Make a rectangular iron box, 5cm long, 4.5cm wide, 4cm high, Do 100 such cartons into a large carton, the volume of the carton is how many cubic decimeters

Make a rectangular iron box, 5cm long, 4.5cm wide, 4cm high, Do 100 such cartons into a large carton, the volume of the carton is how many cubic decimeters

5cm=0.5dm 4.5cm=0.45dm 4cm=.4dm
0.5x0.45x0.4 = 0.090.09x100 = 9 cubic decimeter

It is known that a and B are real numbers, and the system of equations y = x3 − AX2 − bxy = ax + B with respect to X and y has integer solutions (x, y). Find the relationship satisfied by a and B

Substituting y = ax + B into y = x3-ax2-bx, eliminating a and B, we get y = X3 XY, then (x + 1) y = X3, if x + 1 = 0, that is, x = - 1, then the left side of the above formula is 0, and the right side is - 1, which is impossible, so x + 1 ≠ 0, then y = x3x + 1 = x2-x + 1-1x + 1 ∵ x, y are all integers ∵ x + 1 = ± 1, that is, x = - 2 or x = 0 ∵ y = 8 or y = 0, so x = - 2

Given that m.n is reciprocal and x.y is opposite, the absolute value of a is 1
A & # 178; - (x + y + Mn) a + (x + y) to the power of 2012 + (- Mn) to the power of 2013

The absolute value of a is 1
∴mn=1,x+y=0,a^2=1,a=±1
A & # 178; - (x + y + Mn) a + (x + y) 2012 power + (- Mn) 2013 power = 1-A + 0-1 = - a = ± 1

The known function f (x) = (2a + 1) / A-1 / ax, constant a > 0
Let m.n > 0, it is proved that the function f (x) monotonically increases 0 < m < 0 on [M, n], and the domain of definition and value of F (x) are the range of constant a in [M, n]

It is proved that: ∵ f (x) = (2a + 1) / A-1 / AX = (- 1 / a) / x + (2a + 1) / A and a ∵ 0 ∵ 1 / a ∵ 0 ∵ 1 / a ∵ 0 ∵ 0 ∵ 1 / a ∵ 0 ∵ 0 ∵ 1 / a ∵ 0 ∵ 1 / a ∵ 0 ∵ 1 / a ∵ 0 ∵ 0 ∵ 1 / a ∵ 0 (this problem is similar to the inverse proportional function y

If | A-1 | + | B-2 | = 0, find the value of a + B

According to the meaning of the question, A-1 = 0, B-2 = 0, the solution is a = 1, B = 2, so a + B = 3

Complete problem: we know the quadratic functions Y1 and Y2 of X, when x = k, y2 = 17, and the image symmetry axis of quadratic function Y2 is x = - 1, Y1 = a (x-k) ^ 2 (k > 0), Y1 + y2 = x ^ + 6x + 10
(1) Finding the value of K
(2) Find the expression of function Y1, Y2
(3) In the same coordinate system, ask if there is an intersection between the image of function Y1 and the image of function Y2, and explain the reason

(1) When x = k, Y1 = a (x-k) ^ 2, so Y1 = 0 because Y1 + y2 = x ^ 2 + 6x + 10: 17 = k ^ 2 + 6K + 10K = 1 or K = - 7 (rounding off) (2) because the image symmetry axis of Y2 is x = - 1, let y2 = t (x + 1) ^ 2 + by1 + y2 = a (x-1) ^ 2 + T (x + 1) ^ 2 + B = (a + T) x ^ 2 + (2t-2a) x + A + T + B because Y1 + y2 = x ^ 2 + 6x + 10

It is known that f (x) is an even function on R. when x ≥ 0, f (x) = 2x-2x & nbsp; 12, and a is the positive zero of the function g (x) = ln (x + 1) - 2x, then the size relation of F (- 2), f (a), f (1.5) is ()
A. f（1.5）＜f（a）＜f（-2）B. f（-2）＜f（1.5）＜f（a）C. f（a）＜f（1.5）＜f（-2）D. f（1.5）＜f（-2）＜f（a）

When a ＞ 0, it is easy to know that G (x) is an increasing function, and G (2) = ln3-1 ＞ 0, G (1.5) = ln2.5-43 ＜ lne-1 = 0. Then from the existence theorem of zeros, we can know that G (x) has zeros in the interval (1.5, 2). Then from the monotonicity, we can know that a is this zeros, so there is 1.5 ＜ a ＜ 2. When x ≥ 0, we can get f ′ (x) = 2xln2 − 1x by direct derivation, so when x ＞ 1, we can get the following results If f '(x) > 2ln2-1 = ln22-1 > lne-1 = 0, we can see that f (x) increases monotonically on (1, + ∞), we can see that there must be f (1.5) < f (a) < f (2), and f (1.5) < f (a) < f (- 2) because f (x) is even function

If f (x) has f (a + b) = f (a) + F (b) for any AB and X is greater than 0, f (x) > 0, the increasing function on R is proved
If the function f (x) defined on R is not always a zero function, f (a + b) = f (a) + (b) holds for any AB, and if x is greater than 0, f (x) > 0 holds. 1 prove that f (x) is an increasing function on R. if f (4) = 1 / 4, solve the inequality f (x-3) + F (5-3x) ≤ 1 / 2 about X

Let X1 > X2, then x1-x2 > 0, f (x1-x2) > 0
f(x1)=f(x1-x2+x2)=f(x1-x2)+f(x2)>f(x2)
So f (x) is an increasing function over R

If a represents a rational number and the absolute value of - 3-A is equal to the absolute value of 3 + A, then a is a rational number

The absolute value of - 3-A is equal to the absolute value of 3 + a
A is any real number!

Gerund phrase as subject, predicate with singular, why? What other similar phrases?

Yes, because gerund can only be regarded as abstract noun, and abstract noun is uncountable. Of course, singular is used. For example, "neighborhood" is abstract noun, gerund is abstract noun. This is true for infinitive as subject or subject clause