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.
The team has lost only 6 games (according to the comments). Let's say "x" is the amount of losses and "y" is the amount of wins.
For every loss, they have 3 wins, so in the end, this would be the equation:
or
Although the first one would be the easier and most understandable to read.
So:
y = 3(6)
would equal
18
<h2><em>we can write (3x^2-5y^2) as (3x-5y)^2</em></h2><h2><em>(
3
x
−
5
y
)
2 as (
3
x−
5
y
)
(
3
x−
5
y
)</em></h2><h2><em>3
x
(
3
x
−
5
y
)
−
5
y
(
3x
−5
y
)</em></h2><h2><em>3
x
(
3
x
−
5
y
)
−
5
y
(3
x
−
5
y
)</em></h2><h2><em>3
x
(
3
x
)
+
3
x
(
−
5y
)
−
5
y
(
3
x
)
−
5
y(
-5
y
)</em></h2><h2><em>9
x
2
−
15
x
y
−
15y
x
+
25
y
2
</em></h2><h2><em> Subtract 15
y
x from −
15
x
y
.</em></h2><h2><em>9
x
2
−
30
xy
+
25
y
2</em></h2><h2><em> HOPE IT HELPS(◕‿◕✿) </em></h2><h2><em> SMILE!! </em></h2>
Answer:
She used 4 test tubes for the last experiment
Step-by-step explanation:
Total test tubes needed = 14
Experiment 1 = 2 test tubes
Total test tubes remaining = total - used
= 14 - 2
= 12 test tubes
Remaining experiment = 4 - 1
= 3
Test tubes used for the remaining experiment = test tubes remaining / Experiment remaining
= 12 / 3
= 4 test tubes
How many test tubes did she use for the last experiment?
She used 4 test tubes for the last experiment
