Junior 1 math summer homework, 7th grade math summer homework

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A person walks home at more than 6 o'clock, the angle of the hour hand and minute hand is 110 degrees, and when he goes home, it is not yet 7 o'clock, and the angle of the hour hand and minute hand is still 110 degrees.

Solution: Obviously, from 6 o'clock to 7 o'clock, the hour and minute hands can reach an angle of 110 degrees twice.

Set 6 o'clock x to go out.

If it is not more than half past six.

180 degrees - (x 60) * 360 + (360 12) * (x 60) = 110 degrees.

The solution x = 140 11 is approximately equal to the minute.

If it is more than half past six.

x-30) 60]*360-(360 12)*(x60)=110 degrees.

The solution x = 580 11 is approximately equal to the minute.

In the same way, there are two possibilities when he arrives home.

Set it home at 7:y.

If before half past seven.

180 degrees - (60) * 360 + (360 12) * (60) + (360 12) = 110 degrees.

Solution x = 200 11 approximations.

If after half past seven.

y-30) 60]*360-(360 12)*(y 60)-(360 12)=110 degrees.

Solution y = 640 11 approximation.

The start time is 6:120 11 a.m.

Later it was 6:600 11 a.m.

Over 480 11 points.

The process is not clear... But it should be right.

2xa-7xb+3a-8b-8x-10=0（2a-7b-8)x=10-3a+8b

To be true for a real number x, then.

2a-7b-8=0

10-3a+8b=0

This is equivalent to 0*x=0, and x can be taken as a real number.

So b=-4 5, a=6 5

2.For triangle ABC and triangle DAC, because the angle DAC = angle B, and the angle ACD = angle ACB (the same angle), the angle ADC = the angle BAC

1.(2a-7b)x+(3a-8b)=8x+10(2a-7b-8)x+(3a-8b-10)=0 is true for any x.

So: 2a-7b-8=0, 3a-8b-10=0a=6 5

b=-4/5

2.In the triangle ADC: angle ADC = 180 degrees - angle c - angle DAC In the triangle ABC: angle BAC = 180 degrees - angle C - angle B and: angle DAC = angle B, so: angle ADC = angle BAC

1 Let the equation x=0 hold, 3a-8b=10

2a-7b)x+(3a-8b)=8x+10 -2a-7b)x=8x,x≠0 when 2a-7b=8 is solved a=,b=

2.∠dac+∠c+∠adc=180°，①b+∠bac+∠c=180°②

, Angular ADC = Angular BAC

Little brother, this is only right:

2a-7b）x+(3a-8b）=8x+102xa-7xb+3a-8b-8x-10=0（2a-7b-8)x=10-3a+8b

To be true for a real number x, then.

2a-7b-8=0

10-3a+8b=0

This is equivalent to 0*x=0, and x can be taken as a real number.

So b=-4 5, a=6 5

2.For triangle ABC and triangle DAC, because the angle DAC = angle B, and the angle ACD = angle ACB (the same angle), the angle ADC = the angle BAC

The first question is to equal the equations on both sides.

2a-7b=8,3a-8b=10, the two-formula simultaneous equation is solved, a=, b=The second problem is in the triangle ABC and the triangle ADC.

Because dac= b

And because acd= acb (common angle).

So adc = bac (the sum of the inner angles of the triangle is one hundred and eighty degrees).

1) When panning, s is 3 cm; At this time, the overlap is rectangular, horizontal, 2 wide, and 3

2) When 2 t 4, the area of the figure swept diagonally by a small square is 4 cm; At this time, the overlapping part is the small square, and the small square happens to be in the whole big square, so the area of the overlapping part at this time is the area of the small square is 4

3) When s=2cm, the distance of the small square translation is 1 or 5 cmAt this time, the area is less than three in question 1, indicating that the distance of movement is less than that, then the distance moved is 2 divided by 2 = 1, and there is also a case that when the small square moves out of the large square, the overlapping part is 2, and the length of the overlapping part can be calculated to be 1, so the distance moved at this time is 2+(4-1)=5

I hope it can help you, you can't ask questions, satisfied, this kind of problem has to learn to draw by hand, and it is easy to solve it by drawing the right image according to different exercise times.

(1) translation, then the bottom of the overlapping part is, the height is 2cm, 2) 2 t 4 this range, the distance of the small square is between 2cm and 4cm, and you can get a diagonal line just swept through a small square + a triangle, so the area is equal to 2 2 + 2 2 2 = 6 (cm2).

3) When s=2cm2, the distance = 2 2=1 (cm),

s=2t (0=6) does not overlap the large square.

1) s=2*

2)s=43) 2=2t,t=1 or 12-2t=2,t=5

Looking at the screenshot, the first half of the fifth question is estimated to have been copied incorrectly.

1、（2n-1）/（n^2)

2. The selling price is RMB, and the profit is 20% A RMB. -m^2

x^2+2y^2=2(x^2+y^2)=2[(x+y)^2-2xy]=2(9^2-2*14)=106

I'm just doing the task, please ignore me.

Solution: (1) according to the meaning of the topic: (2) according to the meaning of the topic:

ac=10 steps, cd=10 steps, ed=45 steps, bac= cde=90° in bac and cde bac= cde (verified) ac=cd (verified) acb= dce (opposite apex angle) bac cde(asa) ab=de=45 steps and a step about 40cm ab=45 40=1800cm= that is, the distance between Xiao Ming and the tree is about when he is at point A.

Analysis: abc and cde are congruent triangles, ab=de

He has walked a total of 45 steps == de=45-10-10=25 steps, (1) please draw a diagram by yourself;

2) If Ming's step is about 40 cm, the distance between Ming and the tree at point A is =25*40=1000cm=10m

1 (2) according to the title: ac=10 steps, cd=10 steps, ed=45 steps, bac= cde=90° in bac and cde bac= cde (verified) ac=cd (verified) acb= dce (opposite apex angle) bac cde(asa) ab=de Xiao Ming took a step of about 40cm, and walked a total of 45 steps ab=de=45-10-10=25 steps ab=25 40=1000cm=10m That is, the distance between Xiao Ming and the small tree is about 10m when he is at point A

1.Drawing shows that there are three kinds of regular polygons: regular triangles, squares, and regular hexagons, and the sides are the same length, so the solution can be obtained.

then b = 5 6

In order to meet the conditions otherwise it is the only solution.

4.Yes, because 1 a+1 b>1 (a+b), the result may appear.

2 There are an infinite number of solutions according to the system of equations, for the case of the solution of the system of equations (where the different times are a1, a20, b1, b2 are not at the same time 0):

If a1 a2=b1 b2=c1 c2, the system of equations has an infinite number of solutions.

So a=-2, b=1 3

x=3/4

4 unknown variables, 2 equations.

n*x+m*y=1575

x+2y=205

Even if it is a generation, it is a ternary equation, and there are many solutions.

At most, you can only give an interval range, and don't think about the correct solution.

In addition, the teacher who seriously despises this kind of topic for the first year of junior high school students is brain-dead and a waste of children's time.