Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim (6x – In(x)) XStep 1 As x → [infinity], In(x) Step 2 Therefore, lim (6x - In(x)) is indeterminate of type [infinity] – [infinity] . We will change x → [infinity] it to a product by factoring out 4x to get lim x → [infinity] 6x 6x(1 - In(x)/6xStep 3 We now have lim x → [infinity] (6x − ln(x)) = lim x → [infinity] 6x 1 − ln(x) 6x. Let's first focus on lim x → [infinity] ln(x)/6x . Since ln(x) → [infinity] as x → [infinity], then this limit is indeterminate of type [infinity]/[infinity]. Using l'Hospital's Rule, we find: lim x → [infinity] ln(x)/6x = lim x → [infinity] 1/6x = 0Step 4 Since lim x → [infinity] In(x)/4x = 0, then we can conclude that: x → [infinity] lim (6x - In(x)) = lim 6x(1 - In(x)/6x)

the answer is c. 23/3

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation:

twenty-six miles over two gallons because the proportion for the original 13 miles to one gallon would be written 13/1 and, doing the same thing to the top and bottom, i multiplied by 2 and got 26/2