The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are
For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
Y= 5 y =7 thats the smaller slip and larger
Answer: x=−10
Step-by-step explanation:
x−3−2(x+6)=−5
x+−3+(−2)(x)+(−2)(6)=−5(Distribute)
x+−3+−2x+−12=−5
(x+−2x)+(−3+−12)=−5(Combine Like Terms)
−x+−15=−5
−x−15=−5
−x−15+15=−5+15
−x=10
Answer:
y =2
Step-by-step explanation:
Substitute -10 for x:
2(-10)-9y = -38
-20-9y = -38
-9y = -18
9y = 18
y = 2
Answer:
D
Step-by-step explanation:
It's these shapes because:
Rectangle- A quadrilateral with 2 pairs of congruent sides and only 90° angles is a rectangle.
Parallelogram- A quadrilateral with 2 pairs of parallel sides is a parallelogram.
