Hi there!
I believe the answer is $300.
Why?
Because 20 (bird houses) x 15 (money for each one) = $300
I hope this helps.
:)
When we divide the figure in four parts, we obtain four squares with its sides of 5 centimeters of lenght, with quarter circles (two) in each one of them. The limits of the shadded area are arcs of cirncunference
To calculate the area of the shadded part, we must choose one square of 5cmx5cm and divide it in 3 sectors:
a: the area below the shaded area.
b: the area above the shaded area.
c: the shaded area.
The area of the square (As) is:
As=L²
As=(5 cm)²
As=25 cm²
The area of a circunference is: A=πR² (R:radio), but we want the area of the quarter circle, so we must use A=1/4(πR²), to calculate the area of the sectors a+c:
A(a+c)=1/4(π(5)²)
A(a+c)=25π/4
The area of the sector "b" is:
Ab=As-A(a+c)
Ab=25-25π/4
Ab=25(1-π/4)
The area of the sector a+b, is:
A(a+b)=2Ab
A(a+b)=2x25(1-π/4)
A(a+b)=50(1-π/4)
Then, the shadded area (Sector c) is:
Ac=As-A(a+b)
Ac=25-50(1-π/4)
Ac=25-(50-50π/4)
Ac=25-50+50π/4
Ac=(50π/4)-25
Ac=(25π/2)-25
Ac=25(π/2-1)
The area of each shaded part is: 25(π/2-1)
To calculate the perimeter of a shaded part, we must remember that the perimeter of a circunference is: P=2πR. If we want the perimeter of a quarter circle we must use: P= 2πR/4. But there is two quarter circles in the square of 5cmx5cm, so the perimeter of the shaded area is:
P=2(2πR/4)
P=4πR/4
P=πR
P=5π
The perimeter of each shaded part is: 5π
Answer:
- Seat 1: Peter
- Seat 2: Lia
- Seat 3: Kenny
- Seat 4: Jennie
- Seat 5: Harvey
- Seat 6: Joyce
- Seat 7: Rick
- Seat 8: Mark
Step-by-step explanation:
From Rule 6: Jennie is at Seat 4 and next to Harvey; and by rule 1 (Harvey has a higher seat number than Jennie)
- Seat 4: Jennie
- Seat 5: Harvey
From Rule 2: Peter is across from Harvey.
Since Seat 5 is across from Seat 1
From Rule 7: Peter is next to Lia.
Therefore, Lia can either be in Seat 2 or Seat 8, but by Rule 3 (Neither Joyce nor Lia is in seat 8), therefore, Lia is in Seat 2.
By Rule 5: Lia is next to Kenny, therefore:
From Rule 4: Rick is across from Kenny, and Rick is also next to Joyce.
Seat 7 is across from Seat 3, therefore:
Since Rick is also next to Joyce (and by Rule 3, Joyce cannot be on Seat 8)
Therefore, we have:
Seat 1: Peter
Seat 2: Lia
Seat 3: Kenny
Seat 4: Jennie
Seat 5: Harvey
Seat 6: Joyce
Seat 7: Rick
Since Mark has the remaining seat, Mark is on Seat 8.
