Finding an irrational number between which given pair of numbers supports the idea that irrational numbers are dense

Mathematics

1 answer:

Fittoniya [83]3 years ago

8 0

Answer:

3.33 and 1/3

Step-by-step explanation:

"Dense" here means that there are infinite irrational numbers between two rational numbers. Also, there are infinite rational numbers between two rational numbers. That's the meaning of dense. Actually, that can be apply to all real numbers, there always is gonna be a number between other two.

But, to demonstrate that irrationals are dense, we have to based on an interval with rational limits, because the theorem about dense sets is about rationals, and the dense irrational set is a deduction from it. That's why the best option is 2, because that's an interval with rational limits.

You might be interested in

Juanita a talented artist. She makes and sells t-shirts showing her art work.

liq [111]

Answer:

answer 436 tubig is water halimaw is monster ang Ganda mo Sana kaso ang height mo pang kinder

Step-by-step explanation:

answer is 436 I hope it helps

4 0

2 years ago

What is the vertex of the absolute value function defined by ƒ(x) = |x + 2| + 4?

Marrrta [24]

Answer:

C. (-2,4)

Step-by-step explanation:

We have been given a function $f(x)=|x+2|+4$ and we are asked to find the vertex of our absolute value function.

The rules for the translation of a function are as follows:

$f|x-h|=\text{Graph shifted to right by h units}$

$f|x+h|=\text{ Graph shifted to left by h units}$

$f|x|+h=\text{ Graph shifted upward by h units}$

$f|x|-h=\text{ Graph shifted down by h units}$

Upon comparing our absolute function with above transformations we can see that our function is shifted to two units right of the origin(0,0) so x coordinate of our absolute function will be -2.

Our function is shifted upward from origin by 4 units, therefore, y-coordinate of our absolute value function will be 4.

$f(x)=|x--2|+4$