answer:First, we transform the given circle equation x^{2}+y^{2}-4x+2y-4=0 into its standard form. To do that, we complete the squares for x and y: 1. For the x terms: Take half of -4, square it, and add/subtract it inside the parentheses: (frac{-4}{2})^{2}=4. Add and subtract 4 to get: (x^{2}-4x+4)-4+(y^{2}+2y-4)=0 2. For the y terms: Take half of 2, square it, and add/subtract it inside the parentheses: (frac{2}{2})^{2}=1. Add and subtract 1 to get: (x^{2}-4x+4)+(y^{2}+2y+1)-4-1=0 Now, we can rewrite the equation in the standard form: (x-2)^{2}+(y+1)^{2}=9 We know that the standard form of a circle's equation is given by: (x-h)^{2}+(y-k)^{2}=r^{2} where (h,k) represents the center of the circle and r is its radius. Comparing our equation with the standard form, we notice that the center of the circle is at (2,-1), and its radius r is equal to sqrt{9}=3. Therefore, the radius of the circle is 3. The answer is: boxed{A}

answer:Solution: (I) Let the common difference of {a_n} be d, and the common ratio of {b_n} be q, then according to the problem, we have q > 0 and begin{cases} 1+2d+q^4=21 1+4d+q^2=13 end{cases} Solving these, we get d=2, q=2. Therefore, a_n=1+(n-1)d=2n-1, b_n=q^{n-1}=2^{n-1}. (II) frac{a_n}{b_n}= frac{2n-1}{2^{n-1}}, S_n=1+ frac{3}{2^1}+ frac{5}{2^2}+ldots+ frac{2n-3}{2^{n-2}}+ frac{2n-1}{2^{n-1}}, frac{1}{2}S_n= frac{1}{2}+ frac{3}{2^2}+ frac{5}{2^3}+ldots+ frac{2n-3}{2^{n-1}}+ frac{2n-1}{2^n}, Subtracting the second equation from the first, we get frac{1}{2}S_n=1+2left( frac{1}{2}+ frac{1}{2^2}+ldots+ frac{1}{2^{n-1}}right)- frac{2n-1}{2^n}, Thus, S_n=2+2+ frac{2}{2}+ frac{2}{2^2}+ldots+ frac{2}{2^{n-2}}- frac{2n-1}{2^{n-1}}=2+2timesleft(1+ frac{1}{2}+ frac{1}{2^2}+ldots+ frac{1}{2^{n-2}}right)- frac{2n-1}{2^{n-1}}=2+2times frac{1- frac{1}{2^{n-1}}}{1- frac{1}{2}}- frac{2n-1}{2^{n-1}}=6- frac{2n+3}{2^{n-1}}. Therefore, the sum of the first n terms S_n of the sequence left{ frac{a_n}{b_n} right} is boxed{6- frac{2n+3}{2^{n-1}}}.

answer:To factor 45y^2 + by + 45, write it as: [ 45y^2 + by + 45 = (cy + d)(ey + f) ] where c, d, e, f are integers. 1. **Equating Coefficients**: Upon expanding (cy + d)(ey + f), we get: [ ce y^2 + (cf + de) y + df = 45y^2 + by + 45 ] From this, equate: - ce = 45 - cf + de = b - df = 45 2. **Finding Possible Values for c, d, e, f**: The pairs (c, e) and (d, f) must multiply to 45, so considering the divisors of 45 we have: - (1, 45), (3, 15), (5, 9), (9, 5), (15, 3), (45, 1), and their negatives. 3. **Checking Parity**: Since 45 is odd, both c, e and d, f must be odd to preserve odd products. This implies cf + de = b must be even, due to the sum/product pattern of odd and even numbers. 4. **Example Calculation**: Selecting pairs (c, e) = (5, 9), (d, f) = (5, 9): [ cf + de = 5 times 9 + 9 times 5 = 45 + 45 = 90 ] So, here b = 90 is even. Similarly, evaluating other combinations will yield specific even values for b. 5. **Conclusion**: The integer b must be even for the expression to properly factor into the required form with integer coefficients. Final conclusion: [ btext{ must be some even number} ] The final answer is boxed{D) some even number}

answer:Let the common ratio of the geometric sequence be q. Thus, we can express a_4 and a_7 as follows: a_4 = 16q^3 quad text{and} quad a_7 = 16q^6. Since the arithmetic mean of a_4 and a_7 is frac{9}{8}, we have: frac{a_4 + a_7}{2} = frac{9}{8}, which simplifies to: a_4 + a_7 = frac{9}{4}. Substituting the expressions for a_4 and a_7, we get: 16q^3 + 16q^6 = frac{9}{4}. Dividing both sides by 16, we obtain: q^3 + q^6 = frac{9}{64}. Factoring out q^3, we get: q^3(1 + q^3) = frac{9}{64}. Let x = q^3. We have a quadratic equation: x + x^2 = frac{9}{64}. Setting the equation to zero, we have: x^2 + x - frac{9}{64} = 0. Solving this quadratic equation, we discard the negative root (because the sequence has positive terms), giving us x = q^3 = frac{1}{2}. Therefore, the common ratio q is: q = sqrt[3]{frac{1}{2}} = frac{1}{2}. Now, using the formula for the sum of the first n terms of a geometric sequence, we have: S_5 = frac{a_1(1 - q^5)}{1 - q}. Substituting a_1 = 16 and q = frac{1}{2}: S_5 = frac{16(1 - (frac{1}{2})^5)}{1 - frac{1}{2}}. Calculating the sum: S_5 = 16 times frac{1 - frac{1}{32}}{frac{1}{2}} = 16 times frac{frac{31}{32}}{frac{1}{2}} = 16 times 2 times frac{31}{32} = 16 times frac{31}{16} = 31. Thus, the value of S_5 is boxed{31}.