How many natural numbers from 1 to 100 are not multiples of 6 or 7?

How many natural numbers from 1 to 100 are not multiples of 6 or 7?

There are 70, 100-16 (multiples of 6 in 100) - 14 (multiples of 7 in 100) = 70

Fill in the natural numbers within 100 as required. The factor of 60 is the multiple of 11 and the multiple of 13

The factor of 60: 1,2,3,4,5,6,10,12,15,20,30,60
Multiple of 11: 11,22,33,44,55,66,77,88,99
Multiple of 13: 13, 26, 39, 52, 65, 78, 91
I wish you progress in your study

Factors and multiples from 1 to 100

1:1 2:1,2 3:1,3 4:1,2,4 5:1,5 6:1,2,3,6 7:1,7 8:1,2,4,8 9:1,3,9 10:1,2,5,10 11:1,11 12:1,2,3,4,6,12 13:1,13 14:1,2,7,14 15:1,3,5,15 16:1,2,4,8,16 17:1,17 18:1,2,3,6,9,18 19:1,19 20:1,2,4,5,10,20 21:1,...

Find the derivative of the function determined by the parametric equation y = 2Sin & # 178; t, x = 3cos & # 178; t,

y/2=sin²t
x/3=cos²t
y/2+x/3=sin²t+cos²t=1
y=-2/3x+1
y'=-2/3

To solve the equations: a + B + C + D = 10, 4A + 3B + 2C + D = 20, A-B + C-D = - 2, 2a-b + 4c-3d = 0, once every four elements, I was a junior one,
On analogical reasoning,

A = 1, B = 2, C = 3, d = 4, first mark ①, ②, ③, ④ + ④ to get 3A + 5c-2d = 10, ⑤, ② + ③ × 3 to get 7a + 5c-2d = 14, ⑥ - ⑤ to get 4A = 4, a = 1 to substitute a = 1 into ①, ③ to get B + C + D = 9, ⑦ B-C + D = 3, Ⅷ - Ⅷ to get 2C = 6, C = 3 to substitute a = 1, C = 3 into ⑤ to get 2D = 8, d = 4 to substitute a = 1, C = 3, d = 4 into ① to get b = 2, a = 1

2a-3b-4c = 2, find the value of 4A ^ / 8 ^ b * (1 / 16) ^ C-8

Let 4A ^ / 8 ^ b * (1 / 16) ^ C-8 = f (x), and take the logarithm with 2 as the base on both sides. We obtain log2 (4 ^ A * (1 / 16) ^ C / 8 ^ b) = log2 (f (x) + 8), a * log24 - (b * log28) + (- C * log216) = log2 (f (x) + 8), 2A - 3B - 4C = log2 (f (x) + 8), that is log2 (f (x) + 8) = 2

Can 2 (2a-b-2c) (4a ^ 2 + B ^ 2 + 4C ^ 2) be decomposed?

2 (2a-b-2c) (4a ^ 2 + B ^ 2 + 4C ^ 2) must not be decomposed in the real range
It's another matter to prove it

The second power of B + the second power of C + 2Ab + 2Ac + 2BC factorization
There is no quadratic power of A

b ² + c ² + 2 a b + 2 a c + 2 b c
= （b ² + 2 b c + c ²）+ 2 a b + 2 a c
= （b + c）² + 2 a（b + c）
= （b + c）（b + c + 2 a）

It is known that A.B.C is the three sides of a triangle respectively. Try to explain that b square - a square - C square - 2Ac is less than 0

b²－a²－c²－2ac
＝b²－（a²＋2ac＋c²）
＝b²－（a＋c）²
＝（b－a－c）（b＋a＋c）
In a triangle, the sum of two sides is greater than the third side
b－a－c＝b－（a＋c）＜0 a＋b＋c＞0
B square - a square - C square - 2Ac is less than 0

It is known that a, B and C are the three sides of triangle respectively. Try to explain that the square of B-the plane of a-the square of c-2ac is less than 0

b^2-a^2-c^2-2ac
=b^2-(a+c)^2
Due to B