answer:1. **Rotation**: The original equation ( y = (x - 3)^2 + 4 ) rotated 180 degrees about its vertex changes the concavity and equation becomes ( y = -(x - 3)^2 + 4 ). 2. **Translation Right**: Shifting this equation 4 units to the right changes the equation to ( y = -(x - 7)^2 + 4 ). 3. **Translation Up**: Shifting it 3 units up changes the equation to ( y = -(x - 7)^2 + 7 ). 4. **Finding Zeros**: To find the zeros, set the equation to zero: [ 0 = -(x - 7)^2 + 7 (x - 7)^2 = 7 x - 7 = pmsqrt{7} x = 7 pm sqrt{7} ] Therefore, the zeros are ( x = 7 + sqrt{7} ) and ( x = 7 - sqrt{7} ). 5. **Sum of Zeros**: Summing these values gives: [ (7 + sqrt{7}) + (7 - sqrt{7}) = 14 ] Conclusion: The sum of the x-coordinates of the zeros of the resulting parabola is (boxed{14}).

answer:To simplify the given expression frac{1}{{a-3}}-frac{6}{{{a^2}-9}}, let's follow the steps closely related to the solution provided: 1. Recognize that the denominator a^2 - 9 can be factored into (a+3)(a-3), which helps in combining the fractions: frac{1}{{a-3}}-frac{6}{{{a^2}-9}} = frac{1}{{a-3}}-frac{6}{(a+3)(a-3)}. 2. Rewrite the first fraction with a common denominator: =frac{a+3}{(a+3)(a-3)}-frac{6}{(a+3)(a-3)}. 3. Combine the fractions by subtracting their numerators: =frac{a+3-6}{(a+3)(a-3)}. 4. Simplify the numerator: =frac{a-3}{(a+3)(a-3)}. 5. Cancel out the (a-3) term in the numerator and the denominator: =frac{1}{a+3}. Therefore, the simplified form of the given expression is boxed{frac{1}{a+3}}, which corresponds to choice boxed{A}.

answer:We need to determine the largest possible value of ( N ) such that the product of three natural numbers that sum to 1003 ends with ( N ) zeros. 1. **Choose the numbers:** Let's identify three natural numbers whose sum is 1003. Consider the numbers 625, 250, and 128. Clearly: [ 625 + 250 + 128 = 1003 ] 2. **Calculate their product:** The product of these numbers is: [ 625 cdot 250 cdot 128 ] 3. **Factorize each number:** Factorizing the numbers into their prime factors, we get: [ 625 = 5^4 ] [ 250 = 2 cdot 5^3 ] [ 128 = 2^7 ] 4. **Multiply and simplify using prime factors:** [ 625 cdot 250 cdot 128 = 5^4 cdot (2 cdot 5^3) cdot 2^7 ] [ = 5^4 cdot 5^3 cdot 2 cdot 2^7 ] [ = 5^{4+3} cdot 2^{1+7} ] [ = 5^7 cdot 2^8 ] [ = 2 cdot 2^7 cdot 5^7 ] [ = 2 cdot 10^7 ] [ = 20000000 ] The product is ( 20000000 ), which ends with 7 zeros. 5. **Check if it's possible to get more than 7 zeros:** Let’s assume there are three natural numbers whose product ends with at least 8 zeros. This implies that the product is divisible by ( 10^8 = 2^8 cdot 5^8 ). Since the sum of those three numbers, 1003, is not divisible by 5, at least one of these numbers must not be divisible by 5. Let's label the three numbers as ( a, b, ), and ( c ) such that ( a + b + c = 1003 ). For their product to be divisible by ( 5^8 ), at least one of these numbers should contribute ( 5^8 ) or the product of some other two should meet this requirement. 6. **Analyze the conditions:** - If a number is divisible by ( 5^5 ), it becomes ( geq 3125 > 1003 ). So, not feasible. - If two numbers each provide at least ( 5^4 ), then their sum must be even larger than the feasible total: [ 2 cdot 5^4 = 625 + 625 = 1250 , text{which is} > 1003 ] Therefore, it’s impossible to have the product end with more than 7 zeros due to the restrictions imposed by adding to 1003 while individually being divisible by high powers of 5. # Conclusion: The largest possible value of ( N ) is ( 7 ). [ boxed{7} ]

answer:1. The derivative of f(x) is f'(x) = e^x, so the slope k = f'(0) = 1. Therefore, the equation of the tangent line is y = x + 1. 2. Let g(x) = e^x - x - 1. Then, g'(x) = e^x - 1. Since g'(x) > 0 for x > 0 and g'(x) < 0 for x < 0, g(x) is decreasing on (-infty, 0) and increasing on (0, +infty). Therefore, at x = 0, g(x) reaches its minimum value, which is also the least value, i.e., g(x) geq g(0) = 0. Hence, e^x geq x + 1. The final answers are: 1. The equation of the tangent line is boxed{y = x + 1}. 2. The inequality is proven as boxed{e^x geq x + 1}.