On the polynomial (A-4) x3-xb + X-B of X is a quadratic trinomial, then a=______ ，b=______ ．

On the polynomial (A-4) x3-xb + X-B of X is a quadratic trinomial, then a=______ ，b=______ ．

∵ the polynomial (A-4) x3-xb + X-B is a quadratic trinomial, ∵ (1) does not contain X3 term, i.e. A-4 = 0, a = 4; (2) the degree of the highest term is 2, i.e. B = 2

It is known that the polynomial (a + b) X4 + (B + 2) x3-2 (A-1) x2 + ax-3 about X does not contain X3 and X2 terms, then a=______ ,b=________ .

Excluding X3 and X2 items
So these two coefficients are 0
So B + 2 = 0, - 2 (A-1) = 0
So a = 1, B = - 2

Simplified evaluation: 4x3 - (3x2 + 5x-2) + 2 (3x + 3 / 2x2-2x) - 1, where x = 2007

4X & # 179; - (3x & # 178; + 5x-2) + 2 (3x + 3 / 2x & # 178; - 2x & # 179;) - 1 where x = 2007
=4x³-3x²-5x+2+6x+3x²-4x³
=x+2
=2007+2
=2009

The volume of a cone with a base circumference of 6.28 cm is 25.12 cubic cm. What is the height of the cone?

Radius = perimeter (2 π) = 6.28 (2 * 3.14) = 1cm
Volume = 1 / 3 * π * square of radius * height
Height = 3 * volume (square of π * radius) = 3 * 25.12 (3.14 * 1 * 1) = 24 cm
A: slightly

The volume of a cone is 113.04 cubic centimeters, the circumference of its bottom is 12.56 centimeters, and what is its height?
The main formula

The perimeter of the bottom surface is 12.56 cm, radius = 12.56 / (2 * 3.14) = 2
Bottom area = 3.14 * 2 ^ 2 = 12.56
56 = 9
The height is nine centimeters

The volume of a cone is 942 cubic centimeters, and the circumference of its bottom surface is 125.6 centimeters. What's the height of the cone? Let's use the formula

What is the base radius of the cone
125.6 △ 3.14 △ 2 = 20 (CM)
What is the base area of the cone
20 × 20 × 3.14 = 1256 (cm2)
Cone height
942 × 3 △ 1256 = 2.25 (CM)

Given that the sequence {a (n)} satisfies the following conditions, write out its first five terms and induce a general term formula of the sequence (no proof required)
（1）a（1）=0,a（n+1）=a(n)+(2n-1)；
（2）a（1）=1,a（n+1）=( 2·a(n) )/( a(n)+2 )
remarks:
"1", "n + 1" and "n" in a (1), a (n + 1) and a (n) denote the number of serial numbers
In question (1), a (1) = 0, a (2) = 1, a (3) = 4, a (4) = 9, a (5) = 16
In question (2), a (1) = 1, a (2) = 2 / 3, a (3) = 1 / 2, a (4) = 2 / 5, a (5) = 1 / 3

A (1) = (1-1) square, a (2) = (2-1) square, a (3) = (3-1) square, a (4) = (4-1) square, a (5) = (5-1) square, and so on: a (n) = (n-1) square;
A (1) = 1 = 2 / 2, a (2) = 2 / 3, a (3) = 1 / 2 = 2 / 4, a (4) = 2 / 5, a (5) = 1 / 3 = 2 / 6, and so on
a(n)=2/n+1.

The general term formula of sequence {an} is an = (2n-9) / (4n-26). Find the maximum term and minimum term in the sequence and prove it

It is reduced to an = 1 / 2 + 2 / (2n-13)
So the minimum term is n = 6, A6 = - 3 / 2
The maximum term is n = 7, a7 = 5 / 2

Sum: (A-1) + (A2-2) + (A3-3) + +（an-n）．

S=（a-1）+（a2-2）+（a3-3）+… +（an-n）=（a+a2+a3+… +an）-（1+2+3+… +n) When a = 0, s = - (1 + 2 + 3 +...) +n) When a = 1, s = n − N22; when a ≠ 1 and a ≠ 0, a + A2 + a3 + +an=a(1−an)1−a，∴S=a(1−an)1−a−n(n+1)2...

What are the general term formula and summation formula of the equal ratio sequence? What are the letters A1, Q and so on?

The first term of the equal ratio sequence is A1, and the common ratio is Q
The general formula is
It's very simple when q = 1, all the terms are equal to the first term A1, and nature is A1 * n
When q is not equal to 1
an=a1*q^(n-1)
The sum of the first n terms is
Sn=a1(1-q^n)/(1-q)