Multiply. Write your answer as a fraction in simplest form. Three divided by sixteen times six
1 answer:
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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)
A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.
Let's look at your function.
is in the numerator. It is defined for every value of x. There is no problem there.
is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is:
Let's look at your function.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is: