WILL AWARD BRAINLIEST! A control tower observes the flight of an aircraft.
At 09:23 the aircraft is 580 km away on a bearing of 043º.
At 09:25 the aircraft is 360 km away on a bearing of 016º.
What is the speed and the course of the aircraft?
(Use a scale of 1 cm to 50 km.)

27465 km/h

Step-by-step explanation:

From the figure, let us use cosine formula to calculate the resultant displacement.

B^2 = C^2 + A^2 - 2(A)(C) cosØ

Where C = 580km

A = 360 km

Ø = 153 degree

Substitute all the parameters into the formula

B^2 = 580^2 + 360^2 - 2(360)(580)cos153

B^2 = 466000 - ( - 372084.32 )

B^2 = 466000 + 372084.32

B^2 = 838084.32

Square root both sides

B = 915.5 km

You are told to use a scale of 1 cm to 50 km.

Therefore, B = 915.5/50 = 18.3 cm

The time given are: 09:23 and 09:25.

The time difference = 25 - 23 = 2 minute.

Convert minutes to hours

2 minute = 2/60 = 1/30 hours

Speed = distance/time

Speed = 915.5 ÷ 1/30

Speed = 915.5 × 30

Speed = 27465 km/h

Or

Convert 2 minute to second

2 minute = 2 × 60 = 120 seconds

Speed = distance/time

Speed = 18.3/120

Speed = 0. 1525 cm/s

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: