Asap and i will give brainliest

using the numbers 5, 8, and 24, create the problem, using no more than four operations
(addition, subtraction, multiplication, division, square, square root, cube, cube root) where the solution will be an irrational number. explain why the result of your operations is an irrational number.

this is a write out question so make it long and it is worth 5 !

answer: all you need to do to get an irrational is take the square root of a number that is not a perfect square. none of the numbers are perfect squares. after you do that you can do any closed operation you want with the resulting irrational and the two other rationals and you will end up with an irrational.

8*24*√5 is an example.

8+24+√5 is another

5*√8 + 24 is another.

if you take the cube root of any number that isn't a perfect cube you get an irrational, so you could take the cube root of 5 or 24 instead of the square root and do the same thing.

can i get brainliest thx

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!