - 5400 meters (∴ unit rate = -450 meters/per minute)

At this rate, it would require 12 minutes for the plane's attitude to change by -5,400 meters.

Because -450 × 12 = - 5400 meters

Therefore, it would require 12 minutes for the plane's attitude to change by -5,400 meters.

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Sana bought a dress for $32.99 and two purses. The total cost for the three items was $57.97. What was the cost for one purse?

faust18 [17]

Answer:

$12.49

Step-by-step explanation:

cost of two purses = $57.97 - cost of dress = $57.97 - $32.99 = $24.98

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3 0

3 years ago

How do I solve this?

Verdich [7]

Consider the point P where the gears meet. When the smaller gear rotates clockwise, the larger one will rotate counterclockwise.

Through one rotation of the smaller gear, P will have traveled the circumference of the smaller gear, which is $8\pi$ in.

At the same time, a point P' on the larger gear traverses the same distance along the larger gear's circumference. This point traces out an arc that is subtended by some angle $\theta$ . The arc is as long as the smaller gear's circumference.

The measure of a circle's interior angle subtended by an arc is proportional to a complete revolution, i.e. an angular displacement of $2\pi$ radians:

$\dfrac{14\pi\,\mathrm{in}}{2\pi\,\mathrm{rad}}=\dfrac{8\pi\,\mathrm{in}}\theta\implies\theta=\dfrac{8\pi}7\,\mathrm{rad}\approx205.7^\circ$

For part 2, we apply the same reasoning to the larger gear. In one full rotation of the larger gear, the point P' traverses the circumference $14\pi$ in, and so does the point P on the smaller gear.