Answer: The shaded area is 99.55 units squared.
Step-by-step explanation:
r = 3.2
Now, we can see that the sides of the square are equal to two times the diameter of the circles (or four times the radius of the circles), so the length of the sides of the square is:
L = 2*(2*3.2) = 12.8
The area of the square is A1 = L^2 = 12.8*12.8 = 163.84 units squared.
the shaded semicircle has a diameter of 4 times r (so the radius is 2 times r), and the area is equal to half the area of a circle:
A2 = (1/2)*pi*(2r)^2 = (1/2)*3.14*(6.4)^2 = 64.31 units squared.
And now we must subtract the area of the four smaller circles inside the square, the area of each one is:
A3 = pi*r^2 = 3.14*(3.2)^2 = 32.15 units squared.
Then the shaded area is:
A = A1 + A2 - 4*A3 = 163.84 + 64.31 - 4* 32.15 = 99.55 units squared.
The equation of the vertical line that satisfies the condition is x=0.
A-2: Project D.
b: Project A.
Not sure about a-1, hope I helped.
Answer:
AY = 16
IY = 9
FG = 30
PA = 24
Step-by-step explanation:
<em>The </em><em>centroid </em><em>of the triangle </em><em>divides each median</em><em> at the ratio </em><em>1: 2</em><em> from </em><em>the base</em>
Let us solve the problem
In Δ AFT
∵ Y is the centroid
∵ AP, TI, and FG are medians
→ By using the rule above
∴ Y divides AP at ratio 1: 2 from the base FT
∴ AY = 2 YP
∵ YP = 8
∴ AY = 2(8)
∴ AY = 16
∵ PA = AY + YP
∴ AP = 16 + 8
∴ AP = 24
∵ Y divides TI at ratio 1: 2 from the base FA
∴ TY = 2 IY
∵ TY = 18
∴ 18 = 2
→ Divide both sides by 2
∴ 9 = IY
∴ IY = 9
∵ Y divides FG at ratio 1:2 from the base AT
∴ FY = 2 YG
∵ FY = 20
∴ 20 = 2 YG
→ Divide both sides by 2
∴ 10 = YG
∴ YG = 10
∵ FG = YG + FY
∴ FG = 10 + 20
∴ FG = 30
Answer:
Yes because there are no repeated dependent variables.
Step-by-step explanation:
