How does Mathematica implement cross computing? How to realize cross computing? For example: Input: S1 = {1,2}, {3,4}, {5,6,7}; s2={{4,7},{24,36}}; Output: S3 = {24,28,30,36} Note: S1 is the multiplication table, S2 is the interval table. The following "multiplication" means that any two elements in the two tables are multiplied Firstly, take {1,2} and {3,4} in S1 as "multiplication" equal to {3,4,6,8}, and take the first interval value {4,7} in S2 to get {4,6} Multiply {4,6} and {5,6,7} to get {20,24,28,30,36,42}, take the second interval value {24,36} in S2 to get {24,28,30,36}, and so on

"Multiply {4,6} by {5,6,7}" to get {20,24,28,30,36,42} "
Times @@@ Tuples@{{4,6},{5,6,7}}
Take the second interval value {24,36} in S2 to get {24,28,30,36}
Select[%,IntervalMemberQ[Interval@{24,36},#] &]

Why can't we figure out what's wrong in Mathematica
DSolve[3 == v[x]v'[x] + [v[x]]^2/x,v[x],x]

Dsolve [3 = = v [x] V '[x] + (v [x]) ^ 2 / x, v [x], x], that's right

1. On tree planting day, the school plans to plant 240 trees, and the teacher has planted 100 trees. The remaining trees are planted by the students of grade 6, grade 5 and grade 4, and the students of grade 3
The ratio of the number of species is 5:4:1. How many trees did the students of the three grades grow?

240-100 = 140
5+4+1=10
6： 140 × 5 / 10 = 70 trees
5： 140 × 4 / 10 = 56 trees
4： 140 × 1 / 10 = 14 trees

Given a ^ 2 + A-1 = 0, find the value of the algebraic formula A ^ 5-5a + 3

a^5-5a+3
=a³﹙a²＋a-1﹚－a^4＋a³-5a+3
=a³﹙a²＋a-1﹚－a²﹙a²＋a-1﹚＋2a³-a²-5a+3
=2a﹙a²＋a-1﹚-3a²-3a+3
=-3﹙a²＋a-1﹚
=0

How many bottles of drinks can you drink for 20 yuan?

For 20 yuan, you can buy 10 bottles first. Among 10 bottles, there are 10 caps and 10 empty bottles. 1) if 10 caps are replaced by 5 bottles, there are 5 caps and 5 empty bottles (15 empty bottles in total). A) if 5 caps are replaced by 2 bottles, the remaining 1 cap will become 2 caps (3 caps in total) and 2 empty bottles (17 empty bottles in total). B) 3 bottles

Decomposition factor: 1 A ^ 2 / 4 + 1 AB / 3 + 1 B ^ 2 / 9

1A ^ 2 / 4 + 1ab / 3 + 1b ^ 2 / 9
=1/36(9a^2+12ab+4b^2)
=1/36(3a+2b)^2

Passenger cars and freight cars run 80 kilometers per hour and freight cars 68 kilometers per hour from the two places at the same time. The two cars meet at 30 kilometers from the midpoint. How many kilometers is the distance between the two places?

A: the distance between a and B is 740 km

Given that x = 2 is the solution of the equation 3-ax / 2 = 2 / 3 + 2A (a is a known number), find the value of A

Substitute x = 2 into the equation
3-2a/2=2/3+2a
3-a=2/3+2a
3a=7/3
a=7/9

When they set out, their speed ratio was 3:2. After their first meeting, a's speed increased by 20%, and B's speed increased by 30%. In this way, when a arrives at B, B is still 14 kilometers away from a, so how many kilometers is the distance between a and B?

Suppose the distance between a and B is SKM, and the speed of a and B is 3x and 2x respectively. When a and B meet for the first time, the distance they take is 3s5 = 0.6skm and 2s5 = 0.4skm respectively. According to the time series equation of a to B after meeting: 0.4s3x (1 + 20%) = 0.6s − 142x (1 + 30%), s = 45km. A: the distance between a and B is 45km

Solve the equation 2x = X-1 / 3 = 3-x + 5 / 5. Thank you

2x=x－1／3＝3－x＋5／5 2x=2/3=3-2x 4x=2/3 x=1/6

Why can't we figure out what's wrong in Mathematica DSolve[3 == v[x]v'[x] + [v[x]]^2/x,v[x],x]