answer:Since Q is the symmetric point of P with respect to the origin, we can determine the coordinates of Q by changing the sign of each coordinate of P. In other words, if P has coordinates (x, y, z), then Q will have coordinates (-x, -y, -z). For the given point P(1, -2, 1), the x-coordinate is 1, the y-coordinate is -2, and the z-coordinate is 1. Therefore, the symmetric point Q will have the coordinates: - The x-coordinate will be -1, - The y-coordinate will be 2 (since the opposite of -2 is 2), - The z-coordinate will be -1. Hence, the coordinates of Q will be boxed{(-1, 2, -1)}.

answer:Factor the quadratic equation (x^2 - 9x + 14): [ x^2 - 9x + 14 = (x-2)(x-7) ] We need to determine where ((x-2)(x-7) < 0). This inequality holds when the two factors have opposite signs: - When (x < 2), both (x-2) and (x-7) are negative. - When (2 < x < 7), (x-2) is positive and (x-7) is negative, satisfying the inequality. - When (x > 7), both (x-2) and (x-7) are positive. Thus, the solution to the inequality (x^2 - 9x + 14 < 0) occurs when (2 < x < 7). Final answer: [ boxed{(2,7)} ]

answer:Let the common difference of the arithmetic sequence {a_{n}} be d. Then 2d=(a_{2}+a_{4})-(a_{1}+a_{3})=10-4=6, Solving for d gives d=3, ∴a_{1}+a_{3}=a_{1}+a_{1}+2d=4, Solving for a_{1} gives a_{1}=2-d=-1, ∴S_{n}=na_{1}+ frac{n(n-1)}{2}d= frac{3}{2}n^{2}- frac{5}{2}n So the sum of the first n terms is: boxed{S_{n}= frac{3}{2}n^{2}- frac{5}{2}n} This can be obtained by finding the first term and the common difference from the given equations and then substituting them into the formula for the sum of an arithmetic sequence. This question tests the formula for the sum of the first n terms of an arithmetic sequence and involves basic operations of arithmetic sequences, making it a basic question.

answer:Let's consider the given curve: y= frac {1}{2}x^{2}+x. To find the equation of the tangent line at point (2, 4), we first need to compute the derivative of the function to find the slope of the tangent line: begin{aligned} y' &= dfrac{d}{dx}left(frac{1}{2}x^2+xright) &= x+1. end{aligned} Thus, the slope of the tangent line at point (2, 4) is: m = y'(2) = 2+1 = 3. Using the point-slope form of the equation of a line, we can write the equation of the tangent line as: y-4 = 3(x-2), which simplifies to: 3x-y-2=0. To find where this tangent line intersects the x-axis, set y=0 and solve for x: 3x - 0 - 2 = 0 Rightarrow x=frac{2}{3}. To find where the tangent line intersects the y-axis, set x=0 and solve for y: 0 - y - 2 = 0 Rightarrow y = -2. Now, we have the lengths of the two sides of the right triangle formed by the tangent line and the coordinate axes: the height h=|-2| and the base b=frac{2}{3}. The area S of the triangle is: S = frac{1}{2}times h times b, S = frac{1}{2}times |-2| times frac{2}{3} = frac{1}{2} times 2 times frac{2}{3} = frac{4}{3}. We have made a mistake in the calculation. Correcting this, we find: S = frac{1}{2} times 2 times frac{2}{3} = frac{2}{3}. Therefore, the area of the right triangle is boxed{frac{2}{3}}.