1: a) -51
b) -461
2: a) 18078.415936
b) <span>61715.95</span>
Answer:
Problem B: x = 12; m<EFG = 48
Problem C: m<G = 60; m<J = 120
Step-by-step explanation:
Problem B.
Angles EFG and IFH are vertical angles, so they are congruent.
m<EFG = m<IFH
4x = 48
x = 12
m<EFG = m<IFH = 48
Problem C.
One angle is marked a right angle, so its measure is 90 deg.
The next angle counterclockwise is marked 30 deg.
Add these two measures together, and you get 120 deg.
<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.
Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.
m<G = 180 - 120 = 60
9514 1404 393
Answer:
34.5 square meters
Step-by-step explanation:
We assume you want to find the area of the shaded region. (The actual question is not visible here.)
The area of the triangle (including the rectangle) is given by the formula ...
A = 1/2bh
The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...
A = (1/2)(11 m)(9 m) = 49.5 m^2
The rectangle area is the product of its length and width:
A = LW
The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...
A = (5 m)(3 m) = 15 m^2
The shaded area is the difference between the triangle area and the rectangle area:
shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2
The shaded region has an area of 34.5 square meters.
Answer: 126.48x+109.56 would be the largest possible perimeter if the area is 1000x + 750.
Step-by-step explanation: Not 100% certain of this answer.
find square root...
31.62x+27.39
that would give us 4 sides and since perimeter of rectangle is 2l+2w...
4(31.62x+27.39)=
126.48x+109.56 would be the largest possible perimeter if the area is 1000x + 750.
