Answer:
23% of Michael's rock collection is metamorphic rock.
Step-by-step explanation:
We are given the following in the question:
Percentage of igneous rock in the collection = 27%
Percentage of sedimentary rock in the collection = 50%
Remaining rocks are metamorphic rock.
Total collection = 100%
Percentage of metamorphic rocks =
Thus, 23% of Michael's rock collection is metamorphic rock.
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: The correct option is B-He left out the place holder under the 4.
Step-by-step explanation:
Consider the provided multiplication.
Jared need to multiply 36.47 with 12
The correct multiplication should be:
36.47
× 12
7294
36470
43764
Now consider the Jared's work he just left out the place holder under the 4.
Because it is not just 1×36.47 it is actually 10×36.47, so he needs to add a zero under the number 4.
