There's not enough information given to find a definite solution for the two numbers. Instead there will be many solutions. Two variable problems require either a second equation, a value for one of the variables or a relation given between them. (like if they're consecutive integers or one is twice the other)
So, what you can do is write out the equation and "guess and test" till you find two integers that work.
x^2 - y^2 = 9^2 - 1
x^2 - y^2 = 80
y = sqrt( x^2 - 80)
putting in integer values for x starting at 9.
1 = sqrt(81 - 80) (9, 1) works likewise so would (-9,1) (-9,-1) as well as (12, 8)
1 less than square of a number is 3.
[tex]\sqrt{x}[/tex] -1
Step-by-step explanation:
simplified into words. lol
Plz mark me brainliest!
It's ok if u don't...lol
I hope this helped!!
[tex]x1[/tex]
Step-by-step explanation:
Let the unknown number be x
[tex]1 < x^{2} \\\\Switch\:sides\\\\x^21\\\\\mathrm{For\:}u^n\:\:a\mathrm{,\:if\:}n\:\mathrm{is\:even}\mathrm{\:then\:}u\:\:\sqrt[n]{a}\\\\x1[/tex]
There's not enough information given to find a definite solution for the two numbers. Instead there will be many solutions. Two variable problems require either a second equation, a value for one of the variables or a relation given between them. (like if they're consecutive integers or one is twice the other)
So, what you can do is write out the equation and "guess and test" till you find two integers that work.
x^2 - y^2 = 9^2 - 1
x^2 - y^2 = 80
y = sqrt( x^2 - 80)
putting in integer values for x starting at 9.
1 = sqrt(81 - 80)
(9, 1) works likewise so would (-9,1) (-9,-1) as well as (12, 8)
The y-value is 1 less than the square of the x-value,
Step-by-step explanation:
if we follow the letters and convert them into numbers we will get:
y-value=y
1 less than=-1
square of x value=x^2
so
we put the number in order according to the words:
y value=1 less than the square of x
so
y=x^2-1
hope this helps!
c² - 1
Step-by-step explanation:
c² - 1
If factorization required:
(c)² - (1)²
(c - 1)(c + 1)
if c was 9 then it would be
8 squared.
this question kind of bland