answer:1. **Simplify the equation**: Recognize the equation and rewrite it using substitutions: [ a = x - frac{1}{3}, quad b = x + frac{1}{6} ] Substituting, the equation becomes: [ a^2 + ab = 0 ] 2. **Factor the equation**: Factor out the common term: [ a(a + b) = 0 ] Substituting back for a and b, we have: [ (x - frac{1}{3})((x - frac{1}{3}) + (x + frac{1}{6})) = 0 ] Simplifying the expression inside the parentheses: [ (x - frac{1}{3})(2x - frac{1}{3} + frac{1}{6}) = 0 ] Simplifying further: [ (x - frac{1}{3})(2x - frac{1}{6}) = 0 ] 3. **Solve for (x)**: Setting each factor to zero: [ x - frac{1}{3} = 0 quad text{or} quad 2x - frac{1}{6} = 0 ] Solving these equations: [ x = frac{1}{3} quad text{and} quad 2x = frac{1}{6} Rightarrow x = frac{1}{12} ] 4. **Identify the smaller root**: Comparing (frac{1}{3}) and (frac{1}{12}), we find that (frac{1}{12}) is smaller. The smaller root of the equation is (frac{1{12}}). The final answer is boxed{textbf{(A)} frac{1}{12}}

answer:To calculate the total number of cakes Ginger makes in 10 years, we break down the calculation based on each recipient: 1. For her 2 children, she bakes 4 cakes each year (birthdays, Christmas, Easter, and Halloween). Therefore, the total number of cakes for her children each year is: [ 2 text{ children} times 4 text{ cakes/child/year} = 8 text{ cakes/year} ] 2. For her husband, she makes cakes for 6 occasions each year (birthdays, Christmas, Easter, Halloween, their anniversary, and Valentine's Day). Thus, she makes: [ 6 text{ cakes/year for her husband} ] 3. For her parents, she makes a cake for each on their birthdays, totaling: [ 2 text{ parents} times 1 text{ cake/parent/year} = 2 text{ cakes/year} ] Adding these amounts together gives the total number of cakes she makes in a year: [ 8 text{ cakes/year for children} + 6 text{ cakes/year for husband} + 2 text{ cakes/year for parents} = 16 text{ cakes/year} ] To find out how many cakes she makes in 10 years, we multiply the annual total by 10: [ 16 text{ cakes/year} times 10 text{ years} = 160 text{ cakes} ] Therefore, in 10 years, Ginger makes a total of boxed{160} cakes.

answer:The given proposition is a universal statement. The correct negation of a universal statement "For all x in mathbb{R}, P(x)" is a particular statement of the form "There exists an x in mathbb{R} such that P(x) is not true". In formal logic, the negation of "P(x) is true" is "P(x) is not true", which can be expressed equivalently as "P(x) is false or P(x) is equal to the threshold where the truth might change". For the proposition e^x > x, the negation would be e^x leq x. Therefore, the negation of the given statement is "There exists an x in mathbb{R} such that e^x leq x". The correct translation and negation lead us to option D. [boxed{text{D: There exists an } x text{ in } mathbb{R} text{ such that } e^x leq x}]

answer:To solve the problem, we start by defining the common ratio of the geometric sequence {a_{n}} as q. Given the information, we have two equations based on the sums of the first few terms: 1. The sum of the first three terms, S_{3} = a_{1} + a_{2} + a_{3} = 21 2. The sum of the first two terms, S_{2} = a_{1} + a_{2} = 9 From these, we can express the ratio of S_{3} to S_{2} in terms of q: [ frac{S_{3}}{S_{2}} = frac{a_{1} + a_{2} + a_{3}}{a_{1} + a_{2}} = frac{1 + q + q^{2}}{1 + q} = frac{21}{9} ] Simplifying the fraction on the right side gives us frac{7}{3}. Thus, we have: [ frac{1 + q + q^{2}}{1 + q} = frac{7}{3} ] Cross-multiplying to solve for q gives us a quadratic equation: [ 3(1 + q + q^{2}) = 7(1 + q) ] [ 3 + 3q + 3q^{2} = 7 + 7q ] [ 3q^{2} - 4q - 4 = 0 ] Solving this quadratic equation for q, we find that q = 2 or q = -frac{2}{3}. However, since all terms of the sequence are positive, we discard the negative value, leaving us with q = 2. Next, we use the sum of the first two terms to find a_{1}: [ S_{2} = a_{1} + a_{2} = a_{1} + qa_{1} = 9 ] Substituting q = 2 into the equation gives us: [ a_{1} + 2a_{1} = 9 ] [ 3a_{1} = 9 ] [ a_{1} = 3 ] Therefore, the value of a_{1} is boxed{3}, which corresponds to choice C.