Let b > a > 0 be a constant and f (x) = 2x-x ^ 2. It is known that when x belongs to [a, b] and the range of F (x) is [1 / B, 1 / a], the values of a and B can be obtained

Let b > a > 0 be a constant and f (x) = 2x-x ^ 2. It is known that when x belongs to [a, b] and the range of F (x) is [1 / B, 1 / a], the values of a and B can be obtained

(2) f(x)=2x-x^2
x> 1 single minus
When x ∈ [a, b], G (x) = f (x) and the range of G (x) is [1 / B 1 / a]
g(a)=1/a g(b)=1/b
2a-a^2=1/a
a^3-a^2-a^2+a-a+1=0
(a-1)(a^2-a-1)=0
A1 = 1 A2 = (1 + radical 5) / 2 A3 is directly rounded off
g(a)=1/b g(b)=1/a (a

Two identical rectangles are 9 cm long and 6 cm wide. If they are stacked together, what is the perimeter of this figure?
One on the left, one on the right, one on the top of the right

48

4.5.7.9 calculation 24
Who would use 4.5.7.9 to calculate 24?
Note: don't copy!

4×7＋5－9＝24

Let the first term A1 = 1 of the sequence {an}, the first n terms and Sn satisfy the relation. 3tsn - (2t + 3) sn-1 = 3T (where t > 0, n = 2, 3, 4,...) (1) Prove that the sequence {an} is an equal ratio sequence. (2) let the common ratio of the sequence {an} be f (T), and make the sequence {BN}, so that B1 = 1, BN = f (1bn-1) (n = 2, 3, 4...) (3) sum Sn = b1b2-b2b3 + b3b4 - +（-1）n-1bnbn+1．

(1) ∵ 3tsn - (2t + 3) sn-1 = 3T ∵ 3tsn-1 - (2t + 3) sn-2 = 3T (n ＞ 2) by subtracting the two formulas, 3T (sn-sn-1) - (2t + 3) (sn-1-sn-2) = 0 can be obtained, and 3tan = (2t + 3) an-1 (n ≥ 3) ∵ Anan − 1 = 2T + 33T ∵ A1 = 1 ∵ A2 = 2T + 33T, that is, a2a1 = 2T + 33T sequence {an} is

A classroom is 8.5 meters long, 4.5 meters wide, and paved with 5 decimeter long square bricks. How many pieces do you need?

Total number of square bricks required = (8.5 * 4.5) / 0.5 = 17 * 9 = 153

Calculation: 4 / 15 △ 2 / 3 + 11 / 15 △ 2 / 3

＝4/15×3/2+11/15×3/2
＝（4/15+11/15）×3/2
＝1×3/2
＝3/2

How does 2sina + root 3cosa = root 7 transform into 1-sin square a = root 7 under 2sina + root 3 times root

Use the formula that the sum of squares of sine and cosine equals one
Let me show you another way: the X under the root sign can be expressed as sqtr (x)
The nth power of X can be expressed as x ^ n
As your original equation: 2 * Sina + sqrt (3 * COSA) = sqrt (7)
2*sina+sqrt(3)*sqrt(1-(sina)^2）=sqrt(7）
(sina)^2+(cosa)^2=1

In the triangle ABC, the angles B = 90 degrees, ab = 6cm, BC = 12cm. Point P starts from point a, moves 1cm / s along edge AB to point B, and point Q starts from point B, moves 2cm / s along edge BC to point C. if point P and Q start at the same time, the area of triangle PBQ is equal to 8 square centimeters in a few seconds?

5

When x = 2, the minimum value of function y is 1. When x = 3, y = 3 / 2
2 the symmetry axis of the image is a straight line x = 2, passing through points (0,3), (1,6)
We know that the maximum value of the quadratic function y = ax & # 178; + bx-c is 2, the vertex of the image is on the line y = x + 1, and the image passes through the point (3, - 6)
Urgent

From the minimum value, let y = a (X-2) ^ 2 + 1, x = 3, y = 3 / 2 = a + 1, get: a = 1 / 2, so y = (1 / 2) (X-2) ^ 2 + 1, symmetry axis X = 2, pass through point (0,3), then symmetry point (4,3), from these two points, let y = ax (x-4) + 3, substitute (1,6), get 6 = a (- 3) + 3, get: a = - 1, so y = - x (x-4) + 3 = - x ^ 2 + 4x 3, vertex ordinate is y = 2, then horizontal

∫ (upper 1, lower-1) 2 + SiNx / 1 + X & # 178; DX

(2+sinx)/(1+x²)=2/(1+x^2)+sinx/(1+x^2)
2 / (1 + x ^ 2) is even function
SiNx / (1 + x ^ 2) is an odd function, symmetric interval integral is 0
So the original formula = 2 ∫ (upper 1, lower 0) 2 / (1 + X & # 178;) DX = 4arctanx | (upper 1, lower 0) = π