Physics, 19.05.2021 18:50, natalie407888

How do astronomers describe the scale of the universe?

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answered: hey9691

No one can know the size of the universe, it may in fact be infinitely large.

Explanation:

No one knows the size of the universe because we can't see the entire universe. We can only see the part that is close enough to us that there has been time for the light from it to arrive here. What that means is any part of the universe that it will take more than 14.5 billion years for light to travel to us can not bee seen. It is possible that the universe has no size, that is it is infinite.

We can measure what we call the "visible universe" and that is a sphere around us with a radius of about 50 billion light years, the reason it is more than the age of the universe (14.5 billion years) is because the universe is expanding more rapidly than the speed of light.

The way scientists measure the size of the visible universe is with what is called red shift. Similar to the way a car horn's pitch changes as it passes you due to doppler shift, light too has a shift in is frequency or color depending on if it is moving away or toward you. As light moves away from you it shifts for red and the faster it is moving away the more it shifts. We can estimate the distance objects are away by how much they are red shifted. When we look at the most distance objects we can see they are about 50 million light years away.

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .