3. Factor.

1 answer:

5 0

It seems that some the work is already here, but I'd be glad to!! So for #3 which is 9x^2+15x, we can factor out both a 3 and an x (3x) so we know that 3x * 3x =9x^2 and 3x * 5 = 15x so once we take the 3x out of the equation, we are left with 3x(3x+5) and that's as far as you can factor.
For #4, we see that the common factor is 10m because 10m * 2n = 20mn and 10m * 3 = 30m so once we take 10m out of the original, it becomes 10m(2n-3)
For #5, this one the common factor is 4xy because 4xy * 2xy=8x^2y^2 and 4xy*x= 4x^2y and 4xy*3=12xy so once we take the 4xy out of the equation, it becomes 4xy(2xy-x-3)
Hope this helps!

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Find the indefinite integral by using the substitution x = 4 sec(θ). (Use C for the constant of integration.) x2 − 16 x d

guapka [62]

Answer:

$\frac{x^2-16}{2} + 16ln\frac{4}{x} +16C$

Step-by-step explanation:

Given the indefinite integral $\int\limits{\frac{x^2-16}{x} } \, dx$ , using the substitute

x = 4 sec(θ)...1

The integral can be calculated as thus;

First let us diffrentiate the substitute function with respect to θ

dx/dθ = 4secθtanθ

dx = 4secθtanθdθ... 2

Substituting equation 1 and 2 into the integral function we will have;

$\int\limits{\frac{(4sec \theta)^2-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{16sec^2 \theta-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{(16(sec^2 \theta-1)}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\from \ trig \ identity,\ sec^2 \theta - 1 = tan^\theta\\\\\int\limits{\frac{16 tan^2 \theta}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\\int\limits 16 tan^3 \theta d \theta\\\\$