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$$
\boxed{\frac{3875}{5304}} \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{50} - \frac{1}{52} \right) \right) $$
$$
Then $ x^4 = (x^2)^2 = (y - 1)^2 = y^2 - 2y + 1 $.
So the remainder is $ -2x - 2 $.
Similarly, $ f(\omega^2) = \omega^2 + 3\omega + 1 = a\omega^2 + b $
\frac{(x - 2)^2}{\frac{60}{9}} - \frac{(y - 2)^2}{\frac{60}{4}} = 1 e 1 $, and $ \omega^2 + \omega + 1 = 0 $.
$$
\boxed{\frac{21}{2}} Most terms cancel, leaving:
$$
f(x) = (x^2 + x + 1)q(x) + ax + b \frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652} \Rightarrow a(\omega - \omega^2) = (\omega - \omega^2)(1 - 3) = -2(\omega - \omega^2) $$

Question: Given $ h(x^2 + 1) = 2x^4 + 4x^2 + 3 $, find $ h(x^2 - 1) $.
$$ \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right) Add the two expressions:

$$

Question: Compute $ \sum_{n=1}^{50} \frac{1}{n(n+2)} $.
$$ $$

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\boxed{(2, 2)} Now substitute $ y = x^2 - 1 $:
\frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \left( \frac{3978 - 103}{2652} \right) = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304} $$ Divide both sides by 60 to get standard form:
9(x - 2)^2 - 4(y - 2)^2 = 60 $$
$$

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h(x^2 - 1) = 2(x^2 - 1)^2 + 1 = 2(x^4 - 2x^2 + 1) + 1 = 2x^4 - 4x^2 + 2 + 1 = 2x^4 - 4x^2 + 3 $$
$$ g(3) = 3^2 - 3(3) + 3m = 9 - 9 + 3m = 3m $$
$$
$$

\frac{1}{2} \left( \frac{3}{2} - \frac{1}{51} - \frac{1}{52} \right) = \frac{1}{2} \left( \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right) \right)

$$
9(x - 2)^2 - 36 - 4(y - 2)^2 + 16 = 44 \frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2} $$ $$
AreaQuestion: A microbiome researcher studying gut health models bacterial growth with the function $ f(x) = x^2 - 3x + m $, and models immune response with $ g(x) = x^2 - 3x + 3m $. If $ f(3) + g(3) = 42 $, what is the value of $ m $?
$$ 9(x^2 - 4x) - 4(y^2 - 4y) = 44 $$ $$ $$ (9x^2 - 36x) - (4y^2 - 16y) = 44 Set equal to 42:
$$ \boxed{2x^4 - 4x^2 + 3} This is a hyperbola centered at $ (2, 2) $.
-2\omega + b = \omega + 3\omega^2 + 1 \Rightarrow b = 3\omega + 3\omega^2 + 1 = 3(-1) + 1 = -2 $$
$$ Now solve the system:
Subtract (1) - (2):
$$
$$ Evaluate $ f(3) $:
Evaluate $ g(3) $:
h(y) = 2(y^2 - 2y + 1) + 4(y - 1) + 3 = 2y^2 - 4y + 2 + 4y - 4 + 3 = 2y^2 + 1 a\omega + b = \omega + 3\omega^2 + 1 \quad \ ext{(1)}
Then:
Solution: Let $ y = x^2 + 1 \Rightarrow x^2 = y - 1 $.
$$ f(\omega) = \omega^4 + 3\omega^2 + 1 = \omega + 3\omega^2 + 1 = a\omega + b $$ Distribute and simplify:

Question: An urban mobility engineer designing EV charging stations models traffic flow with $ f

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- In the second: $ -x + y = 4 $, from $ (-4, 0) $ to $ (0, 4) $.
$$
$$ Let $ f(x) = x^4 + 3x^2 + 1 $. The remainder when dividing by a quadratic will be linear: $ ax + b $.
$$

a(\omega - \omega^2) = (\omega - \omega^2) + 3(\omega^2 - \omega) $$
- Third: $ -x - y = 4 $, from $ (-4, 0) $ to $ (0, -4) $.
$$
$$ So $ h(y) = 2y^2 + 1 $.

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Solving gives $ A = \frac{1}{2}, B = -\frac{1}{2} $, so:
\Rightarrow a = -2 $$ $$

Substitute into the expression:
$$ Solution:
$$ $$
Solution: Perform polynomial long division or use the fact that the roots of $ x^2 + x + 1 = 0 $ are the non-real cube roots of unity, $ \omega $ and $ \omega^2 $, where $ \omega^3 = 1 $, $ \omega \ \frac{1}{n(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right) Now compute the sum:
$$
$$ Complete the square:
$$
- Fourth: $ x - y = 4 $.
$$
f(3) = 3^2 - 3(3) + m = 9 - 9 + m = m Plug in $ x = \omega $:
\boxed{-2x - 2} $$ The vertices are $ (4, 0), (0, 4), (-4, 0), (0, -4) $.
$$ $$
Compute the remaining:
Solution: To find the center, we complete the square for both $ x $ and $ y $ terms.
$$ - In the first quadrant: $ x + y = 4 $, from $ (4, 0) $ to $ (0, 4) $.
$$

4m = 42 \Rightarrow m = \frac{42}{4} = \frac{21}{2} $$
$$
Group terms:
$$ Solution: Use partial fractions to decompose the general term:
Find common denominator for $ \frac{1}{51} + \frac{1}{52} $:
\sum_{n=1}^{50} \frac{1}{n(n+2)} = \frac{1}{2} \sum_{n=1}^{50} \left( \frac{1}{n} - \frac{1}{n+2} \right)

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$$ So:
This diamond has diagonals of length 8 (horizontal) and 8 (vertical).
$$ $$
a\omega^2 + b = \omega^2 + 3\omega + 1 \quad \ ext{(2)}
9[(x - 2)^2 - 4] - 4[(y - 2)^2 - 4] = 44 f(3) + g(3) = m + 3m = 4m Solution: The equation $ |x| + |y| = 4 $ represents a diamond (a square rotated 45 degrees) centered at the origin.
Substitute $ a = -2 $ into (1):

Question: Find the area of the region enclosed by the graph of $ |x| + |y| = 4 $.
In each quadrant, the equation simplifies to a linear equation. For example:
$$
This is a telescoping series:
$$ Factor out leading coefficients:

Question: Find the remainder when $ x^4 + 3x^2 + 1 $ is divided by $ x^2 + x + 1 $.