answer:Since overrightarrow{a}=(m,2) and overrightarrow{b}=(2,-1) are perpendicular to each other (overrightarrow{a} perp overrightarrow{b}), we have: overrightarrow{a}cdotoverrightarrow{b} = 2m - 2 = 0 Solving this equation for m gives m=1. Thus, the vector overrightarrow{a} is: overrightarrow{a}=(1,2) Calculating 2overrightarrow{a}-overrightarrow{b} we get: 2overrightarrow{a}-overrightarrow{b} = 2(1,2)-(2,-1) = (0,5) Then, we find overrightarrow{a}+overrightarrow{b} as: overrightarrow{a}+overrightarrow{b} = (1,2) + (2,-1) = (3,1) Now we can determine |,2overrightarrow{a}-overrightarrow{b},|, which is the magnitude of the vector (0,5): |,2overrightarrow{a}-overrightarrow{b},| = sqrt{0^2 + 5^2} = 5 The dot product of overrightarrow{a} with (overrightarrow{a}+overrightarrow{b}) is: overrightarrow{a} cdot (overrightarrow{a}+overrightarrow{b}) = 1 cdot 3 + 2 cdot 1 = 5 Finally, we have: frac{|,2overrightarrow{a}-overrightarrow{b},|}{overrightarrow{a} cdot (overrightarrow{a}+overrightarrow{b})} = frac{5}{5} = boxed{1} Therefore, the correct answer is: B: 1.

answer:1. Given: - Point ( A(0,3) ) - Line ( L: y = 2x - 4 ) - Circle ( C ) with radius ( 1 ) and center on line ( L ) - Point ( M ) on circle ( C ) such that ( |MA| = 2|MD| ) 2. Let the center of the circle ( C ) be denoted by ( (a, 2a - 4) ) because the center lies on the line ( L ). 3. We must analyze the condition ( |MA| = 2|MD| ). 4. Consider points ( B(0, -3) ), ( D(0, 1) ), and ( E(0, -1) ): - These points satisfy the property such that ( |MD| ) is the perpendicular distance of ( M ) from the y-axis. - Since ( |MA| = 2|MD| ), point ( M ) must lie on circle ( E ), centered between points ( B ) and ( D ). 5. For the circle ( E ) centered at ( (0, -1) ) and radius ( 2 ) (since the distance between ( B ) and ( D ) is ( 4 )), ( |MD| ) must be such that ( 1 leq |EC| leq 3 ). 6. Using the coordinates of ( C(a, 2a - 4) ) and considering the center condition: - The distance from the origin ((0, -1)) to ((a, 2a - 4)) must satisfy ( 1 leqslant |EC| leqslant 3 ). 7. Distance ( |EC| = sqrt{a^2 + (2a - 3)^2} ): [ sqrt{a^2 + (2a - 3)^2} ] Simplify ( (2a-3)^2 ): [ (2a - 3)^2 = 4a^2 - 12a + 9 ] 8. Thus, [ |EC| = sqrt{a^2 + 4a^2 - 12a + 9} = sqrt{5a^2 - 12a + 9} ] We need ( 1 leq sqrt{5a^2 - 12a + 9} leq 3 ). 9. Square both sides: [ 1 leq 5a^2 - 12a + 9 leq 9 ] 10. We analyze both inequalities separately: - ( 5a^2 - 12a + 9 geq 1 ): [ 5a^2 - 12a + 8 geq 0 ] This inequality holds true for ( a leq 0 ) or ( a geq frac{8}{5} ). - ( 5a^2 - 12a leq 0 ): [ 5a(a - frac{12}{5}) leq 0 ] This inequality holds true for ( 0 leq a leq frac{12}{5} ). 11. Combining both conditions: [ 0 leq a leq frac{12}{5} ] This represents the range for ( a ). # Conclusion: [ boxed{left [0, frac{12}{5} right ]} ]

answer:To find 75^{-1} pmod{76}, observe that the relationship 75 equiv -1 pmod{76} simplifies the calculation. Squaring both sides of the equivalence: [ 75^2 equiv (-1)^2 equiv 1 pmod{76} ] This equivalency establishes that 75^2 equiv 1 pmod{76}, indicating that multiplying 75 by itself gives a product congruent to 1 modulo 76. Thus, 75 is its own inverse under modulo 76. Therefore, [ 75^{-1} equiv boxed{75} pmod{76} ]

answer:1. **Identify the centers and radii of the circles:** Let (O_1) and (O_2) be the centers of the circles with radii (O_1A_1) and (O_2A_2) respectively. 2. **Draw perpendiculars from (O_1) and (O_2) to the line (O_1O_2):** Erect perpendiculars from the points (O_1) and (O_2) to the line segment connecting these points, (O_1O_2). 3. **Mark equal segments on the perpendicular lines:** [ text{On the perpendicular from } O_1, text{ mark a point } B_1 text{ such that } O_1B_1 = O_2A_2. ] [ text{On the perpendicular from } O_2, text{ mark a point } B_2 text{ such that } O_2B_2 = O_1A_1. ] 4. **Construct the midpoint (M) of the segment (B_1B_2):** [ M = text{Midpoint of the segment } B_1B_2. ] 5. **Erect a perpendicular at point (M) to (B_1B_2):** At the midpoint (M), erect a perpendicular to the segment (B_1B_2). This perpendicular will intersect the line (O_1O_2) at point (N). 6. **Prove that (N) lies on the radical axis:** To verify, note the distance relations based on the circles' definitions: [ O_1N^2 + O_1B_1^2 = O_2N^2 + O_2B_2^2. ] Since (O_1B_1 = O_2A_2) and (O_2B_2 = O_1A_1), rewrite as: [ O_1N^2 + O_2A_2^2 = O_2N^2 + O_1A_1^2. ] Rearrange this equation to identify that [ O_1N^2 - O_1A_1^2 = O_2N^2 - O_2A_2^2. ] This indicates that point (N) satisfies the radical axis equation, and therefore (N) lies on the radical axis. 7. **Construct the radical axis:** Finally, erect a perpendicular to the line (O_1O_2) at point (N). This line is the radical axis of the two circles. Conclusion: [ boxed{} ]