Prove (x^2)+1/(x^2 )≥x + (1/x )≥2 for x> 0

First of all, let's prove that

We can rewrite the fraction as

And the inequality as

Again, we can rewrite as

The numerator is a square, so it is always greater than or equal to zero.

The denominator is positive, because we're assuming x>0.

So, this is a fraction between positive numbers, and it is always positive, and the first part is finished.

As for the second part, we have

The numerator is not the product of a square (always greater than or equal to zero) and a quadratic equation with no solutions and positive leading coefficient (always positive), so it's always greater than or equal to zero. The denominator is x^2, and since x>0 by hypothesis, it is positive. This ends the proof.

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

happy to

pls mark as brainliest

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.