answer:Alyssa had 3 roses in the vase initially. After cutting more roses from her garden, there are now 14 roses in the vase. To find out how many roses she cut, we subtract the initial number of roses from the final number of roses. 14 roses (final number) - 3 roses (initial number) = 11 roses Alyssa cut boxed{11} roses from her flower garden.

answer:Step 1: From the problem statement, we have the expression for profit y as follows: [ y = left(4 + frac{20}{P}right)P - x - 6left(P + frac{1}{P}right), ] where P is given by P = frac{x + 2}{4}. Step 2: Substituting P into the expression for y, we get: [ y = left(4 + frac{20}{frac{x+2}{4}}right)frac{x+2}{4} - x - 6left(frac{x+2}{4} + frac{4}{x+2}right). ] Step 3: Simplifying the expression for y, we obtain: [ y = 19 - frac{24}{x+2} - frac{3}{2}x, quad text{where} 0 leq x leq a. ] Step 4: We express y in a different form to analyze its maximum: [ y = 22 - frac{3}{2}left(frac{16}{x+2} + x + 2right). ] Step 5: Applying the AM-GM inequality (Arithmetic Mean - Geometric Mean inequality), we have: [ y leq 22 - 3sqrt{frac{16}{x+2} cdot (x+2)} = 10, ] and equality holds when frac{16}{x+2} = x + 2, which occurs at x = 2. Step 6: For the case when a geq 2, the promotional costs investment that maximizes profit is when x = 2 ten thousand yuan, making the maximum profit: [ boxed{y = 10} text{ ten thousand yuan}. ] Step 7: To verify whether this is the maximum profit for a < 2, we find the derivative of y with respect to x: [ y' = frac{24}{(x+2)^2} - frac{3}{2}. ] Step 8: Since y' is always positive when a < 2, the function is monotonically increasing on the interval [0, a]. Step 9: Hence, when a < 2, the maximum profit occurs at x = a. The company's maximum profit, when promotional costs are invested to a ten thousand yuan, is: [ boxed{y = 19 - frac{24}{a+2} - frac{3}{2}a}. ]

answer:To solve the problem, we'll break it down into two parts: the division of space stickers and the division of cat stickers. **Step 1: Dividing the space stickers** Paige has 100 space stickers to share among her 3 friends. To find out how many stickers each friend gets and how many are left over, we perform the division: - 100 , text{stickers} div 3 , text{friends} = 33 , text{stickers per friend with} , 1 , text{sticker remaining} **Step 2: Dividing the cat stickers** Next, Paige has 50 cat stickers to share among her 3 friends. Again, we perform the division to find out the distribution: - 50 , text{stickers} div 3 , text{friends} = 16 , text{stickers per friend with} , 2 , text{stickers remaining} **Step 3: Calculating the total remaining stickers** Finally, to find out how many stickers Paige will have left after sharing, we add the remaining stickers from both the space and cat stickers: - 1 , text{remaining space sticker} + 2 , text{remaining cat stickers} = 3 , text{stickers left} Therefore, Paige will have boxed{3} stickers left after sharing with her friends.

answer:1. **Rewrite the expression**: Rewrite ( left(1 - frac{1}{a}right)^7 ) using the property of exponents: ( frac{(a-1)^7}{a^7} ). 2. **Expand using the Binomial Theorem**: Using the theorem, the expansion of ( (a-1)^7 ) is: [ (a-1)^7 = sum_{k=0}^7 binom{7}{k} a^{7-k} (-1)^k ] This results in terms for coefficients ( binom{7}{k}(-1)^k ) for ( k = 0, 1, 2, ldots, 7 ). 3. **Identify the last three terms**: The last three terms in ( (a-1)^7 ) for ( k = 0, 1, 2 ) are: - For ( k=0 ): ( binom{7}{0} a^7 (-1)^0 = 1 ) - For ( k=1 ): ( binom{7}{1} a^6 (-1)^1 = -7a^6 ) - For ( k=2 ): ( binom{7}{2} a^5 (-1)^2 = 21a^5 ) 4. **Sum up coefficients of the last three terms**: The coefficients are 1, -7, and 21. Their sum is: [ 1 - 7 + 21 = 15 ] Conclusion with the boxed answer: [ 15 ] The final answer is boxed{**B) 15**}