a. Since GC bisects FGH and the angle FGH = 122°, we know that the angle FGC = 122/2 = 61°
b. Since GC bisects FGH and the angle CGH = 42°, we know that the angle FGH = 2*42 = 84°
Answer:
see attached
Step-by-step explanation:
Each spot in the triangle is the sum of the two numbers immediately above. The middle number of row 2 will be 1+1 = 2. The two numbers in row 3 will be 1+2 = 3 and 2+1 = 3. It continues like this. The second number in each row is the row number. Each row is symmetrical about the center.
Answer:
1.5 unit^2
Step-by-step explanation:
Solution:-
- A graphing utility was used to plot the following equations:

- The plot is given in the document attached.
- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).
- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.
We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).
The double integration formulation can be written as:

Answer: 1.5 unit^2 is the amount of area bounded by the given curve f ( x ) and the boundary equations.
Step-by-step explanation:
63.5*0.088 =5.588 tax for your answer