Replace x with π/2 - x to get the equivalent integral

but the integrand is even, so this is really just

Substitute x = 1/2 arccot(u/2), which transforms the integral to

There are lots of ways to compute this. What I did was to consider the complex contour integral

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

and it follows that

Think of factor pairs. She could buy 1 pair for $54, 54 pairs for $1, 2 pairs for $27, 27 pairs for $2, 3 pairs for $18, 18 for $3, 6 pairs for $9, or 9 pairs for $6
First, distribute the minus sign across the second set. So you have:
x - y + 1 - x - y + 1
Now, we combine like terms:
x-x=0
-y-y=-2y
1+1=2
So we have -2y+2
Hope that helps.
PEMDAS
2.5x + 0.1x = 2.6x
2.6x - 0.01x = 2.59x
Answer:
Roy is 10 years old at present and Joan is 5 years old at present
Step-by-step explanation:
Let
x----> Roy's age
y----> Joan's age
we know that
x=2y ----> equation A
(x+3)+(y+3)=21 ----> equation B
substitute equation A in equation B
(2y+3)+(y+3)=21
solve for y
3y+6=21
3y=21-6
3y=15
y=5 years
Find the value of x
x=2y ----> x=2(5)=10 years
therefore
Roy is 10 years old at present
Joan is 5 years old at present