Ignore the mess... I need help with #20 please and thank you
1 answer:
Because the triangle is isosceles, angle P and angle R are the same. Thus, your work would be:
. You then simplify this to 6x+4 + 196 - 8x = 180. Combine like terms, you get - 2x = - 20. x = 10.
Plug x = 10 back into the equation 98 - 4x to get 98 - 4(10) which becomes 98 - 40 = 58.
Angle P should be equal to 58 degrees.
To check your work add up the angles to see if they equal 180:
58 + 58 + 6(10) + 4 = 116 + 64 = 180.
Plug x = 10 back into the equation 98 - 4x to get 98 - 4(10) which becomes 98 - 40 = 58.
Angle P should be equal to 58 degrees.
To check your work add up the angles to see if they equal 180:
58 + 58 + 6(10) + 4 = 116 + 64 = 180.
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Answer:
- P = 6750n +7800r
- n ≤ 19
- r ≤ 31
- 160n +620r ≤ 8110
- n = 19
- r = 8
- 190,650 people
Step-by-step explanation:
This linear programming problem is described by an objective function and constraints on the variables. A graphical solution works well.
<h3>Objective function</h3>The goal is to maximize the number of people exposed to the company's ad(s). The number of people reached is the sum of the products of the number of ads and the number reached per ad.
For n newspaper ads, we are told that 6750n people are reached.
For r radio ads, we are told that 7800r people are reached.
The total number of people reached is ...
P = 6750n +7800r . . . . . . . the function we wish to maximize
<h3>Newspaper ads</h3>We can run at most 19 newpaper ads:
n ≤ 19
<h3>Radio ads</h3>We can run at most 31 radio ads:
r ≤ 31
<h3>Budget</h3>The cost of n newspaper ads will be $160n.
The cost of r radio ads will be $620r.
We must stay within a budget for the ads, $8100:
160n +620r ≤ 8110
<h3>Solution</h3>The white area in the first quadrant of the attached graph represents the feasible solution space. (We reversed the inequality symbol in each inequality so the solution space would be white, not triple-shaded.) The corners of the solution space represent possible (n, r) pairs where the objective function might be maximized.
The solid red line on the graph shows the maximum value the objective function might have, and the (n, r) pairs that would give that maximum value. The value of the objective function increases the farther the line is from the origin. Drawing the line on the graph lets us readily identify the (n, r) coordinate pair that will place this line as far as possible from the origin, maximizing P. We find that to be (n, r) = (19, 8).
- the number of newspaper ads to run is 19
- the number of radio ads to run is 8
- the group exposure is (6750)(19) +(7800)(8) = 190,650
