One half minus one sixth minus one twelfth. What's one fifty-six Quick. There's a process. There's a formula Better have an explanation 1. 1 / 2 - 1 / 6 - 1 / 12 - 1 / 20 - 1 / 30 - 1 / 42 - 1 / 56 1 / 2 + 1 / 6 + 1 / 12 + 1 / 20 +. + 1 / 72 + 1 / 9 11 × 13 / 1 + 13 × 15 / 1 + 15 × 17 / 1 + 17 × 19 / 1 + 19 × 21 / 1 4. (1-half) × (1-third) × (1-quarter) ×. (1-one hundredth) 5. (1 + 2 / 1 + 3 / 1 + 4 / 1) × (2 / 1 + 3 / 1 + 4 / 1 + 5 / 1) × (1 + 2 / 1 + 3 / 1 + 4 / 1 + 5 / 1) × (2 / 1 + 3 / 1 + 4 / 1) seek

One half minus one sixth minus one twelfth. What's one fifty-six Quick. There's a process. There's a formula Better have an explanation 1. 1 / 2 - 1 / 6 - 1 / 12 - 1 / 20 - 1 / 30 - 1 / 42 - 1 / 56 1 / 2 + 1 / 6 + 1 / 12 + 1 / 20 +. + 1 / 72 + 1 / 9 11 × 13 / 1 + 13 × 15 / 1 + 15 × 17 / 1 + 17 × 19 / 1 + 19 × 21 / 1 4. (1-half) × (1-third) × (1-quarter) ×. (1-one hundredth) 5. (1 + 2 / 1 + 3 / 1 + 4 / 1) × (2 / 1 + 3 / 1 + 4 / 1 + 5 / 1) × (1 + 2 / 1 + 3 / 1 + 4 / 1 + 5 / 1) × (2 / 1 + 3 / 1 + 4 / 1) seek

1. Original formula = 1 / 2 - (1 / 2-1 / 3) - (1 / 3-1 / 4) - (1 / 4-1 / 5) - (1 / 5-1 / 6) - (1 / 6-1 / 7) - (1 / 7-1 / 8) = 1 / 82, original formula = 1 / 2 + (1 / 2-1 / 3) + (1 / 3-1 / 4) + (1 / 4-1 / 5) + (1 / 5-1 / 6) + (1 / 6-1 / 7) + (1 / 7-1 / 8) = 1 / 2 + 1 / 2-1 / 8 = 7 / 83, original formula = 1 / 2 [(...)

88,24,56,40,48, (), 46, how to do numerical reasoning

The sum of every two numbers is twice that of the next one, so 46 * 2-48 = 44

The three sides of a right triangle are 6 cm, 8 cm and 10 cm respectively. The area of this triangle is () square cm

24

It is known that the solution of the equation KX = 4-x about X is a positive integer

The original equation is transformed into KX + x = 4, that is, (K + 1) x = 4, the solution of the equation KX = 4-x about X is a positive integer, and the product of K + 1 and X is 4, then K + 1 = 4 or K + 1 = 2 or K + 1 = 1 can be obtained, and the solution is k = 3 or K = 1 or K = 0. Therefore, the integer solution of K can be obtained as 0, 1, 3

The circumference of a circular sheet iron is equal to that of a rectangular sheet iron. It is known that the length of the rectangular sheet iron is 10 decimeters and the width is 5.7 decimeters. What is the area of the circular sheet iron

The perimeter of rectangle is 2 (10 + 5.7) = 31.4
So the radius of the circle is 31.4 / 2 / 3.14 = 5
So the circle area is 3.14 * 5 ^ 2 = 78.5

Simple calculation of 37 * 15 + 45 * 21

37*15+45*21
=37*15+15*3*21
=37*15+63*15
=(37+63)*15
=100*15
=1500

As shown in the figure, BP and CP are bisectors of ∠ B and ∠ C in any △ ABC. It can be seen that ∠ BPC = 90 ° + 12 ∠ A. if △ ABC in the figure is changed into quadrilateral ABCD in the figure, BP and CP are still bisectors of ∠ B and ∠ C. It is conjectured that the quantitative relationship between ∠ BPC and ∠ A and ∠ D is linear______ ．

The intersection of Ba and CD is extended at the point E. according to the known conclusion, ∠ BPC = 90 °+ 12 ∠ BEC, and ∠ e = ∠ bad - ∠ ade = ∠ bad - (180 ° - ∠ ADC). The ∠ BPC = 90 °+ 12 ∠ bad-90 ° + 12 ∠ ADC, i.e. ∠ BPC = 12 ∠ bad + 12 ∠ ADC

Let B: B1, B2, B3,..., Br be linearly expressed by a: A1, A1,..., as (B1, B2,..., BR) = (A1, A2,..., as) K,
Where k is a s * r matrix and vector group A is linearly independent. It is proved that vector group B is linearly independent if and only if R (k) = R

Vector group B is linearly independent
(B1, B2,..., BR) x = 0 has only zero solution
(A1, A2,..., as) KX = 0 has only zero solution
--Because vector group A is linearly independent
--So
KX = 0 has only zero solution
R (k) = R (the number of columns of K)

The top and bottom of the trapezoid are 2 and 8, and the diagonal is 6 and 8. Calculate the area of the trapezoid

24
Through the vertex to do a diagonal parallel line, extend the bottom, there will be a right triangle 6,8,10
So the area is 24, because there is no map, so you need to try it yourself

Given the function y = LG (AX2 + 2aX + 1): (1) if the domain of definition of the function is r, find the value range of a; (2) if the domain of definition of the function is r, find the value range of A

(1) When a ≠ 0, there should be a ＞ 0 and △ = 4a2-4a ＜ 0, and the solution is a ＜ 1. Therefore, the value range of a is [0,1). (2) if the value range of the function is r, then AX2 + 2aX + 1 can take all positive integers, a ＞ 0 and △ = 4a2-4a ≥ 0. The solution is a ≥ 1, so the value range of a is [1, + ∞)