4.50 should be the right answer.
Answer:
65º
Step-by-step explanation:
- The angle of a straight line is 180º, so ∠ABD=180º and ∠ABC=(180-6x)º
- The sum of the interior angles of a triangle is 180, so (x+40)º+(3x+10)º+(180-6x)º=180
- We can solve from there, x+40+3x+10+180-6x=180
- Combine like terms, -2x+230=180
- Subtract 230, -2x=-50
- Divide by -2, x=25
- m∠CAB=(x+40)º=(25+40)º=65º
- m∠ABC=(180-6x)º=(180-150)º=30º
- m∠BCA=(3x+10)º=(75+10)º=85º
Answer:
Maximize C =


and x ≥ 0, y ≥ 0
Plot the lines on graph




So, boundary points of feasible region are (0,1.7) , (2.125,0) and (0,0)
Substitute the points in Maximize C
At (0,1.7)
Maximize C =
Maximize C =
At (2.125,0)
Maximize C =
Maximize C =
At (0,0)
Maximize C =
Maximize C =
So, Maximum value is attained at (2.125,0)
So, the optimal value of x is 2.125
The optimal value of y is 0
The maximum value of the objective function is 19.125
Let U={q, r, s, t, u, V, W, X, Y, Z},
trapecia [35]
Answer:
Ooh is theis unions I did it but