The answer is 5.13 in²
Step 1. Calculate the diameter of the circle (d).
Step 2. Calculate the radius of the circle (r).
Step 3. Calculate the area of the circle (A1).
Step 4. Calculate the area of the square (A2).
Step 5. Calculate the difference between two areas (A1 - A2) and divide it by 4 (because there are total 4 segments) to get <span>the area of one segment formed by a square with sides of 6" inscribed in a circle.
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Step 1:
The diameter (d) of the circle is actually the diagonal (D) of the square inscribed in the circle. The diagonal (D) of the square with side a is:
D = a√2 (ratio of 1:1:√2 means side a : side a : diagonal D = 1 : 1 : √2)
If a = 6 in, then D = 6√2 in.
d = D = 6√2 in
Step 2.
The radius (r) of the circle is half of its diameter (d):
r = d/2 = 6√2 / 2 = 3√2 in
Step 3.
The area of the circle (A1) is:
A = π * r²
A = 3.14 * (3√2)² = 3.14 * 3² * (√2)² = 3.14 * 9 * 2 = 56.52 in²
Step 4.
The area of the square (A2) is:
A2 = a²
A2 = 6² = 36 in²
Step 5:
(A1 - A2)/4 = (56.52 - 36)/4 = 20.52/4 = 5.13 in²
Answer:
A
Step-by-step explanation:
Answer:
y=4x+7
Step-by-step
(-2,1) and (1,-11)
1)Rule you should use is run over rise aka slope
y²-y^1=
x^2-x^1=
(-11)-1=12
1-(-2)=3
12/3=4
slope =4
2)Rule of point slope intercept
y - y1 = m(x - x1)
y+1=4(x+2)
y+1=4x+8
-1 -1
y=4x+7
Answer:
The algebraic expression is v = 2x
v is the volume of the sugar syrup and
x is the mass of sugar in grams.
Step-by-step explanation:
Let x be the mass of sugar in grams and v be the volume of sugar syrup.
So, mass of sugar in grams/volume of sugar syrup × 100 % = 50 %
x/v × 100 % = 50 %
x/v = 50/100
x/v = 1/2
v = 2x
So, the algebraic expression required is v = 2x where v is the volume of the sugar syrup and x is the mass of sugar in grams.
Answer:
86.5 ns
Step-by-step explanation:
The speed in the original direction (horizontally) is unchanged by the vertical force the field exerts. The travel time is ...
time = distance/speed = (4.5×10^-2 m)/(5.20×10^5 m/s) = 8.65×10^-8 s
_____
An engineer would express this time using the SI prefix nano- for 10^-9. The time is 86.5 ns.
