answer:Let the center of the circle be (x,0). We know the distance from the center to (2,5) and from the center to (3,4) must be the same. Using the distance formula, we have: [ sqrt{(x-2)^2 + (0-5)^2} = sqrt{(x-3)^2 + (0-4)^2} ] Simplifying the equations: [ sqrt{(x-2)^2 + 25} = sqrt{(x-3)^2 + 16} ] Squaring both sides to remove the square roots: [ (x-2)^2 + 25 = (x-3)^2 + 16 ] Expanding and simplifying: [ x^2 - 4x + 4 + 25 = x^2 - 6x + 9 + 16 ] [ x^2 - 4x + 29 = x^2 - 6x + 25 ] [ -4x + 29 = -6x + 25 ] [ 2x = 4 ] [ x = 2 ] The center of the circle is (2,0). The radius can be found using the distance from the center to one of the points: [ sqrt{(2-2)^2 + (0-5)^2} = sqrt{0 + 25} = sqrt{25} = boxed{5}. ]

answer:Start by solving the equation 2x - 7 = 8x - 1: 1. Subtract 2x from both sides: -7 = 6x - 1. 2. Add 1 to both sides: -6 = 6x. 3. Divide both sides by 6: x = -1. Now substitute x = -1 into the expression 5(x-3): 1. Substitute -1 for x: 5(-1-3) = 5(-4) = -20. The solution is boxed{-20}.

answer:Let's denote the number of students in grade 1 as G1, in grade 2 as G2, and in grade 5 as G5. According to the problem, the total number of students in grades 1 and 2 is 30 more than the total number of students in grades 2 and 5. This can be written as: G1 + G2 = G2 + G5 + 30 To find out how much lesser the number of students in grade 5 is compared to grade 1, we need to isolate G5 and G1 in the equation. First, we can simplify the equation by subtracting G2 from both sides: G1 = G5 + 30 Now, to find out how much lesser G5 is compared to G1, we subtract G5 from G1: G1 - G5 = 30 This means that the number of students in grade 5 is boxed{30} less than the number of students in grade 1.

answer:(I) Let d denote the common difference of the arithmetic sequence {a_n}, with a_3 = -5, and S_4 = -24. We have the following system of equations: begin{cases} a_1 + 2d = -5 & 4a_1 + 6d = -24 & end{cases} Solving the system, we obtain a_1 = -9 and d = 2. Therefore, the general term formula for the sequence is a_n = -9 + 2(n-1) = 2n - 11. (II) First, let's find when 2n - 11 > 0, which gives n > frac{11}{2}. So, a_n < 0 for n leq 5, and a_n > 0 for n geq 6. Now we can calculate T_{20}: begin{aligned} T_{20} &= |a_1| + |a_2| + |a_3| + ldots + |a_5| + |a_6| + |a_7| + ldots + |a_{19}| + |a_{20}| &= -a_1 - a_2 - a_3 - a_4 - a_5 + a_6 + a_7 + a_8 + ldots + a_{19} + a_{20} &= -2(a_1 + a_2 + a_3 + a_4 + a_5) + (a_1 + a_2 + ldots + a_{19} + a_{20}) &=-2left[5 times (-9) + frac{5 times 4}{2} times 2right] + left[20 times (-9) + frac{20 times 19}{2} times 2right] &= (-2) times (-25) + 200 &= boxed{250}. end{aligned}