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
P = d - 5.50
The cost of buying materials will be subtracted from what he earns. Therefore, what’s left over is the profit
9514 1404 393
Answer:
-2018
Step-by-step explanation:
The n-th term is ...
an = a1 +d(n -1)
So, the given terms are ...
-53 = a1 +12d
-128 = a1 +37d
Subtracting the first from the second gives ...
(a1 +37d) -(a1 +12d) = (-128) -(-53)
25d = -75
d = -3
The 668th term will be ...
a668 = a1 +d(668 -1) = a1 +667d = (a1 +37d) +630d
a668 = -128 +630(-3) = -128 -1890 . . . . substitute for a38
a668 = -2018
Answer:
The first 5 terms are;
-2, 2, 13,38 and 91
Step-by-step explanation:
Here, we want to write the first 5 terms of the sequence.
We already have the first term as 1
Now, we need the 2nd term
Putting two in place of n, we have ;
2f(1) + 3n
= 2(-2) + 3(2) = -4 + 6 = 2
For the 3rd term, put 3 in place of n
2f(2) + 3(3)
sine f(2) = 2, we have
2(2) + 9 = 4+ 9 = 13
For the fourth term, put 4 in place of n, we have
2f(3) + 3(4)
since f(3) = 13
we have; 2(13) + 12 = 26 + 12 = 38
For the 5th term, put 5 in place of n, we have
2f(4) + 3(5)
since f(4) = 38, we have
2(38) + 15 = 76 + 15 = 91
