answer:1. Start with the equation 4x + 5y = 20. 2. Rearrange the equation to solve for y: [ 5y = -4x + 20 ] 3. Divide all terms by 5: [ y = -frac{4}{5}x + 4 ] This equation is now in the slope-intercept form, where y = mx + b, with m representing the slope. 4. From the equation y = -frac{4}{5}x + 4, the slope (m) is boxed{-frac{4}{5}}.

answer:Label the center of the larger circle as O and the points of contact between the larger circle and the smaller circles as A and B. Draw the radius OA of the larger circle. [asy] size(120); import graph; filldraw(Circle((0,0),2.67),mediumgray); filldraw(Circle((-1.33,0),1.33),white); filldraw(Circle((1.33,0),1.33),white); draw((-2.67,0)--(0,0)); label("A",(-2.67,0),W); label("O",(0,0),E); label("B",(2.67,0),E); [/asy] 1. The radius of the larger circle is 8 cm, therefore, the area of the larger circle is: [ text{Area}_{text{large}} = pi times 8^2 = 64pi text{ cm}^2. ] 2. The radius of each smaller circle is half of the radius of the larger circle, hence 4 cm. The area of each smaller circle is: [ text{Area}_{text{small}} = pi times 4^2 = 16pi text{ cm}^2. ] 3. The total area of the two smaller circles is: [ 2 times text{Area}_{text{small}} = 2 times 16pi = 32pi text{ cm}^2. ] 4. The area of the shaded region is the area of the larger circle minus the combined area of the two smaller circles: [ text{Shaded Area} = 64pi - 32pi = boxed{32pi text{ cm}^2}. ] Conclusion: The shaded area is boxed{32pi text{ cm}^2}.

answer:First, expand the left side of the equation: [ (2010 + 2y)^2 = 2010^2 + 4 cdot 2010 cdot 2y + 4y^2 = 2010^2 + 8040y + 4y^2. ] Set this equal to the right side: [ 2010^2 + 8040y + 4y^2 = 4y^2. ] Simplify the equation by canceling out the 4y^2 terms: [ 2010^2 + 8040y = 0. ] Solve for y: [ 8040y = -2010^2, ] [ y = -frac{2010^2}{8040}. ] Simplify further: [ y = -frac{2010^2}{8040} = -frac{2010 cdot 2010}{8040} = -frac{4040100}{8040} approx -502.5. ] However, since the exact calculation must be maintained: [ y = -frac{2010 cdot 2010}{8040} = -frac{2010}{2} = -1005. ] Therefore, the solution is: [ y = boxed{-1005}. ]

answer:To solve the problem, let's consider the equation: [ P(x+Q(y))=Q(x+P(y)) ] for all real numbers (x) and (y). 1. **When (P(x) equiv Q(x)) (for all (x in mathbb{R}))**: - If (P(x)) and (Q(x)) are identical polynomials, the equation (P(x+Q(y)) = Q(x+P(y))) is satisfied. 2. **Assuming (P(x)) and (Q(x)) are two different polynomials**: - First, let's consider the case where (P(x)) is a constant polynomial. If (P(x) = c), then (Q(x + P(y)) = Q(x + c)) is a constant. Hence, (Q(x)) must also be a constant polynomial. Therefore, (P(x) equiv Q(x)), which contradicts our assumption that (P) and (Q) are different polynomials. 3. **Consider (P(x)) and (Q(x)) are not constant polynomials**: - Suppose (P(x) = a x^{n} + b x^{n-1} + R(x)) with (deg(R) < n-1), and (Q(x) = c x^{m} + d x^{m-1} + S(x)) with (deg(S) < m-1), where (a , c) are nonzero real numbers, (n, m in mathbb{Z}_{+}). 4. **Highest Degree Terms Analysis**: - Consider (P(x+Q(y))) as a polynomial in (x): [ P(x + Q(y)) = a (x + Q(y))^n + b (x + Q(y))^{n-1} + R(x + Q(y)) ] - The highest degree term would be (a x^n), and the coefficient of (x^{n-1}) would be (a n Q(y) + b). - Similarly, for (Q(x + P(y))) as a polynomial in (x): [ Q(x + P(y)) = c (x + P(y))^m + d (x + P(y))^{m-1} + S(x + P(y)) ] - The highest degree term would be (c x^m), and the coefficient of (x^{m-1}) would be (c m P(y) + d). 5. **Equating the Highest Degree Terms**: - Since (P(x+Q(y)) = Q(x+P(y))) for all ((x, y) in mathbb{R}^2), we equate the highest degree terms: [ a x^n = c x^m ] - This implies that (m = n) and (a = c). 6. **Equating the Coefficients of (x^{n-1})**: - For (n = m) and (a = c), we get: [ a n Q(y) + b = c m P(y) + d ] Since (a = c) and (n = m), [ a n Q(y) + b = a n P(y) + d ] [ a n (Q(y) - P(y)) = d - b ] [ Q(y) = P(y) + frac{d - b}{a n} ] 7. **Considering (frac{d - b}{a n} neq 0)**: - Let (t = frac{d - b}{a n}). Hence, [ Q(y) = P(y) + t ] 8. **Substituting back into the original equation**: - Substitute (Q(x) = P(x) + t) into (P(x + Q(y)) = Q(x + P(y))): [ P(x + (P(y) + t)) = (P(x) + t) + P(y) ] [ P(x + P(y) + t) = P(x) + P(y) + t ] - Since (P) and (Q) are polynomials, this equality must hold for all (x) and (y). - Therefore, [ P(x) = x + u text{ and } Q(x) = x + v ] where (u, v) are arbitrary real numbers, and (u neq v). # Conclusion: boxed{ begin{array}{c} P(x) equiv Q(x) quad text{for all} , x in mathbb{R} , text{or} P(x) = x + u quad text{and} quad Q(x) = x + v end{array} }