answer:The set P={x|x=a^2+1, xin mathbb{R}}={x|xgeq 1}, and the set Q={x|y=log(2-x), xin mathbb{R}}={x|x<2}. Therefore, Pcap Q={x|xgeq 1}cap{x|x<2}=[1,2). Hence, the answer is boxed{[1,2)}.

answer:To find the image of 2i under the dilation centered at -1 + 4i with a scale factor of -2, we follow these steps: 1. Let z be the image of 2i under the dilation. The formula for dilation centered at a point c with scale factor k is given by: [z - c = k(w - c),] where w is the original point, and z is the image of w. 2. In this case, c = -1 + 4i, k = -2, and w = 2i. Substituting these values into the formula, we get: [z - (-1 + 4i) = -2(2i - (-1 + 4i)).] 3. Simplifying the right side of the equation: begin{align*} z + 1 - 4i &= -2(2i + 1 - 4i) &= -2(-3i + 1). end{align*} 4. Multiplying through by -2: begin{align*} z + 1 - 4i &= 6i - 2. end{align*} 5. Solving for z: begin{align*} z &= 6i - 2 - 1 + 4i &= -3 + 10i - 6i &= -3 + 8i. end{align*} Therefore, the image of 2i under the given dilation is boxed{-3 + 8i}.

answer:To find the smallest positive integer ( n ) such that no arithmetic progression of 1999 reals contains exactly ( n ) integers, we follow these steps: 1. **Arithmetic Progression Definition**: - Consider an arithmetic progression (AP) with terms given by ( a + kd ) where ( a ) is the first term, ( d ) is the common difference, and ( k ) is an integer. 2. **Setting Up the Problem**: - We want to examine a specific AP that includes 1999 terms: ( { a, a + d, a + 2d, ldots, a + 1998d } ). 3. **Integer Terms in AP**: - For an AP to contain integer terms, the common difference ( d ) must be such that the differences between consecutive terms are integer-spaced precisely under the constraint of reals involved. - For simplicity, let's set ( a = 0 ) (without loss of generality). 4. **Rational vs. Irrational Differences**: - Choose ( d = frac{1}{n} ) where ( n ) is not a divisor of 1999 since we want ( d neq frac{k}{1999} ) for any integer ( k ). 5. **Forming Sequences**: - Consider the sequence ( { frac{0}{n}, frac{1}{n}, frac{2}{n}, ldots, frac{1998}{n} } ). 6. **Calculations for Integer Entries**: - The number of integer entries in ( { frac{k}{n} : k = 0, 1, 2, ldots, 1998 } ) depends on the values of ( k ) such that ( frac{k}{n} ) is an integer. - The number of such integers ( k ) is given by (leftlfloor frac{1998}{n} rightrfloor + 1). 7. **Identifying Critical Points**: - To find ( n ) such that the gap between ( leftlfloor frac{1998}{n} rightrfloor ) and (leftlfloor frac{1998}{n+1} rightrfloor) is significant (3 or more), calculate (leftlfloor frac{1998}{n} rightlfloor) for various values. - Checking specific ( n ): - ( n = 31 rightarrow leftlfloor frac{1998}{31} rightrfloor = 64 ) - ( n = 30 rightarrow leftlfloor frac{1998}{30} rightrfloor = 66 ) - ( n = 29 rightarrow leftlfloor frac{1998}{29} rightrfloor = 68 ) - ( n = 28 rightarrow leftlfloor frac{1998}{28} rightrfloor = 71 ) 8. **Determine the Gap**: - The gap between integer counts for these ( n )'s influences whether they can contain precisely 70 integers. - Increment ( n ) until the gap exceeds 70. 9. **Final Verification**: - At ( n = 29 ): the set ( left{ frac{k}{29} right} ) has 68 integers. - At ( n = 28 ): the set ( left{ frac{k}{28} right} ) has 71 integers. - Thereby, we observe the absence of exactly 70 integers between ( n = 30 ) and ( n = 29 ). # Conclusion: Thus, the smallest positive integer ( n ) such that no arithmetic progression of 1999 reals contains exactly ( n ) integers is: [ boxed{70} ]

answer:Given the equation sqrt{x + 12} = 10, we want to find the value of x that satisfies this condition. Following the steps: 1. **Square both sides** to eliminate the square root: [ (sqrt{x + 12})^2 = 10^2 ] [ x + 12 = 100 ] 2. **Solve for x**: [ x = 100 - 12 ] [ x = 88 ] Thus, the solution to the equation sqrt{x + 12} = 10 is x = boxed{88}.