In order to indicate that "a and B are not equal to 0", the C language expression that should be used is () A）（a!=0） || （b!=0） B）a || b C）!（a=0）&&（b!=0） D）a && b But I can't see what's wrong with C. can you point it out,

In order to indicate that "a and B are not equal to 0", the C language expression that should be used is () A）（a!=0） || （b!=0） B）a || b C）!（a=0）&&（b!=0） D）a && b But I can't see what's wrong with C. can you point it out,

C. ! (a = 0) is to take the result of a = 0 as not. When a is not equal to 0, the result of a = 0 should be 0. Add a! In front of it to indicate not 0
B! = 0 doesn't need to be parsed, does it?

Proof: for any natural number n, (n + 5) - (n + 2) (n + 3) must be divisible by 6
Speed as the title

(n + 5) - (n + 2) (n + 3) = 6 N is meaningless here. It should be that "n * (n + 5) - (n-3) * (n + 2)" can be divisible by 6... N * (n + 5) - (n-3) * (n + 2) = n ^ 2 + 5N - (n ^ 2-n-6) = 6N + 6, so it can be divisible by 6

（101+102+103+…… +399）—（91+92+93+…… +389）

（101+102+103+…… +399）-
（91+92+93+…… +389）
=10+10+…… +10+10
=10*（399-101+1）
=2990

When two cylinders with the same bottom surface are combined into a one meter long cylinder, the surface area is reduced by 56.52 square decimeters, and the surface area of the combined cylinder is () square meters

When two cylinders with the same bottom surface are combined into a one meter long cylinder, the surface area of the cylinder is reduced by 56.52 square decimeters. The surface area of the cylinder is () square meters. Two bottom areas: 56.52 square decimeters = 0.5652 square meters. The perimeter of the bottom surface: (radius square: 0.5652 △ 2 △ 3.14 = 0.09, radius: 0.3m) 0.3 × 2

80% x-1.2x12 = 1.6 to solve the equation

0.8x=1.6+1.44
0.8x=3.04
=38

Known quadrilateral is a square, the area of blank triangle is 56 square centimeter, ED length is 7 centimeter

56 × 2 △ 7, = 112 △ 7, = 16 (CM). 16 × 16-56, = 256-56, = 200 (cm 2). Answer: the area of shadow is 200 cm 2

125 * 48, 26 * 25 * 12 how can these be easily calculated

125*48
=125*8*6
=1000*6
=6000
26*25*12
=26*3*4*25
=78*100
=7800

As shown in the figure, in △ ABC, D is the midpoint of AC, e is a point of the extension line of segment BC, the parallel line passing through point a as be intersects with the extension line of segment ed at point F, connecting AE and cf. (1) verification: AF = CE; (2) if AC = EF, and ∠ ACB = 135 °, try to judge what kind of quadrilateral afce is, and prove your conclusion

(1) It is proved that: ∵ AF ‖ EC, ∵ DFA = ∵ Dec, ∵ DAF = ∵ DCE, ∵ D is the midpoint of AC, ∵ Da = DC, ≌ DAF ≌ DCE, ≌ AF = CE; (2) quadrilateral afce is a square. The reasons are as follows: