Calculate (5 × (36) + 15 × (8) + 5 × (4 × (25) × 9 (50) Fast, fast
1、385.5
2、360
RELATED INFORMATIONS
- 1. 1. (- 2x-1) (3x-2) 2.2x (x-3) (1 / 2x + 2) 3. (a + 3) (A-1) + a (A-2) 4. (a square + 3) (A-2) - A (a square-2a-2)
- 2. How much is 1 / 3 plus 1 / 15 plus 1 / 35 plus 1 / 63 plus 1 / 99
- 3. 1-3n=-8 0.2x+12=0.3x+9 1/2x+1=1/3x-5
- 4. 666 * 778 + 333 * 444 (ingenious calculation)
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- 7. 1. If a > b, E > F, C > 0, then F-AC < e-bc
- 8. 2 meters = how many decimeters
- 9. Seven and a half minus five sixths
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- 11. The general solution to the x power of the differential equation y "+ 3Y + 2Y = e
- 12. For a division with remainder, the quotient is 14 and the remainder is 6. Given that the sum of the divisor and the divisor is 126, what are the divisor and the divisor respectively?
- 13. Given the function f (x) = x2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant and the range of a is obtained Can we use the method of variable separation?
- 14. What is 3 plus 5 divided by 2 times 4?
- 15. (a + b) 2 power * (B + a) 3 power fast
- 16. How to round? Take a few examples to see if 3561 is about 3600 or 4000 Give a few more examples and concepts
- 17. The function y = sin ^ 2 x-cosx + 3, X belongs to (2 π / 3, π / 6] and the range is
- 18. As shown in the figure, the side length of square ABCD is 2, point E is the midpoint of BC side, BG ⊥ AE is made through point B, the perpendicular foot is g, extend BG to intersect AC at point F, then CF=______ .
- 19. As shown in the figure, e is a point on the edge BC of rectangle ABCD, EF ⊥ AE, EF intersects AC respectively, CD intersects at points m, F, BG ⊥ AC, perpendicular foot is C, BG intersects AE at point h. (1) (1) Verification: △ Abe ∽ ECF; (2) Find the triangle similar to △ ABH and prove it; (3) If e is the midpoint of BC, BC = 2Ab, ab = 2, find the length of EM. BG ⊥ AC, perpendicular to g, wrong number How to solve without trigonometric function
- 20. As shown in the figure, e is a point on the edge BC of rectangle ABCD, EF ⊥ AE, EF intersects AC respectively, CD intersects at points m, F, BG ⊥ AC, perpendicular foot is C, BG intersects AE at point H (1) Verification: △ Abe ∽ ECF; (2) Find the triangle similar to △ ABH and prove it; (3) If e is the midpoint of BC, BC = 2Ab, ab = 2, find the length of EM BG ⊥ AC is g, wrong number