Determine the solution for x2 + 36 > 12x

solution for x^2 + 36 > 12x is x > 6 or x< 6

Step-by-step explanation:

We need to solve the equation:

x^2 + 36 > 12x

Subtract 12x from both sides

x^2 + 36 - 12x > 0

Rearranging

x^2 -12x + 36 > 0

Now factorizing:

x^2 -6x -6x + 36 >0

x(x-6)-6(x-6) > 0

(x-6)(x-6) > 0

(x-6)^2 > 0

For, x^n > 0 and n is even then x >0 or x<0

so,

x-6 >0 or x-6 <0

x > 6 or x< 6

So, solution for x^2 + 36 > 12x is x > 6 or x< 6

For this case we have the following inequality:

We subtract 12x on both sides of the inequality:

We convert to an equation:

To factor the equation, we must find two numbers that when multiplied give +36 and when added together obtain -12. These numbers are -6 and -6.

So, we have:

The solution is

We create test intervals:

We test a value in the interval, such as x = 0:

Is fulfilled!

We test a value in the interval, such as x = 10.

Is fulfilled!

Thus, the solution is given by:

or

or

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.