How to know what multiples are in a certain range? For example, how many multiples of 3 are in 100? It's better to be regular What about 4, 5, 6, 7 and so on? The rules are similar to those on the third floor

How to know what multiples are in a certain range? For example, how many multiples of 3 are in 100? It's better to be regular What about 4, 5, 6, 7 and so on? The rules are similar to those on the third floor

Suppose that all natural numbers are a, B, N, and there are m multiples of N in the range of [a, b]. The minimum value of M is: (B-A) / n-1, and the maximum value is: (B-A) / N + 1, taking integer

How many of the numbers from 1 to 100 are not multiples of 5 or 7?

There are 20 multiples of 5
The multiple of 7 is 100 △ 7 = 14 2. There are 14
[5,7] = multiple of 35 100 △ 35 = 2 30, there are two
Therefore, the numbers that meet the requirements of the topic are as follows:
100-20-14 + 2 = 68

Among the numbers 1-100, how many are not multiples of 5 or 7
This is a math problem. It's due in 30 minutes

100-20-14+2=68

Given the function f (x) = e ^ ax * (A / x + A + 1), where a is greater than or equal to - 1, find the monotone interval of F (x)

When a = 0, f = 1, there is no monotone interval. When a = - 1, f = - 1 / Xe ^ x, let the derivative be greater than 0, x greater than - 1, let the derivative be less than 0, X less than - 1. When a is greater than 0, find the derivative f = AE ^ ax (a ^ 2 / x + A ^ 2 + A-A / x ^ 2), let the equation = 0, get X1 = - 1, X2 = 1 / (a + 1), let the derivative function be greater than 0, get x greater than 1 / (a + 1) or less than - 1, let X be less than 0, get x belonging to (- 1,1 / (a + 1)), In the same way, when a is greater than - 1 and less than 0, let the derivative be greater than zero, then x belongs to
(- 1,1 / (a + 1)), the derivative is less than 0, the solution is x greater than 1 / (a + 1) or X less than - 1
To sum up, when a is greater than 0, the monotone increasing interval is (1 / (a + 1), positive infinity and (negative infinity, - 1), monotone decreasing interval is (- 1,0) and (0,1 / (a + 1)); when a is less than 0, it is greater than - 1, monotone increasing interval (- 1,0) and (0,1 / (a + 1)), monotone decreasing interval is (1 / (a + 1), positive infinity and (negative infinity, - 1); when a = 0, there is no monotone interval, a = - 1, monotone increasing interval is (- 1,0) and (0,1), The monotone decreasing interval is (negative infinity, - 1)

Given the function f (x) = x ^ 2-ax + A / e ^ x, where e is about 2.7128, find the monotone interval of the function

When f (x) = (x ^ 2-ax + a) / e ^ x (x ∈ R) f '(x) = [(2x-a) e ^ X - (x ^ 2-ax + a) e ^ x] / e ^ (2x) = - [x ^ 2 - (a + 2) x + 2A] / e ^ x = - (x-a) (X-2) / e ^ x a = 2, f' (x) = - (X-2) & # 178 / e ^ x ≤ 0, f '(x) > 0 = = = > 20 = = > A, f (x) = - ∞, + ∞ a > 2

The derivative y'of implicit function determined by ysinx + LNY = 1
The answer in the book is y'sinx + ycosx + 1 / y times y'sinx ycosx and 1 / y times y '?

For example, the derivative y of ysinx can be regarded as a function of X
This follows the rule of multiplication and derivation
Y is y'sinx, and the derivative is cosx
The derivation of ysinx is y'sinx + ycosx
LNY derivation means that the derivative of a composite function is y '/ y
Ysinx + LNY = 1 will be both sides of the equation at the same time x derivative
So y'sinx + ycosx + (y '/ y) = 0
If there is any unclear place, please ask

Find the derivative of the following implicit function y = cos (x + y)

y'=--sin(x+y)/(1+sin(x+y))

X ^ 2-y ^ 2 = 4 finding the second derivative of implicit function y

If f (x) is an even function with continuous first derivative, is f (0) an extreme point
If there is no second derivative at x = 0, is it an extreme point

It must be, as long as f (x) is even and differentiable
Because the main condition above is that f '(x) = Lim f (x) / x = lim - f (x) / x x approaches to 0
So we can get f '(x) = 0. The definition of extreme point is not this

If x = 1 / (1 + T); y = t / (1 + T); find the second derivative of the parametric equation?

dx/dt=-1/(1+t)²
dy/dt=[(1+t)-t]/(1+t)²=1/(1+t)²
dy/dx=1/(1+t)²/[-1/(1+t)²]=-1
therefore
d²y/dx²=0