Because the asymptotes are at x = 1 and x = 2, the factors in the denominator is (x-1)(x-2) = x^2 - 3x + 2.
So, a = -3 and b = 2, which has a sum of -1.
As n approaches infinity, n^4 is significantly bigger than n^2. So, we can simplify this as \(\frac{2n^4}{6n^4} = \boxed{\frac{1}{3}}.\)
y varies directly as x, so the fraction x/y is constant. We get
\(\frac{x}{y} = \frac{-6}{10} = \frac{6}{y}\)
So, y = -10.
Simplify:
\(2.2x - 3.1y = -3.2\)
\(0.4x + y = 8.8\)
Solving, we get \((x, y) = \boxed{(7, 6)}.\)
You have selected the correct answer, no worries.
Let Class 2 have x house points. Therefore,
\(2x + x = 675\)
\(x = \boxed{225}\)
Class 2 has 225 points.
The formula is \(A = P(1 + \frac{r}{n})^{nt}\), where:
P = principal (starting amount)
r = annual rate
n = compound type (annually)
t = time in years
Plug in P = 10,000, r = 6%, n = 1, and t = 4 to get A = $12624.
We can complete the square. After simplifying, we get x^2 + 9x.
This simplifes to \(x^2 + 9x = (x + \frac{9}{2})^2 - \frac{81}{4}\). So, the constant is 81/4.
We have \(4 - n \equiv 10 \quad (\mod 14)\)
So, \(n \equiv 8 \quad (\mod 14)\)
Therefore, \(m+n \equiv 4+8 \equiv \boxed{12} (\mod 14)\)
We have \(2 < \sqrt{t} < 35\)
\(4 < t < 1225\)
There are 1224 - 5 + 1 = 1220 values of t.
