SOLVE THE EQUATION!!!!!!!!
2 answers:
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Answer:
Step-by-step explanation:
The given polynomial function is
According to the Rational Roots Theorem, the ratio of all factors of the constant term expressed over the factors of the leading coefficient.
The potential rational roots are
9514 1404 393
Answer:
1. asymptotes: x=-6, x=6, y=0; zero: x=0
2. 20 blue fish
Step-by-step explanation:
1. The vertical asymptotes are where the denominator is zero. It factors as (x-6)(x+6), so will have zeros at x=-6 and x=6.
The horizontal asymptote is at the limit when x goes to infinity. Here, the ratio of highest-degree terms is 6/x, so the value goes to zero as x gets large.
The zero is where the numerator is zero, at x=0.
In summary:
- vertical asymptotes: x = ±6
- horizontal asymptote: y = 0
- zero: (0, 0)
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2. 60% are red, so 40% are blue. The difference in percentages is 60%-40% = 20%. The percentage of blue fish (40%) is twice the difference (=2×20%), so the number of blue fish is twice the difference in numbers.
blue fish = 2×10 = 20 fish
<em><u>Check</u></em>
There are then 20+10=30 red fish, so 20+30=50 fish total. Reds are 30/50 = 60% of the total.
<u><em>Additional comment</em></u>
I find it often works well to work with ratios, which I can often do in my head. Other folks like to see equations. Here, we could write equations like ...
r = 60%(r + b) . . . . reds are 60% of the total
r - b = 10 . . . . . . . . 10 more reds than blues
We want the value of b, so we can eliminate r by substituting for it:
r = 10 + b . . . . from the second equation above
(10 +b) = 0.60((10 +b) + b) . . . . substitute for r
10 +b = 6 + 1.2b . . . simplify
4 = 0.2b . . . . subtract b+6
20 = b . . . . . . multiply by 5 (there are 20 blue fish)
X=29 and y=10
You would isolate one of the variables and then plug the expression into the other equation to find the value of one variable. Then you would plug this value into the other equation to determine the value of the remaining variable.
You would isolate one of the variables and then plug the expression into the other equation to find the value of one variable. Then you would plug this value into the other equation to determine the value of the remaining variable.