we have

we know that
<u>The Rational Root Theorem</u> states that when a root 'x' is written as a fraction in lowest terms

p is an integer factor of the constant term, and q is an integer factor of the coefficient of the first monomial.
So
in this problem
the constant term is equal to 
and the first monomial is equal to
-----> coefficient is 
So
possible values of p are 
possible values of q are 
therefore
<u>the answer is</u>
The all potential rational roots of f(x) are
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Answer:
x = 2
y = 1
Step-by-step explanation:
3x - y = 5..............(1)× 1
2x + y = 5...….......(2)× 1
3x - y = 5..............(3)
2x + y = 5.............(4)
Cancel y
3x + 2x = 5 + 5
5x = 10
Divide both sides by 5
x = 2
Substitute for the value of y
3x - y = 5
3(2) - y = 5
6 - y = 5
6 - 5 = y
1 = y
y = 1
Answer:
s=P-b/2
Step-by-step explanation:
P-b=2s
P-b/2=s
The answer is the one you put that and it says its correct