Complete question:
A circle with radius 3 has a sector with a central angle of 1/9 pi radians
what is the area of the sector?
Answer:
The area of the sector = 
square units
Step-by-step explanation:
To find the area of the sector of a circle, let's use the formula:
Where, A = area
r = radius = 3
Substituting values in the formula, we have:
The area of the sector = square units
The third term was likely to be 25. You can see a pattern -6 when it goes up a term.
Answer:
2x + 3y = 13
4x - y = -2
In the second equation, subtract 4x from both sides.
-y = -2 - 4x
Divide both sides by -1.
y = 2 + 4x
Put this into the first equation in place of y.
2x + 3(2 + 4x) = 13
Multiply everything in the parenthesis by 3.
2x + 6 + 12x = 13
Combine like terms.
14x + 6 = 13
Subtract 6 from both sides.
14x = 7
Divide 14 on both sides.
x = 7 / 14
x = 0.5
Put this into the second equation in place of x.
4(0.5) - y = -2
2 - y = -2
Subtract 2 from both sides.
-y = -4
Divide both sides by -1.
y = 4
So x = 0.5 and y = 4.
Step-by-step explanation:
Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
