After the earthquake in Yunnan Province, teachers and students of our school enthusiastically donated money. In the past six years, more than 160 yuan was donated by female students, accounting for 40% of the whole class, while 23 yuan was donated by male students?

After the earthquake in Yunnan Province, teachers and students of our school enthusiastically donated money. In the past six years, more than 160 yuan was donated by female students, accounting for 40% of the whole class, while 23 yuan was donated by male students?

160 (1-40% - 22 + 3) = 160 (60% - 25), = 160 (20%), = 800 yuan; a: a total of 800 yuan for the disaster area

After the earthquake in Yunnan Province, teachers and students of our school enthusiastically donated money. In the past six years, more than 160 yuan was donated by female students, accounting for 40% of the whole class, while 23 yuan was donated by male students?

160 (1-40% - 22 + 3) = 160 (60% - 25), = 160 (20%), = 800 yuan; a: a total of 800 yuan for the disaster area

5.12 "after the Wenchuan earthquake, teachers and students of a school enthusiastically donated money. 60% of the girls in a class of grade 6 donated 32 yuan less than 40% of the whole class, while the boys donated 32 yuan less than the girls

Boys' donation is two-thirds of girls' donation. According to the equation, if girls donate X Yuan, boys donate two-thirds X Yuan
Donation for the whole class * (1 + 2 / 3) X Yuan
Equation: 40% * (1 + 2 / 3) X-60% x = 32
The solution is x = 320
So the whole class donated: (1 + 2 / 3) * 320 = 1600 / 3 yuan

If the second power of 2 × the nth power of 8 = the 11th power of 2, then n = 1
The n power of [(a-b) to the m power] and the 4 power of (B-A)=
For positive integer n, the result of calculating the 2n + 1 power of (- 3) + 3 × (- 3) 2n power is
If 2x + 5y-3 = 0, find the value of x power of 4 × 32y
Given that 3M = 4, the nth power of 3 = 5, find the value of 2m + nth power of 3
Acenaphthene know very simple, but do not insult acenaphthene intelligence!

1.2^2*8^n=2^2*2^3n=2^(2+3n)=2^11
n=3
2.[(a-b)^m]^n*(b-a)^4=(a-b)^(mn+4)
3.（－3）^(2n+1)+3×（－3）^2n=-3×（-3)^2n+3×（－3）^2n=0
4. This question should be 32 ^ y
4^x×32^y=2^2x*2^5y=2^(2x+5y)=2^3=8
It's wrong. The m power of 3 = 4
3^(2m+n)=(3^m)^2*3^n=80
Children should use their brains and do more questions~

In the parallelogram ABCD, BD is a diagonal line. The parallel line passing through point C is the extension line of BD and intersects with two points F and e respectively. It is proved that points B, C and D are three
It is proved that the points B, C and D are the midpoint of each side of the triangle AEF

∵BD∥EF,AB∥CD,
The formula BF = CD, BD = CF can also be proved by parallelogram
Similarly, ∵ BD ∥ EF, BC ∥ AE,
∴BC=DE,BD=CE,
And ∵ ad = BC, ab = CD,
∴AB=BF,EC=CF,AD=DE,
That is, points B, C and D are the midpoint of each side of the triangle AEF

Simple calculation of 25 * 2.4/25-2.12 and 2 / 3-1.8 + 1 and 5 / 6 15.73-6.17 + 4.27-3.83

25 * 2.4 / 25-2.1 = 2.4 * 25 / 25 - 2.1 = 2.4-2.1 = 0.3.2 and 2 / 3-1.8 + 1 and 5 / 6 = 3 - 1 / 3 - 2 + 0.2 + 2 - 1 / 6 = 3 - (1 / 3 + 1 / 6) + 1 / 5 = 3 - 1 / 2 + 1 / 5 = 3 - 5 / 10 + 2 / 10 = 3 - 3 / 10 = 2 and 7 / 10.15.73-6.17 + 4.27-3.83 = 15.73 + 4.27 - (6.17 + 3.8

1. The distance between B and B is 10km. Party A and B start from AB and go in the same direction at the same time. If Party A is behind Party B, when Party A overtakes Party B, the equivalent relationship is: the distance that Party A takes = ()

Route B + 10

As shown in the figure, a, B, C and D are the four vertices of the rectangle, ab = 16cm and ad = 6cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until it reaches B, and point Q moves to D at the speed of 2 & nbsp; cm / S. (1) how many seconds does P and Q start from the start? The area of quadrilateral pbcq is 33cm2; (2) from the start of P and Q to a few seconds? The distance between point P and point q is 10 cm

(1) Let the area of the quadrilateral pbcq be 33cm2 from the starting point of P and Q to x seconds, then Pb = (16-3x) cm, QC = 2xcm. According to the trapezoidal area formula, we get 12 (16-3x + 2x) × 6 = 33, and the solution is x = 5. (2) let the distance between P and Q be 10cm from the starting point of P and Q to T seconds

How to calculate 125 * (24 + 8) * 25

125*(24+8)*25=125*32*25=125*8*4*25=(125*8)*(4*25)=1000*100=100000

About the problem of ordinal axis root marking method, please, thank you
What is the highest coefficient to be positive?
Can it be used when the coefficient of X is not 1?
When to wear it? When not to wear it?
When can't the root coefficient method be used?
When does it start from the bottom?

X ^ 2-3x + 2 ≤ 0
General steps:
1. Decomposition factor: (x-1) (X-2) ≤ 0
2. Find the root of equation (x-1) (X-2) = 0: x = 1 or x = 2
3. Draw the number axis and flower the point where the root is
4. Notice that at this time, from the far right, a curve is drawn from the upper right of point 2, passing through point 2, and continue to draw, similar to a parabola, and then passing through point 1, and extending infinitely to the upper left of point 1. That is, from top to bottom, from left to right
5. Look at the problem solving. If you want to find a solution of ≤ 0 in the problem, you only need to look at which segment is on the number axis and below the number axis, and observe to get: 1 ≤ x ≤ 2
For example, an inequality after factoring a factor:
x(x+2)(x-1)(x-3)>0
Similarly, first find the root of the equation x (x + 2) (x-1) (x-3) = 0
x=0,x=1,x=-2,x=3
Mark these points in turn on the number axis. Or, draw a curve from the top right of the rightmost point 3, passing through point 3, which is similar to an opening upward parabola between 1 and 3, passing through point 1; continue to extend to the upper left of point 1, which is similar to an opening downward curve between point 0 and 1, passing through point 0; continue to extend to the lower left of 0, It is similar to a parabola with an opening upward between 0 and - 2, passing through point - 2 and extending infinitely to the upper left of point - 2
The equation requires > 0
Just look at the X range of the curve above the number axis
x