Math problems, solutions and detailed procedures. Hurry, hurry

, the square is sin2x=-24 25,(sinx-cosx) 2=49 25,-pi 20,sinx-cosx<0,sinx-cosx=-7 5

2.From 1 we get sinx = -3 5, cosx = 4 5, tanx = -3 4, (sin2x + 2sin quadratic x -tanx) = -56 25

1.(sinx+cosx) 2=1+2sinxcosx=1 25 gives 2sinxcosx=-24 25

So (sinx-cosx) 2=1-2sinxcosx=49 25 and -pai 2sinx

So sinx-cosx=-7 5

I'm sorry, I didn't understand the second question.

1. sinx cosx=1 5 and sin x cos x=1 to get sinx= 3 5, cosx=4 5.

sinxcosx 2sin x can be calculated, tanx=sinx cosx= 3 4. Just substitute the calculation.

Here's the answer, see for yourself.

I can only lose 50 words, I can't help you.

If the two numbers are added, you can just remove the parentheses, and the number after the plus sign does not need to be changed.

7)-1 3+2 5=-5 15+6 15=1 15 (pass the score first, let the denominator match it at a glance, and then calculate).

8) First turn it into a false fraction, and then pass the score, so that the denominator is consistent, and then calculate. After the calculation, remember that if the group can be divided, it should be divided to -13 4-13 12=-39 12-13 12=-52 12=-13 3

p=a1+a2=9

a2=7p=a2+a3=9

Concealment only calls socks a3 = 2

Therefore, the number of macrochain cultivation is listed as: 2 7 2 7 ...

s13=(2+7) 6+2=56 choose b

1) (1-1 6) (1 8 + 1 12) = 42) 6 (1-6 9) = 18 days (1-4 18) (1 9) = 7

3) Team B needs 5 (1 6)=30 days (1-4 20) (1 30)=24

4) AB is 70*4 (4-3)=280 km, car B speed=280 (4+3)=40 km, hour, car A's speed = 280, 4-40=30 km, hour 5) 10 hours=600 minutes, 3 hours, 20 minutes=200 minutes, 200 600=1 3

Journey = 140 (1-1 3-1 12) = 240 km, train speed = 240 3 200 km, min = 24 km, hours.

1. V customer = 1 8, V cargo = 1 12

Then t 8 + t 12 = 1 6 obtain: t = 4 5 (hour) 2, let the efficiency of A, B and C be x, y, z

Then: x+y+z=1 6, x+y=1 9

Get: Z=1 18

B does 4 days and the rest of the work is: 14 18

The rest is made by the two teams of A and B i.e.: (x+y)*t=14 18 gets: t=7 (days).

3. A efficiency x = 1 20, B efficiency 5 * y = 1 6 that is, y = 1 30 A for four days of first repair, that is, the remaining 4 5 projects.

Then: t 30 = 4 5 De: Team B re-repairs a total of t = 24 (days) 4, let the speed of A and B be x, y distance s

Then 4*(x+y)=s,3*(x+y)=s-70 solution s gets: s=840

From the meaning of the question: x=s 7, y=(s-70) 7, we get: x=120km h, y=110km h

6 (1 8 + 1 12) = 4 hours.

2, [1-(1 6-1 9) 4] 1 9 = 7 (day) 3, (1-1 20 4) (1 6 5) = 24 (day) 4, from the title can be seen that the speed of car B is 70 kilometers per hour (draw a picture yourself), then the distance between the two places is 70 7 = 490 kilometers, (490-70) 7 = 60 kilometers, and the speed of car A is 60 kilometers per hour 5

>b a-b>0（=

0<1 a vs. 2 a

Discuss a>0 1 a-2 a==-1 a<01 a<2 a

a<0 1/a-2/a==-1/a>01/a>2/a

With the help of the difference method, sometimes the commercial method can also be used.

A b yields much more negative results and smaller results.

2 A Large.

<> detailed steps are written on the paper slag, and I guess that the line is quiet.

Let the radius of the sector be r arc length l central angle a (radian) because perimeter = 6 so 2r+l=6 area is 2 get (1 2) rl=2 join the above two equations to get r=1 l=4 (rounded off, because the arc length of the sector is more than one week when l=4, one week is <4) or r=2 l=2 a=l r =2 2=1 The radians of the central angle of the circle are 1 radian.