Answer:
15 degrees.
Step-by-step explanation:
First, count from -34.44 degrees to 0 degrees which is (0 - 34.44)=34.44 counts
Next, you determine how many more counts you need to reach the 49.44 degrees ; that would be 49.44 - 34.44 = 15
Therefore, the Antarticas highest recorded temperature was 15 degrees.
The answer is D. 12
4, 3 and 2 are all factors of 12 (least common denominator)
Answer:
$60
Step-by-step explanation:
3 tenths of 100 is 30
1 * 10 is 10
100 - 30 + 10
= 100 - 40
= $60
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
10648, as 0.44 * 24200 is 10648.
