answer:To find the area of a triangle with vertices at (x_1, y_1), (x_2, y_2), and (x_3, y_3), we use the formula: [ text{Area} = frac{1}{2} left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) right| ] In our scenario, let's denote the third vertex coordinates as (x, 0) since it lies on the x-axis. The other vertices are (3,2) and (0,0). Plugging into the area formula, we have: [ 24 = frac{1}{2} left| 3(0 - 0) + 0(0 - 2) + x(2 - 0) right| ] [ 24 = frac{1}{2} left| 2x right| = |x| ] So, x = pm 24. Since we are informed that the vertex is on the negative branch of the x-axis, we select x = -24. Thus, the coordinates of the third vertex are boxed{(-24, 0)}.

answer:1. Let ( P(x) ) be a polynomial with integer coefficients such that ( P(x) = 5 ) for five different integers ( x ). Let these integers be ( a, b, c, d, e ). Therefore, we have: [ P(a) = P(b) = P(c) = P(d) = P(e) = 5 ] 2. Define a new polynomial ( g(x) ) as follows: [ g(x) = P(x) - 5 ] Since ( P(a) = 5 ), ( P(b) = 5 ), ( P(c) = 5 ), ( P(d) = 5 ), and ( P(e) = 5 ), it follows that: [ g(a) = g(b) = g(c) = g(d) = g(e) = 0 ] Therefore, ( g(x) ) has roots at ( x = a, b, c, d, e ). Hence, ( g(x) ) can be factored as: [ g(x) = k(x - a)(x - b)(x - c)(x - d)(x - e) ] where ( k ) is a constant. 3. Since ( P(x) ) has integer coefficients, and ( P(x) = g(x) + 5 ), ( g(x) ) must also have integer coefficients. Therefore, ( k ) must be an integer. 4. We need to show that for any integer ( n ), ( g(n) ) cannot lie in the intervals ( -11 leq g(n) leq 4 ) or ( 6 leq g(n) leq 16 ). This is equivalent to showing that ( P(n) ) cannot lie in the intervals ( -6 leq P(n) leq 4 ) or ( 6 leq P(n) leq 16 ). 5. Consider the values ( g(n) ) can take. Since ( g(x) = k(x - a)(x - b)(x - c)(x - d)(x - e) ) and ( a, b, c, d, e ) are distinct integers, the smallest non-zero product of five distinct integers (in absolute value) is: [ |1 times (-1) times 2 times (-2) times 3| = 12 ] This is because the product of any five distinct integers, none of which are zero, will always be at least 12 in absolute value. 6. Therefore, the possible values of ( g(n) ) (other than 0) must satisfy ( |g(n)| geq 12 ). 7. Since ( g(n) ) cannot be in the range ( -11 leq g(n) leq 4 ) or ( 6 leq g(n) leq 16 ), it follows that ( P(n) = g(n) + 5 ) cannot be in the range ( -6 leq P(n) leq 4 ) or ( 6 leq P(n) leq 16 ). (blacksquare)

answer:To determine how many integers between 1 and 1,000,000 are neither perfect squares, perfect cubes, nor fourth powers, we will make use of the principle of inclusion and exclusion. 1. **Calculation of Perfect Squares:** - Calculate the highest integer ( n ) such that ( n^2 leq 1000000 ). - [ n = lfloor sqrt{1000000} rfloor = lfloor 1000 rfloor = 1000 ] - Therefore, there are 1000 perfect squares between 1 and 1000000. 2. **Calculation of Perfect Cubes:** - Calculate the highest integer ( m ) such that ( m^3 leq 1000000 ). - [ m = lfloor sqrt[3]{1000000} rfloor = lfloor 100 rfloor = 100 ] - Therefore, there are 100 perfect cubes between 1 and 1000000. 3. **Calculation of Sixth Powers:** - Sixth powers are numbers that are both perfect squares and perfect cubes, i.e., numbers of the form ( k^6 ). - Calculate the highest integer ( p ) such that ( p^6 leq 1000000 ). - [ p = lfloor sqrt[6]{1000000} rfloor = lfloor 10 rfloor = 10 ] - Therefore, there are 10 sixth powers between 1 and 1000000. 4. **Calculation of Fourth Powers:** - All fourth powers are also perfect squares. Therefore, we don't need to subtract them separately. We continue our calculations with these already included in the perfect squares. 5. **Applying the Principle of Inclusion and Exclusion:** - The principle of inclusion and exclusion for three sets is given by: [ |A cup B cup C| = |A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| + |A cap B cap C| ] - In this context, we only need to consider the overlaps of squares and cubes (sixth powers). - Therefore, the number of integers which are either perfect squares or perfect cubes or both is: [ 1000 + 100 - 10 = 1090 ] 6. **Final Calculation:** - Subtract the number of integers that are perfect squares, cubes, or both (i.e., sixth powers) from the total number of integers from 1 to 1000000. [ 1000000 - 1090 = 998910 ] # Conclusion: The number of integers between 1 and 1,000,000 that are neither perfect squares, perfect cubes, nor fourth powers is: [ boxed{998910} ]

answer:Given the equation w + frac{1}{w} = s, we can multiply through by w to remove the fraction, resulting in [ w^2 + 1 = ws. ] Rewriting this, we have [ w^2 - sw + 1 = 0. ] Using the quadratic formula to solve for w, [ w = frac{s pm sqrt{s^2 - 4}}{2}. ] Suppose s^2 > 4, then the expression under the square root will be positive and real, leading to [ w = frac{s pm sqrt{s^2 - 4}}{2}. ] Now, if s^2 < 4, this introduces a complex component: [ w = frac{s pm isqrt{4 - s^2}}{2}. ] To find the magnitude of w, calculate [ |w| = sqrt{left(frac{s}{2}right)^2 + left(frac{sqrt{4 - s^2}}{2}right)^2} = sqrt{frac{s^2}{4} + frac{4 - s^2}{4}} = sqrt{1} = boxed{1}. ]