Fourteen less than three fourths of a number is negative eight. Find the number

Answer:

The hourly rate for lectures is $7.33

Step-by-step explanation:

* Let's break down how to tackle the problem.

- For the level 3 course, examination hours are priced at double that of workshop hours.

- Workshop hours cost twice the rate of lecture hours.

- The total includes examination, workshop, and lecture hours.

- Examination lasts 3 hours, workshops 24 hours, and lectures 12 hours.

* Let’s denote the cost of lecture hours as $x per hour.

∴ The lectures cost $x per hour.

∵ Workshop charge is twice that of lectures

∴ Workshop hours cost 2(x) = 2x per hour.

∵ Examination fees are double that of workshop hours

∵ The workshop cost is 2x

∴ Examination fees are 2(2x) = 4x per hour.

- Combining costs for level 3 gives us the total of lecture, workshop, and examination hours.

∵ 12 hours for lectures

∵ 24 hours for workshops

∵ 3 hours for examinations

∵ Thus the total cost for level 3 = 12(x) + 24(2x) + 3(4x).

∴ Total cost for level 3 = 12x + 48x + 12x.

∵ Therefore, total cost = $528.

∴ 12x + 48x + 12x = 528.

∴ 72x = 528; hence we divide both sides by 72.

∴ x = 7.33.

∵ x represents the cost of lecture hours per hour.

∴ Therefore, the hourly price for lectures is $7.33.

To determine the slope, subtract the y coordinates from each other and the x coordinates as well. That gives us: 10000 - 4000 over 0 - 15, which simplifies to 6000 over -15. When divided, it results in -400. So, the slope amounts to -400. I hope this aids you!:D

To formulate the system, it's necessary to consider the slope of each line along with at least one point from each line. The two lines will connect each plane's location to their destination airport. It's important to note that the airport's coordinates represent the intersection of these two lines, corresponding to the solution of the system. First, the slope of the line from airplane one to the airport is: m = 2; this can be observed by plotting the two points. From airplane 1's location, the rise is 8 units while the run is 4 units to reach the airport, making the slope 8 divided by 4 = 2. We then insert the slope and point (2,4) into the point-slope form: y - 4 = 2(x - 4), which can be rearranged to standard form 2x - y = 0. For airplane two, the slope to the airport is obtained by observing the vertical decrease of 3 and a horizontal increase of 9 as we move from the airport to airplane 2. We then substitute the slope and the point (15,9) into the point-slope form: y - 9 = -1/3(x - 15), which can be rearranged to the standard form: x + 3y = 42. Consequently, the system of equations is: 2x - y = 0 and x + 3y = 42. Multiplying the first equation by 3 produces a system of: 6x - 3y = 0 and x + 3y = 42. Adding these equations results in the equation 7x = 42. Thus, x = 6, and by substituting this value back into 2x - y = 0, we determine y = 12. Thus, we demonstrate that the airport's coordinates do indeed comprise the solution to our system.