Change - 18 π / 7 to 2K π + α (0 ≤ α

RELATED INFORMATIONS

• 1. The form of transforming - 1125 ° into α + 2K π (0 ≤ α & lt; 2 π, K ∈ z =) is ()
• 2. The following angles are reduced to 2K π + a (0
• 3. The - 1125 ° is reduced to 2K π + α (0 ≤ α < 2 π, K ∈ z)
• 4. The following angles are reduced to 0 to 2 π plus 2K π (K ∈ z) (1) -25/6π (2)-5π (3)-45° (4)400°
• 5. The form of - 1125 ° to 2K π + α (K ∈ Z, 0 ≤ α < 2 π) is () A. −6π−π4B. −6π+7π4C. −8π−π4D. −8π+7π4
• 6. 16 π of the angle is reduced to α + 2K π (K ∈ Z 0)
• 7. What is the temperature of 4.2K
• 8. Given that point a (0, - 1) and point B are moving points on the curve X ^ 2-y ^ 2 + X + 2Y + 3 = 0, then the trajectory equation of the midpoint of line AB is obtained
• 9. It is known that 2x + y = 7 is a quadratic equation of X and y, and the solutions of the equations are positive integers
• 10. If we keep fixed points F1 (- 2,0) and F2 (2,0), then the locus of the moving point P satisfying the following conditions in the plane is hyperbolic A./PF1/-/PF2/=_ +3 B./PF1/-/PF2/=_ +4 C./PF1/-/PF2/=_ +5 D./PF1/^2-/PF2/^2=_ +4 Note: Pf1 / medium / is absolute value_ +3 is positive and negative
• 11. -Square of a · (- a) · (- a) · (- a) · (- a) · (- a) · (- a) · 2K + 1
• 12. Calculation: - (a-b) of 2K + 1 power × (B-A) of 2K power × (a-b) of 2k-1 power (k is a positive integer) rt
• 13. What is the negative (2k + 1) power of 2 minus the negative (2k-1) power of 2 plus the negative 2K power of 2?
• 14. Simplification: - (a-b) to the power of 2K + 1 * (B-A) to the power of 2K * (a-b) to the power of 2k-3. (k is a positive integer) With a multiply sign, write as clearly as a question
• 15. An exciting Let a = {x | x = 2K, K belongs to Z (integer set)} Let a = {x | x = 2K, K belong to Z (integer set)}, B = {x | x = 2K + 1, K belong to Z}, C = {x | x = 2 (K + 1), K belong to Z}, d = {x | x = 2k-1, K belong to Z}, in a, B, C, D, which sets are equal, which sets are empty, which sets are union Z? Which friend is on the first floor?
• 16. Given M1 = {m | M = x square - y square, X is an integer, y is an integer}, M2 = {m | M = 2K + 1 or M = 4K, K is an integer} to prove M 1 = m 2
• 17. Given a = {x | x = (2k + 1) π, K ∈ Z},}, B = {x | x = (4k-1) π, K ∈ Z}, then the relation between a and B is
• 18. Set M = {XIX = 2K + 1, K ∈ Z}, n = {XIX = 4K ± 1, K ∈ Z} How to prove n ∈ m? It is said in the book that n ∈ M can be explained in detail because m is an odd number set
• 19. Let m = {x = k π K ∈ Z} n = {x | x = 2K π + - π, K ∈ Z}, then the relation between M and N is When m = 0, n is not equal to 0. Why m = n
• 20. We know that M is a set {1, 2, 3 , 2k-1} (K ∈ n *, K ≥ 2), and when x ∈ m, there is 2k-x ∈ M. note that the number of sets m satisfying the condition is f (k), then f (2)=______ ；f（k）=______ ．