If you want to use the linear equation of one variable, you can't learn the linear equation of two variables 14. Xiao Ming's father is riding a motorcycle with Xiao Ming on the road at a constant speed. The following is the mileage Xiao Ming sees every other hour. Can you be sure that Xiao Ming's 12:00 is the milestone he sees? 12: 00 is a two digit number, and the sum of its two digits is 7 13: The ten digit and the one digit numbers at 12:00 are reversed 14: There's a zero in the middle of the two digits you see at 12:00

If you want to use the linear equation of one variable, you can't learn the linear equation of two variables 14. Xiao Ming's father is riding a motorcycle with Xiao Ming on the road at a constant speed. The following is the mileage Xiao Ming sees every other hour. Can you be sure that Xiao Ming's 12:00 is the milestone he sees? 12: 00 is a two digit number, and the sum of its two digits is 7 13: The ten digit and the one digit numbers at 12:00 are reversed 14: There's a zero in the middle of the two digits you see at 12:00


16
Let the two numbers seen at twelve o'clock be x 7-x
The 12 o'clock mileage is 10x + 7-x
The mileage at thirteen is 10 (7-x) + X
The mileage at 14 o'clock is 100x + 7-x
Because it is driving at a constant speed, the mileage of 13:00 - 12:00 = 14:00 - 13:00
Just work it out
It's pretty easy, right



LIM (x tends to 0) f (x) / x = 1, f '' (x) > 0. It is proved that f (x) is greater than or equal to X


Let g (x) = f (x) - x, G '(x) = f' (x) - 1, G "(x) = f (" x)
When G '(x) = 0, f' (x) = 1 = > x = 0, because limx = > 0 [f (x) - f (0)] / (x-0) = 1, G (x) = 0
G "(x) = f (" x) > 0, G '(x) increases monotonically, so x = 0 is the unique solution of G' (x) = 0
g(x) = f(x)-x >=0



Solving an equation ﹥ 8203; 3x = 8 √ 3 + √ 3x


A:
3x=8√3+√3x
Divide both sides by √ 3 to get: √ 3x = 8 + X
Transfer: √ 3x-x = 8
(√3-1)x=8
Multiply both sides by √ 3 + 1:
(√3+1)(√3-1)x=8(√3+1)
2x=8(√3+1)
x=4(√3+1)
x=4√3+4



8+5+2+6+7+8+9+0+1+555-5+5+5-5+5-5+5-5=


It's 601,



lim(n^3+n)/(n^4-3n^2+1)


lim(n^3+n)/(n^4-3n^2+1)
=lim(1/n+1/n^3)/(1-3/n^2+1/n^4)
=(0+0)/(1-0+0)
=0



What is the concept of exponential function?


The general form of exponential function is y = a ^ x (a > 0 and ≠ 1) (x ∈ R). It is one of the elementary functions. It is a monotone, lower convex and differentiable positive function without upper bound defined on the field of real number
When the function applied to the value E is written as exp (x), it can also be written as e, where e is a mathematical constant, which is the base of natural logarithm, approximately equal to 2.718281828, also known as Euler number



Calculate 1.48 ° 33 ′ 28 ″ + 19 ° 56 ′ 42 ″ 2.90 ° - 15 ° 43 ′ 29 ″ 3.32 ° 15 ′ 29 ″ × 44.79 ° 40 ′ 18 ″ 3


1,22o58,58,2,74o16,31,3129o1,56,4,26o33,26. Welcome to inquire or adopt



Find the cosine value of the angle θ between the two tangents at the intersection of the curves f (x) = x ^ 2 + 1 and G (x) = x ^ 3 + X


First of all, the derivation of two functions is: F '(x) = 2x G' (x) = 3x ^ 2 + 1, then the slope of intersection (1,2) passing through (1,2) is 2 and 4 respectively. Let the inclination angle of two tangents be a and B respectively, then the tangent value of the included angle is tan θ = Tan (B-A) = | (TGB TGA) / (1 + TGA * TGB) | = | (2-4) / (1 + 4 * 2) | = 2 / 9, so cos θ =



If a is less than 0, then what is the solution set of the inequality system: X is greater than - 2A; X is greater than - 0.5A


a-2a
x>-0.5a
-2a>-0.5a
So the solution set is x > - 2A



Why do Chinese primary school mathematics textbooks still use vertical multi digit multiplication? I think Arab's "paving the floor brocade" algorithm is better and not easy to make mistakes
Vertical multiplication is actually the evolution of abacus in China. However, this kind of algorithm needs to be memorized in mind, which is obviously prone to mistakes
The algorithm of paving brocade invented by Arabs has the same amount of calculation, but it is not easy to make mistakes


This is a kind of educational habit. Every country has its own educational model. We can't say that any calculation method is the best. For example, there are speed calculation classes everywhere, so it's impossible for schools to use this method to teach calculation. For example, when you say "shop floor brocade", it involves calculation theory and algorithm. The whole textbook needs to be moved, and teachers need to learn it again, There's no need to let everyone give up the correct and not very complicated method, and learn a new unfamiliar method to solve the same problem. Of course, it's OK for individuals to be willing to learn and try, but if all the staff move, LZ can imagine the difficulty