Ask two junior high school math problems, and there should be a detailed process reward

1. If the full water of the whole water tank is 1, then the water injected per minute is 1 5, and the water flowing out per minute is 1 10. Now that x minutes have been injected, the injected water is (1 5-1-10)*x=x*1 10, and the remaining water is 1-x*1 10. Later, the same time tank is injected to be full, then the water injected later is x*1 5, which is equal to 1-x*1 10, and the result can be x=10 3.

The answer to the second question on the third floor is correct.

1.The conditions are not enough.

2.(a-21) (350-10a) 400 and (a-21) 21 less than or equal to 20

Solve.

2.(a-21)(350-10a)=400 solution: a=25 or 31

a<21*(1+20%)

So a <

So a=25

Therefore, the number of items that need to be sold is (350-10a) = 350-10 * 25 = 100 (pieces).

The selling price of each item = a = 25 (yuan).

A: The number of items to be sold is 100 pieces, and the price of each item should be 25 yuan.

If you want to change the words after the last comma in the first question, you can't understand it.

With d as the origin and the straight line where AD is located as the x-axis, a Cartesian coordinate system is constructed.

a(-3,0), c(2,-1), easy to get e(1,2), then.

ae=√｛（3-1）²+0-2）²｝=2√5.

Analysis: According to the question, if EF AD is F, DG BC is G, and it is proved that CDG is equal to EDF, then the value of AE can be obtained

Answer: Solution: As shown in the figure, as EF AD is f, dg bc is in g, according to the nature of rotation, we can know that de=dc, de dc, cdg= edf, cdg edf, df=dg=1, ef=gc=2, ae= root number (16+4) = 2 root number 5

It is known that in the trapezoidal ABCD, AD BC, B=90, AD=3, BC=5, AB=1, the line CD is rotated 90 to the DE position around the point D, and AE is connected, then the length of AE is (2 5).

Pass E for EH AD, cross AD extension line at H, pass D for DF BC at F

EH=fc=BC-AD=5-3=2DH=DF=AB=1 from known demonstrable EDH CDF

In the right triangle EAH, EH=2, Ah=AD+DH=3+1=4, so AE= (2 2+4 2)=2 5

Solution 2: First, find that DC length is equal to root number 5 and AC is equal to root number 26, and then use the triangle cosine theorem to find COS ADC

, so that the value of the sin adc can be obtained, and then the ae value can be found using the formula of the triangle cosine theorem (you should know). In the process of bounding, it is necessary to convert cos ade to sin adc [cos( ade)=cos(2 -1 2 - adc)=sin adc] to solve that ae is equal to the root number 20

1. Let these two equations be equal, that is, x2+2ax+b2=x2+2cx-b2 solve x=b2 (c-a) Let the same root be x1, and use Weida's theorem x1*x2=b2 x1+x2=-2a Another equation x1*x2=-b2 x1+x2=-2c can find that the x2 of the two equations are opposite to each other (compare the two multiplication equations, one of the roots is equal, understand no) and substitute the x2 that is opposite to each other into the additive formula, that is, x1+x2=-2a x1+(-x2)=-2c Add the two formulas to get x1=-(a+c)=b2 (c-a) to get b2+c2=a2, so it's a right triangle (see it, please say that it's hard to hit up).

2. a2+2a=b2+2b, phase shift, a2-b2=2(b-a) (a+b)(a-b)=2(b-a) when a≠b, a+b=-2, then a2+2a=2 b2+2b=2, add the two formulas, i.e., a2+b2+2(a+b)=4 (a+b)2+2(a+b)-2ab=4 ab=-2 1 a+1 b=(a+b) ab=1

When a=b, a=-1 3 1 a+1 b=2 a, you can bring the value of a.

You can't copy a topic, it's a tragedy. Let's copy it all.

First of all, it is very simple to exclude C, A is very simple, think for yourself.

Because of the EGC BGF

So exclude d

The answer is b

1.There are two equal real roots, that is.

i.e. 4sin4 3sin

Let sin = t

So 4t 2-4 3t+3=0

The solution yields t= 3 2

So =60

2.Because of (2,-3a).

So there is -3a = 4a + 2b-3

Because the axis of symmetry is x=1

So -b 2a = 1 i.e. 2a = -b

From a=-1

b=2, so the analytic formula for the parabola is.

y=-x^2+2x-3

Solution 1, -3-2m+4n=-3-(2m-4n)=-3+4=12,a-b+c=2a +2b -3c -(3a -b -2c )+c +2a -3b

2a +2b -3c -3a +b +2c +c +2a -3b =a so it does not include b, c, so it is adopted.

1. m-2n=-2, so -2m+4n=4 (the left and right sides of the equation are multiplied by -2 at the same time), so -3-2m+4n=-3+4=1

2. It is redundant.

Because a-b+c=(2a +2b -3c)-3a -b -2c )+c +2a -3b )=a

So, as long as you know the value of a, you can calculate the value of "a-b+c".

1，-3-2m+4n=-3-2(m-2n)=-3+4=1

2,a-b+c=2a +2b-3c-(3a-b-2c)+c+c+2a-3b=a2=1,b,c cancel each other, so this condition is redundant and makes sense.

-3-2（m-2n）=-3+4=1

It makes sense that the formula can be eliminated by directly substituting BC, which is only related to A.

1.There are a total of x kg of apples and y apple boxes, 25y+40=x

30(y-20)=x

Solution: x=3240

y=1282。Isn't there a unit price for two bulbs?

The first type: the difference in the initial investment is.

32 2 30 (yuan).

The difference in hourly electricity rates is:

Meta so used.

30 hours), the savings can make up for the initial price difference in the second way: or using a column equation:

Set for more than x hours.

Get x=1000

Ready-to-buy energy-saving lamps require a lifespan of more than 1000 hours.

1 If there are x empty boxes, then the apples have (25x+40)kg

From the title, 30 (x-20) = 25x + 40 x = 128 25x + 40 = 3240

2 The title is wrong It may be that the landlord has less than two kinds of lights**, so it is impossible to compare who is more cost-effective.

Set: There are x boxes.

25x+40=30*(x-20)

Solution x = 128 2900kg apples.

2 How much energy-saving lamps and incandescent lamps**?

1。If there are x boxes, then 25x+40=30(x-20), x=128,30(x-20)=3240kg

2。If it is more than x hours, why is there no original price for incandescent lamps and energy-saving lamps, can not be done.

1: There are x chests, which are 25x+40=30*(x-20) to get x=128. 25 * 128 + 40 = 3240 kg.

2: The topic is not complete, right?

Question 1: There are x chests in total.

columnar; 25x+40=30（x-20）

25x-30x=-40-600

5x=-640

x=128.

Substituting oneself = 3240 kg

1. There are 128 boxes and 3240 kilograms of apples.

Let n 2-n = x

Then the original becomes (x+1)(x+3)+1

x^2+4x+3+1

x^2+4x+4

x+2)^2

n^2-n+2)^2

When n is a natural number, (n 2-n + 1) (n 2-n + 3) + 1 is the perfect square.

1:x^3+2x^2+2006=(x^3+x^2-x)+(x^2+x-1)+2007=x(x^2+x-1)+(x^2+x-1)+2007=2007

2:(n^2-n+1)(n^2-n+3）+1=(n^2-n+1)(n^2-n+1）+2(n^2-n+1)+1

Treat (n 2-n+1) as a whole, so that t=(n 2-n+1) is equal to t 2+2t+1

1.x 2+x-1=0 gives x 2=1-x, and x 2+x=1;

x^3+2x^2+2006

x^2(x+2)+2006

1-x)(x+2)+2006

x-1)(x+2)+2006

x^2+x-2)+2006

2.Let n 2-n = x, then the original formula = (x+1)(x+3)+1x 2+4x+4

x+2)^2