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butalik [34]
3 years ago
9

Plane A leaves Tulsa at 2:00 p.m., averaging 300 mph and flying in a northerly direction. Plane B leaves Tulsa at 2:30 p.m., ave

raging 225 mph and flying due east. At 5:00 p.m., how far apart will the planes be?

Mathematics
2 answers:
kherson [118]3 years ago
6 0

Answer:

At 5:00 p.m. the planes will be 1061.323 miles apart.

Step-by-step explanation:

To solve this problem, I add a picture of the situation.

We know that speed is distance over time  ⇒

Speed=\frac{distance}{time} (I)

The first step to solve this exercise is to graph the situation. We can draw a right triangle which vertices will be ''Tulsa'', and the planes ''A'' and ''B'' at 5:00 p.m.

In order to know the measures of the sides, we are going to calculate them using the equation (I)

Plane A leaves Tulsa at 2:00 p.m.

Therefore, at 5:00 p.m. it will have flown 3 hours ⇒

300\frac{mi}{h}=\frac{distance}{3h} ⇒

distance=(300\frac{mi}{h}).(3h)

distance=900mi

At 5:00 p.m. the distance from the plane A to Tulsa is 900 mi

Plane B leaves Tulsa at 2:30 p.m.

Therefore, at 5:00 p.m. it will have flown 2.5 hours ⇒

225\frac{mi}{h}=\frac{distance}{2.5h}

distance=(225\frac{mi}{h}).(2.5h)

distance=562.5mi

At 5:00 p.m. the distance from the plane B to Tulsa is 562.5 mi

Finally, we can find the distance between the plane A and the plane B using the Pythagorean theorem :

(900mi)^{2}+(562.5mi)^{2}=(Distance_{A-B})^{2}

810000mi^{2}+316406.25mi^{2}=(Distance_{A-B})^{2}

(Distance_{A-B})^{2}=1126406.25mi^{2}

Distance_{A-B}=\sqrt{1126406.25mi^{2}}

Distance_{A-B}=1061.322877mi ≅ 1061.323mi

At 5:00 p.m. the planes will be 1061.323 miles apart.

Musya8 [376]3 years ago
4 0

Answer 1061

Step-by-step explanation:

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