answer:We have overrightarrow{m}=(e^{x}+ frac {x^{2}}{2},x) and overrightarrow{n}=(2,a). Then, f(x)= overrightarrow{m}cdot overrightarrow{n}=2e^{x}+x^{2}+ax. Taking the derivative, we get f'(x)=2e^{x}+2x+a. Since f(x) has a monotonically increasing interval in (-1,0), we have f'(x) > 0, which implies a > -2e^{x}-2x text{ on } (-1,0). Let g(x)=-2e^{x}-2x. Then, g'(x)=-2-2e^{x} < 0 text{ on } (-1,0), meaning that g(x) is strictly decreasing on (-1,0). Consequently, g(x) > g(0)=-2 text{ on } (-1,0). This implies that a geq 2. Therefore, the range of values for a is boxed{[-2,+infty)}.

answer:To find the ratio of the increased production to the original production, we first need to determine the original production for the month of March 2020. Assuming March has 31 days, the original daily production of 7000 rolls would result in: Original monthly production = 7000 rolls/day * 31 days = 217000 rolls The increased production during March 2020 was 868000 rolls. To find the increased production, we subtract the original production from the increased production: Increased production = Total production - Original production Increased production = 868000 rolls - 217000 rolls = 651000 rolls Now, we have the increased production (651000 rolls) and the original production (217000 rolls). To find the ratio of the increased production to the original production, we divide the increased production by the original production: Ratio = Increased production / Original production Ratio = 651000 rolls / 217000 rolls To simplify the ratio, we can divide both numbers by the original production (217000 rolls): Ratio = 651000 / 217000 Ratio = 3 Therefore, the ratio of the increased production to the original production is boxed{3:1} .

answer:1. **Substitute ( z = sqrt{xy} ) into the equations:** [ x + y + sqrt{xy} = a ] Let ( x + y = p ) and ( sqrt{xy} = q ). Then the equation becomes: [ p + q = a ] 2. **Simplify the second equation:** [ x^2 + y^2 + z^2 = b^2 ] Using the identity ( x^2 + y^2 = (x + y)^2 - 2xy ), we get: [ x^2 + y^2 = p^2 - 2q^2 ] Substituting ( z = q ), the equation becomes: [ p^2 - 2q^2 + q^2 = b^2 implies p^2 - q^2 = b^2 ] This can be factored as: [ (p + q)(p - q) = b^2 ] Since ( p + q = a ), we have: [ a(p - q) = b^2 implies p - q = frac{b^2}{a} ] 3. **Solve for ( p ) and ( q ):** [ p = frac{a + frac{b^2}{a}}{2} = frac{a^2 + b^2}{2a} ] [ q = frac{a - frac{b^2}{a}}{2} = frac{a^2 - b^2}{2a} ] 4. **Find ( xy ):** [ xy = q^2 = left( frac{a^2 - b^2}{2a} right)^2 = frac{(a^2 - b^2)^2}{4a^2} ] 5. **Form the quadratic equation with roots ( x ) and ( y ):** [ t^2 - pt + xy = 0 ] Substituting ( p ) and ( xy ): [ t^2 - frac{a^2 + b^2}{2a}t + frac{(a^2 - b^2)^2}{4a^2} = 0 ] 6. **Solve the quadratic equation using the quadratic formula:** [ t = frac{frac{a^2 + b^2}{2a} pm sqrt{left( frac{a^2 + b^2}{2a} right)^2 - 4 cdot frac{(a^2 - b^2)^2}{4a^2}}}{2} ] Simplifying inside the square root: [ t = frac{frac{a^2 + b^2}{2a} pm sqrt{frac{(a^2 + b^2)^2 - (a^2 - b^2)^2}{4a^2}}}{2} ] [ t = frac{frac{a^2 + b^2}{2a} pm sqrt{frac{4a^2b^2}{4a^2}}}{2} ] [ t = frac{frac{a^2 + b^2}{2a} pm frac{2b}{2a}}{2} ] [ t = frac{a^2 + b^2 pm 2b}{4a} ] Therefore, the solutions for ( x ) and ( y ) are: [ x, y = frac{a^2 + b^2 pm sqrt{(3a^2 - b^2)(3b^2 - a^2)}}{4a} ] 7. **Conditions for distinct positive ( x, y, z ):** For ( x, y, z ) to be distinct positive numbers, the discriminant must be positive: [ (3a^2 - b^2)(3b^2 - a^2) > 0 ] This implies: [ 3a^2 > b^2 quad text{and} quad 3b^2 > a^2 ] Combining these inequalities, we get: [ sqrt{3} |b| > a quad text{and} quad a > |b| ] The final answer is ( boxed{ x, y = frac{a^2 + b^2 pm sqrt{(3a^2 - b^2)(3b^2 - a^2)}}{4a}, z = frac{a^2 - b^2}{2a} } ) for ( |b| < a < sqrt{3}|b| ).

answer:If there are now 25 rulers in the drawer after Tim placed 14 rulers in it, we can find the original number of rulers by subtracting the number of rulers Tim added from the total number of rulers now in the drawer. So, the calculation would be: 25 rulers (total now) - 14 rulers (added by Tim) = 11 rulers (originally in the drawer) Therefore, there were originally boxed{11} rulers in the drawer.