answer:Let's first find the square of the number 8: 8^2 = 8 * 8 = 64 Now, we know that the relationship between the number 8 and its square results in 72. Since the square of 8 is 64, we need to determine what relationship or operation would take us from 64 to 72. 72 - 64 = 8 The relationship between the number 8 and its square is that you add the original number (8) to its square (64) to get 72. So, the relationship can be expressed as: Number's square + Number = Result 64 + 8 = boxed{72}

answer:To find the average of all the numbers between 18 and 57 that are divisible by 7 but not divisible by any other prime number except 7, we first need to identify those numbers. The numbers divisible by 7 between 18 and 57 are: 21, 28, 35, 42, 49, and 56. Now, we need to eliminate the numbers that are divisible by other prime numbers. The prime numbers less than 57 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53. - 21 is divisible by 3, so it's out. - 28 is divisible by 2, so it's out. - 35 is divisible by 5, so it's out. - 42 is divisible by 2 and 3, so it's out. - 49 is 7 squared, so it's only divisible by 7. - 56 is divisible by 2, so it's out. The only number that fits the criteria is 49. Since there's only one number, the average is the number itself, which is 49. Now, let's find the sum of the square roots of this number. Since there's only one number, the sum of the square roots is simply the square root of 49. The square root of 49 is 7. So, the sum of the square roots of the numbers that are divisible by 7 but not divisible by any other prime number except 7 between 18 and 57 is boxed{7} .

answer:(1) Let's denote the equation of circle C as x^2 + y^2 + Dx + Ey + F = 0. As the points A(6, 2) and B(4, 4) lie on circle C, the following system of equations must be satisfied: [ begin{cases} 36 + 4 + 6D + 2E + F = 0 16 + 16 + 4D + 4E + F = 0 end{cases} ] Moreover, since the origin O(0, 0) is also on circle C, we have: [ F = 0 ] Solving this system of equations, we find: [ begin{cases} 6D + 2E + 40 = 0 4D + 4E + 32 = 0 end{cases} ] Subtracting the first equation from the second one, we get: [ (4D + 4E + 32) - (6D + 2E + 40) = 0 ] which simplifies to: [ -2D + 2E = 8 ] Dividing through by 2 gives us: [ -D + E = 4 ] We also have the equation 4D + 4E = -32, which upon dividing through by 4 yields: [ D + E = -8 ] Using both equations -D + E = 4 and D + E = -8, adding them together eliminates D: [ 2E = -4 ] Thus: [ E = -2 ] Replacing E back into D + E = -8: [ D - 2 = -8 ] Hence: [ D = -6 ] Replacing D and E back into the original equation x^2 + y^2 + Dx + Ey + F = 0, we can now write the equation of circle C as: [ x^2 + y^2 - 6x - 2y = 0 ] Completing the square for both x and y terms to get the standard form of the circle: [ (x^2 - 6x + 9) + (y^2 - 2y + 1) = 9 + 1 ] The equation becomes: [ (x - 3)^2 + (y - 1)^2 = 10 ] which is the equation for circle C. (2) A line passing through point (2, 6) and intersecting C at a chord of length 4 can have two scenarios: either it is vertical, or it has a slope. For the vertical scenario, if the line is of the form x = a and intersects the circle creating a chord of length 4, we check the distance from (2, 6) to the line x - 3 = 0 which corresponds to our circle's center. As (2, 6) is already on the vertical line x = 2 and is 1 unit away from the circle's center (3, 1) in the x-coordinate, this line is indeed a solution. Therefore, one line is: [ x = 2 ] For the scenario with a slope, let's assume a line with the slope-intercept form y - 6 = k(x - 2), or equivalently: [ kx - y - 2k + 6 = 0 ] Since the distance from this line to the center of the circle (3, 1) must be half the chord length, which is 2, we can use the distance formula for a point to a line: [ text{Distance} = frac{|k(3) - 1 - 2k + 6|}{sqrt{k^2 + 1^2}} = frac{|3k + 5 - 2k|}{sqrt{k^2 + 1}} = 2 ] Solving for k: [ frac{|k + 5|}{sqrt{k^2 + 1}} = 2 ] Squaring both sides: [ (k + 5)^2 = 4(k^2 + 1) ] Expanding: [ k^2 + 10k + 25 = 4k^2 + 4 ] Bringing terms to one side: [ 3k^2 - 10k - 21 = 0 ] We can solve this quadratic equation for k. Factoring gives us: [ (k - 3)(3k + 7) = 0 ] From which we find two values for k: [ k = 3 quad text{or} quad k = -frac{7}{3} ] Considering the positive slope results in a chord longer than 4 when visualizing the circle and line placement, the feasible value for k is: [ k = -frac{7}{3} ] We'll use this value of k to write the equation of line l: [ y - 6 = -frac{7}{3}(x - 2) ] Multiplying through by 3 to clear the fraction: [ 3y - 18 = -7x + 14 ] Putting the equation in standard form: [ 7x + 3y - 32 = 0 ] Therefore, the two equations of the lines that satisfy the given conditions are: [ boxed{x = 2} quad text{and} quad boxed{7x + 3y - 32 = 0} ]

answer:Calculate the ratio of length to width for each package: begin{align*} text{X} &= frac{8}{5} = 1.6 text{Y} &= frac{12}{4} = 3 text{Z} &= frac{9}{9} = 1 text{W} &= frac{14}{5} = 2.8 end{align*} Next, compare each ratio to the range 1.5 to 3.0: - Package X: 1.6 (within range) - Package Y: 3 (on the boundary, typically considered within range in such problems) - Package Z: 1 (outside range) - Package W: 2.8 (within range) Count the packages outside the range: - Only Package Z is outside the range. Thus, the answer is boxed{1}.