I think they should keep it around, even if it might not be the most efficient strategy. After all, what if you need to solve a problem using multiple strategies?
Hi there!

To find the indefinite integral, we must integrate by parts.
Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":
u = 4x
du = 4
dv = cos(2 - 3x)
v = 1/3sin(2 - 3x)
Write into the format:
∫udv = uv - ∫vdu
Thus, utilize the solved for expressions above:
4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx
Simplify:
-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx
Integrate the integral:
∫4/3(sin(2 - 3x)dx
u = 2 - 3x
du = -3dx ⇒ -1/3du = dx
-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C
Combine:

I think the domain and range are infinite
Answer:
n = 6
Step-by-step explanation:
Two intersecting chords. The product of the parts of one chord is equal to the product of the parts of the other chord, that is
7(n + 4) = 5(n + 8) ← distribute parenthesis on both sides
7n + 28 = 5n + 40 ( subtract 5n from both sides )
2n + 28 = 40 ( subtract 28 from both sides )
2n = 12 ( divide both sides by 2 )
n = 6
Answer:CASE #1 - Divergent and convergent one-way
residential streets to reduce direct through
routes impacting the neighborhood.
CASE #2 - Alternating one-way streets throughout a
portion of a grid system to gain safety
advantages of one-way operations.
CASE #3 - Creating a one-way couplet by paring a
residential street with a nearby thru street to
create a corridor for thru traffic
Step-by-step explanation: