The solution set of inequality system x + 1 ＞ x − 1 ＜ 0 is______ ．

The solution set of inequality system x + 1 ＞ x − 1 ＜ 0 is______ ．

X + 1 ＞ 0 (1) x − 1 ＜ 0 (2), from (1), X ＞ - 1, from (2), X ＜ 1, so the solution set of this inequality system is: - 1 ＜ x ＜ 1. So the answer is: - 1 ＜ x ＜ 1

(2004 · Sanming) given that the system of inequalities x − a ≥ 0 − 2x ＞ − 4 has solutions, then the value range of a is ()
A. a＞-2B. a≥-2C. a＜2D. a≥2

From (1) we get x ≥ a, from (2) we get x < 2, so the solution set of the original inequality system is a ≤ x < 2, ∵ the inequality system X − a ≥ 0 − 2x ＞ − 4 has a solution, and the value range of ∵ A is a < 2

There is a piece of ice with a mass of 9kg and a density of 0.9 * 10 & sup3; kg / m
1 (volume of ice)
If the ice is endothermic, 3DM & sup3; of ice melts into water
Note: it's wrong for our teacher to say that the second question is 3000g or 3kg. In short, I don't know which question he said. Anyway, there is no 3000g or 3kg

(1) V = m / P = 9 / 0.9 * 1000 kg / m3 = 0.01 M3
(2) 3 cubic decimeter = 0.003 cubic meter
m=V*p=0.003m³*0.9*10³m³=2.7Kg

There are five Wednesdays and four Tuesdays in March of a certain year. What day is November 1 of this year

March has 31 days, that is, four weeks plus three days,
Five Wednesdays, four Tuesdays, so those three days are Wednesdays, Thursdays and Fridays
So may 1st is Wednesday, and October 1st is passing by
31-1 + 30 + 31 + 31 + 30 + 1 = 153 days,
153 △ 7 = 21, more than 6
3+6-7=2
So October 1st is Tuesday

The following relations are represented by letters and formulas
Quick! Quick!
It's math

SUM

A number divided by four to two, divided by five to three, divided by six to four, what is the minimum number? Why reduce two

A number divided by four and two is regarded as a multiple of four minus two
Divide by five and leave three as a multiple of five minus two
Divide by six and four is a multiple of six minus two
So it's the least common multiple of 4 5 6 minus 2
2*2*3*5-2＝60-2＝58

Mathematics formula for Grade Seven

Common mathematical formula table: formula expression
Square difference A2-B2 = (a + b) (a-b)
The square of sum and difference (a + b) 2 = A2 + B2 + 2Ab (a-b) 2 = A2 + b2-2ab
The cube of sum and difference A3 + B3 = (a + b) (A2 AB + B2) a3-b3 = (a-b) (A2 + AB + B2)
Trigonometric inequality | a + B | ≤| a | + | B | A-B | ≤| a | + | B | a | ≤ B-B ≤ a ≤ B
|a-b|≥|a|-|b| -|a|≤a≤|a|
Solutions of quadratic equation of one variable - B + √ (b2-4ac) / 2A - B-B + √ (b2-4ac) / 2A
The relation between root and coefficient X1 + x2 = - B / a X1 * x2 = C / A
Discriminant b2-4a = 0 note: the equation has two equal real roots
B2-4ac > 0 note: the equation has a real root
b2-4ac0
Parabolic standard equation y2 = 2px y2 = - 2px x2 = 2PY x2 = - 2PY
Common mathematical formula table: geometric formula
Side area of straight prism s = C * h side area of oblique prism s = C '* h
Side area of pyramid s = 1 / 2C * h'side area of pyramid s = 1 / 2 (c + C ') H'
The area of the side of the cone s = 1 / 2 (c + C ') l = pi (R + R) l the surface area of the ball s = 4Pi * R2
Cylinder side area s = C * H = 2pi * h cone side area s = 1 / 2 * c * l = pi * r * l
Arc length formula L = a * r (a is the arc number of the center angle R > 0) sector area formula s = 1 / 2 * L * r
Cone volume formula v = 1 / 3 * s * h cone volume formula v = 1 / 3 * pi * r2h
Cylinder volume formula v = s * h cylinder v = pi * r2h
Volume of oblique prism v = s'L (s' is the area of straight section, l is the length of side edge) Note: pi = 3.14159265358979

45 degrees, 18 minutes, 36 seconds + 20 degrees, 43 minutes, 52 seconds___

66 degrees, 2 minutes, 28 seconds
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(the square of a - the square of 2Ab + b) divided by the square of ab-b of a + B

The original formula = (a-b) &# 178; △ B (a-b) / (a + b)
=(a-b)(a+b)/b
=(a²-b²)/b

Solving the problem of quadratic equation with one variable
In 2004, the coverage rate of natural reserves in a city was 4.85%
By the end of 2006, the coverage rate of nature reserves in the city reached 8%. The average annual growth rate of the area of nature reserves in the city should be calculated

Suppose that the average annual growth rate of the area of nature reserves in this city is X
4.85%(1+x)^2=8%
(1+x)^2=8/4.85
(1+x)^2=160/97
1+x=√(160/97)
x=1.28432-1
x=28.4%