Answer:
The velocity of jet in still air is 
and velocity of wind is
.
Step-by-step explanation:
Let x be the rate of the jet in still air and y be the rate of the wind.
Given:
Velocity against wind, a jet travels 2160 miles in 3 hours.
Velocity against wind 
Flying with the wind, same jet travels 6000 miles in 6 hours.
Flying with the wind 
As such its velocity against wind is 
and with wind is
therefore
------------(1)
---------------(2)
Adding equation 1 and 2.
Put x value in equation 2
Hence velocity of jet in still air is and velocity of wind is
.
Answer: the number of minutes of long distance call that one can make is lesser than or equal to 12 minutes.
Step-by-step explanation:
Let x represent the number of minutes of long distance call that one makes.
The first three minutes of a call cost $2.10. After that, each additional minute or portion of a minute of that call cost $0.45. This means that if x minutes of long distance call is made, the total cost would be
2.10 + 0.45(x - 3)
Therefore, the inequality to find the number of minutes one can call long distance for $6.15 is expressed as
2.10 + 0.45(x - 3) ≤ 6.15
2.10 + 0.45x - 1.35 ≤ 6.15
0.75 + 0.45x ≤ 6.15
0.45x ≤ 6.15 - 0.75
0.45x ≤ 5.4
x ≤ 5.4/0.45
x ≤ 12
Answer:
Step-by-step explanation:
Given
See attachment
Required
Determine the slope of the line
From the attachment, we have the following points
and
The slope m is then calculated using:
Substitute values for the x's and y's
<em>Hence, the slope is 0</em>
Answer: 
Step-by-step explanation:
It is important to remember that the fourth digit after the decimal point is in the ten-thousandth place.
Then, given the following number:
You can follow these steps in order ti round it to the nearest ten-thousandth:
1. You can identify that the digit in the ten-thousandth place is:
2. Identify the digit to the right of . This is:
3. Since:
You must round up. Increase the digit by 1. (Notice that 
, then the digit to the left of increases by 1 too). Then:
Answer:
- cos (x) cos (2x) - 2sin(x) sin(2x)
Solution:
(See the solution in the photo)
