Let y = f (x). When x changes from x0 to x0 + △ x, the change of function value △ y is equal to () A. f（x0+△x）B. f（x0）+△xC. f（x0）•△xD. f（x0+△x）-f（x0）

Let y = f (x). When x changes from x0 to x0 + △ x, the change of function value △ y is equal to () A. f（x0+△x）B. f（x0）+△xC. f（x0）•△xD. f（x0+△x）-f（x0）

When x = x0, y = f (x0), x = x0 + △ x, y = f (x0 +), y = f (x0 + △ x), ∧ y = f (x0 + △ x) - f (x0), D

Let function y = f (x), when the independent variable x changes from x0 to other X of x0 +, what is the other y of function change?

His y = f (x0 + his x) - f (x0)

I'd like to ask a question. If you increase the side length of a square by 2.5cm, you will get a new square. The area of the new square is increased by 40.75cm
Find the area of the original square?

Let the side length of the original square be x, then according to the known
X*X+40.75=(X+2.5)*(X+2.5)
We can get x = 6.9
It turns out that the area of a square is x * x = 47.61

How to calculate 24 points with 3,11,2,8

2x8+11-3=24

Let the sum of the first n terms of the sequence {an} be Sn, and an = Sn * sn-1 (n is greater than or equal to 2, Sn is not equal to 0), A1 = 2 / 9
(1) Prove: {1 / Sn} is arithmetic sequence
(2) Finding the set of natural numbers n satisfying an greater than an-1

1.1/S(n)-1/S(n-1)=[S(n-1)-S(n)]/S(n)*S(n-1)=-a(n)/a(n)=-1
The arithmetic sequence of - 1
2. Because 1 / S (1) = 9 / 2, 1 / S (n) = 9 / 2 - (n-1) * 1 = 11 / 2-N = > 2
S(n)=1/(5.5-n) =>
a(n)=1/[(5.5-n)*(6.5-n)]
a(n)>a(n-1) =>
1/[(5.5-n)*(6.5-n)]>1/[(6.5-n)*(7.5-n)] =>
2/[(5.5-n)*(6.5-n)(7.5-n)]>0 =>
(5.5-n)*(6.5-n)(7.5-n)>0
N = {n}

The length of a classroom is 8 meters and the width is 6 meters. How many square bricks do you need at least to pave the floor with 4-decimeter-long square bricks?

8 meters = 80 decimeters, 6 meters = 60 decimeters, 80 × 60 ^ (4 × 4) = 4800 ^ 16 = 300 (pieces) a: at least 300 such square bricks are required

3 out of 5 △ 100% △ 7 out of 9

Original formula = 3 / 5 △ [(3 / 8 + 4 / 8) x9 / 7]
=3/5÷(7/8x9/7)
=3/5÷9/8
=3/5x8/9
=8/15

There are a series of P1 (x1. Y1), P2 (X2, Y2) on the rectangular coordinate plane. PN (xn, yn) for all positive integers n, the point PN is located on the function y = 3x + 13 / 4
In addition, the abscissa of PN takes - 5 / 2 as the first term and - 1 as the tolerance net arithmetic sequence (xn)
1: Finding the coordinates of point PN
2: Let the axis of symmetry of each of the parabola series C1, C2, C3,. CN. Be perpendicular to the X axis. The fixed point of the parabola cn is PN, and the slope of the line passing through the point DN (0, (n ^ 2) + 1) tangent to the parabola cn at the point DN is kn. Find 1 / k1k2 + 1 / k2k3 +. + 1 / kn-1kn

(1) Xn = - 5 / 2 + (n-1) (- 1) = - n-3 / 2yn = 3XN + 13 / 4 = - 3n-5 / 4pn is (- n-3 / 2, - 3n-5 / 4) (2) let CN equation be y = a (x + (2n + 3) / 2) ^ 2 - (12n + 5) / 4 substituted into DN (0, n ^ 2 + 1) a = 1y = x ^ 2 + (2n + 3) x + n ^ 2 + 1y '= 2x + 2n + 3kn = y' (0) = 2n + 31 / (K (n-1) * kn = (1 / 2) * (1 / (2n + 1) - 1 / (2)

Given that the perimeter of the rectangle is 20 cm and the area is 10 square cm, what is the diagonal length of the rectangle

2 (x + y) = 20, xy = 10 (x + y)
Under root sign (x + y) = under root sign [(x + y) - 2XY] = under root sign (100-20) = under root sign 80 = 4 times root sign 5

If the function f (x) = XX2 + 2 (a + 2) x + 3a, (x ≥ 1) can use the mean value theorem to find the maximum value, then it is necessary to add that the value range of a is______ ．

∵ f (x) = XX2 + 2 (a + 2) x + 3A = LX + 3ax + 2 (a + 2) (x ≥ 1), if the function f (x) = XX2 + 2 (a + 2) x + 3a, (x ≥ 1) can use the mean value theorem to get the maximum value, the condition a satisfies is the condition g (x) = x + 3ax (x ≥ 1) can use the mean value theorem to get the minimum value. Obviously, a > 0, from x + 3ax ≥ 23a, if and only if x = 3ax, that is x = 3a, take "="; ∵ x ≥ 1 ∵ 3a ≥ 1, ∵ a ≥ 13 A ≥ 13