There are 100 peach trees in an orchard, and each peach tree bears 1000 peaches on average. Now we are preparing to grow a variety of peach trees to increase the yield. The experiment found that for each variety of peach trees, the yield of each peach tree will decrease by 2, but the number of peach trees can not exceed 100. If we want to increase the yield by 15.2%, how many peach trees should be planted?

Suppose there are many x trees, then (100 + x) (1000-2x) = 100 × 1000 × (1 + 15.2%) (0 ＜ x ＜ 100), the result is: x2-400x + 7600 = 0, (x-20) (x-380) = 0, the solution is X1 = 20, X2 = 380

A B. the distance between the two places is 60km. After a bicycles from a for 4h, B starts from a by motorcycle. The speed of B is three times that of A. It is known that B arrives two-thirds of an hour later than a. the speed of two vehicles can be calculated

Let a's velocity be x, B's velocity be 3x, (60 / x) - 4 + (2 / 3) = 60 / (3x) x = 12km / h, 3x = 36km / h

If the average of a group of data 2, 3, 5, a is 3, the average of data 3, 7, a, B, 8 is 5, and the average of data a, B, C9 is 5, then the variance of data a, B, C, 9 is (the result I calculated is 2.6, but the answer is 6

A = 4 * 3 - (2 + 3 + 5) = 2 is obtained from "the average of a group of data 2,3,5, a is 3", B = 5 * 5 - (3 + 7 + 2 + 8) = 5 is obtained from "the average of data 3,7, a, B, 8 is 5", and C = 4 * 5 - (2 + 5 + 9) = 4 is obtained from "the average of data a, B, C9 is 5", so the variance of a, B, C, 9 is s ^ 2 = (9 + 0 + 1 + 16) / 4 = 6.5

Given the function f (x) = half of the square of X - x = three-thirds, the value of X on [1, b] is the value of [1, b] to find B
Wrong. Wrong title. It should be like this
F (x) = half of the square of x-x + three-thirds of the square of X. the range of X on [1, b] is to find the value of B from [1, b]

f(x)=(1/2)x²-x+3/2
=(1/2)(x²-2x+3)
=(1/2)[(x-1)²+2]
=(1/2)(x-1)²+1
The range of values on X ∈ [1, b] is [1, b]
Obviously, the abscissa of the intersection of this function and X axis is - 1 and 3, the axis of symmetry is x = 1, and the opening is upward
So it is increasing on [1, b]
∴f(1)=1,f(b)=b
F (1) = 1 is established, f (b) = (1 / 2) (B-1) & sup2; + 1 = B
(b-1)²+2=2b
(b-1)²=2（b-1)
∵b＞1
∴b-1=2
∴b=3
This is what we need

If the quadratic function f (x) = X2 - (A-1) x + 5 is an increasing function in the interval (12,1), the value range of F (2) is obtained

The quadratic function f (x) is an increasing function in the interval (12,1). Because the opening of its image (parabola) is upward, its axis of symmetry x = a − 12 or coincides with the straight line x = 12 or lies on the left side of the straight line x = 12, so a − 12 ≤ 12, the solution is a ≤ 2, so f (2) ≥ - 2 × 2 + 11 = 7, that is, f (2) ≥ 7

It is known that the square + BX + C of the function y = 2x is a decreasing function on (- infinity, - 3 / 2) and a decreasing function on (- 3 / 2, positive infinity)
The absolute value of two zeros satisfying x1-x2 is equal to two

From the known conditions,
The symmetry axis of quadratic function image is x = - 3 / 2, and the coefficient of quadratic term is 2
Two zeros satisfy that the absolute value of x1-x2 is equal to two
Then the two zeros are - 3 / 2 + 1 = - 1 / 2 and - 3 / 2-1 = - 5 / 2 respectively
Using double radical
Then y = 2 [(x + 1 / 2) (x + 5 / 2)] = 2x & # 178; + 6x + 5 / 2

Given that f (x) = xsquare-ax-3 is a monotone function in (- infinity, 2), find the value range of F (1) f (2)

If the axis of symmetry of F (x) = x ^ 2-ax-3 is x = A / 2, and f (x) is monotone on (- infinity, 2), then a / 2 > = 2 a > = 4
F (1) f (2) = (- 2-A) (1-2a) = 2A ^ 2 + 3a-2 (a > = 4) the axis of symmetry is a = - 3 / 4
The range of F (1) f (2) is [- 27 / 8, + infinity)
I only agree,

Y = x square + ax + 1 is an increasing function on [2, + infinity], and the value range of a is obtained

If the function f (x) = AX2 + 2 (A-3) x + 1 decreases in the interval [- 2, + ∞), then the value range of a is______ ．

The analytic expression of ∵ function is f (x) = AX2 + 2 (A-3) x + 1 ∵ when a = 0, f (x) = - 6x + 1, is a decreasing function on (- ∞, + ∞), which is in accordance with the meaning of the problem; when a ≠ 0, because the interval [- 2, + ∞) decreases, the image of quadratic function is a parabola with opening downward, about the straight line x = 3 − AA symmetry, we can get a < 03 − AA ≤ − 2, the solution is - 3 ≤ a < 0. In conclusion, we can get the value of A The range is [- 3, 0], so the answer is: [- 3, 0]

The function f (x) = ax squared - (5a-2) x + a squared-4 is an increasing function on [2, + infinity], and the value range of a is obtained

F (x) = ax & # 178; - (5a-2) x + A & # 178; - 4 when a = 0, the first-order function f (x) = 2X-4 is an increasing function, a = 0 satisfies the condition, when a ≠ 0, the second-order function f (x) = ax & # 178; - (5a-2) x + A & # 178; - 4 is an increasing function on [2, + ∞), the opening must be upward, and the axis of symmetry ≤ 2 (because the opening is downward, it recurs on the right side of the axis of symmetry)