21. Let int variables m, N, a, B, C and d all be 0. After executing the expression (M = a = = b) | (n = C = = D), what are the values of M and N?

21. Let int variables m, N, a, B, C and d all be 0. After executing the expression (M = a = = b) | (n = C = = D), what are the values of M and N?

Non zero (- 1)

Ask the master to answer the detailed answer process of 4 + 10 / 6 * 7 / 8mod9 formula in VB
I calculated 6 * 7 = 42
42/8=5.25
10\5.25=1
1mod9=1
4+1=5
But the answer is 6, where am I wrong?
However, \\\\\\\\\\\\\\\\

10 / 6 * 7 / 8 is 1. XX VB rounds it to 2
2 mod 9 is 2
4+2=6

(- 0.5) + three and a quarter + 2.75 + (- five and a half) =

(- 0.5) + three and a quarter + 2.75 + (- five and a half)
=-0.5+3.25+2.75-5.5
=-(0.5+5.5)+(3.25+2.75)
=-6+6
=0

(5 / 8 + 1 / 25) * 17 / 8 + 25 is calculated by simple method

(5 / 8 + 1 / 25) * 8 + 17 / 25
=5/8*8+1/25*8+17/25
=5+(8/25+17/25)
=6

Factorization 1, X & # 178; - (X-Y) &# 178; 2, X & # 178; - ax + ay-y & # 178;
Two or three steps

1)
Original formula = [x + (X-Y)] [x - (X-Y)]
=(2x-y)y
=y(2x-y)
2)
The original formula = x & # 178; - Y & # 178; - ax + ay
=(x+y)(x-y)-a(x-y)
=(x-y)(x+y-a)

What's the product of subtracting 0.75 from 23, dividing by 3 / 4 quotient and multiplying by 5 / 33

（23-0.75÷3／4）×5／33=10／3

A 56cm long wire can be used to weld a 7cm, 3cm wide and () cm high rectangular teaching aid

Because a cuboid is surrounded by four long sides, four wide sides and four heights, all 68 cm or 56 cm are the sum of the 12 edges of the cuboid
1、 (68-10 ×4-5×4)÷4
=(68-40-20)÷4
=8÷4
=2 (CM)
2、（56-7×4-3×4）÷4
=16÷4
=4 (CM)

Arrange the remainder of 1, 2, 3, 4,... Divided by 3 to get a sequence. What is the sum of the first 100 numbers in the sequence?

1. The remainder of 2, 3, 4,... Divided by 3 is 1,2,0,1,2,0
Law: three numbers, one cycle: 1,2,0
And 100 / 3 = 33... 1,
So the 100th remainder is 1
So the sum of the first 100 numbers in this sequence is 33 * (1 + 2 + 0) + 1 = 100

It is known that the first n terms of sequence {an} and Sn = - n ^ 2 + 9N + 2, n belongs to n*
(1) Judge whether {an} is arithmetic sequence
(2) Let RN = | A1 | + | A2 | + +|An |, find RN
(3) Let BN = 1 / [n (12-an)], n belong to n *, TN = B1 + B2 + +BN, whether there is a minimum natural number N0, such that the inequality TN

(1) When n = 1, A1 = 10, and when n > = 2, s (n-1) = - (n-1) ^ 2 + 9 (n-1) + 2
An = SN-S (n-1) = - 2n + 10 does not hold for n = 1
So an is not an arithmetic sequence
(2) When n = 5, an = 0
Rn=S(4)-[Sn-S(4)]=-Sn+2S(4)=n^2-9n-42
(3) When n = 1, B1 = 1 / 2, when n > = 2, BN = 1 / [2n (n + 1)] = 1 / 2 [1 / n-1 / (n + 1)]
When n > = 2, so TN = 1 / 2 + 1 / 2 [(1 / 2-1 / 3) + (1 / 3-1 / 4) + + (1 / n-1 / (n + 1)] = 3 / 4-1 / (2n + 2)
There is always TN

One third is equal to the cycle of zero point three, two thirds is equal to the cycle of zero point six, but why is not three thirds equal to the cycle of zero point nine, but one?

Three thirds of a cycle is equal to zero nine
And the zero nine cycle is one
You can understand this after you learn the limit