When you are given the side length and apothem of a regular polygon and asked for area, you are expected to use the formula
... A = (1/2)Pa
where P is the perimeter of the polygon, "a" is the apothem, and A is the area.
The only additional information you need is that a decagon has 10 sides, so its perimeter is 10×3.2 inches = 32 inches.
The area is ...
... A = (1/2)(32 in)(5 in)
... A = 80 in²
_____
Sometimes the dimensions of a problem like this don't quite add up. The geometry of a regular polygon means that the apothem can be found from the side length and vice versa. In this case, the side length of 3.2 inches means the apothem is about 4.92 inches; or the apothem of 5 inches means the side length is about 3.25 inches. Either way, the area is slightly different from that computed using the given numbers.
This is why I say you're expected to use the above formula. (You're not expected to think about it too much.)
Completing the Square
2. Solve the equation by completing the square. Show your work.
x^2 – 30x = –125
Step 1: Add to both sides of the equation. (2 points)
Add 225 both sides of the equation
x^2 – 30x + 225 = –125
+ 225
x^2 - 30x + 225 = 100
Step 2: Factor the left side of the equation. Show your work. (2 points)
Hint: It is a perfect square trinomial.
x^2 - 30x + 225 = 100
Factor the left side of the equation:
(x - 15)^2 = 100
Step 3: Take the square root of both sides of the equation from Step 2. (1 point)
√(x - 15)^2 = √100
Step 4: Simplify the radical and solve for x. Show your work. (1 point)
x - 15 = + - 10
x - 15 = 10
x = 25
x - 15 = -10
x = 5
Solutions x = 25, 5
Answer:
y=2x+7
Step-by-step explanation:
Answer: About 22.73 days.
Step-by-step explanation: Each person throws away about 4.4 pounds of trash daily. So the way to calculate this is to divide 100 pounds by 4.4 to find out how many days it would take for a person to throw away 10 pounds of trash.
Given:
Q is between P and R, R is between Q and S, 
.
To prove:
Solution:
The two column proof is:
Statements Reasons
1. Q is between P and R 1. Given
2. 
2. Segment addition postulate
3. R is between Q and S 3. Given
4. 
4. Segment addition postulate
5. 5. Given
6. 
6. Substituting property of equality
7. 
7. Subtraction property of equality
8. 8. Simplify
Hence proved.
