30 points ask the master to solve the junior high school math problem

If the quadrilateral EACF is a rhombus, it is a rhombus.

then ac=cf=fe=ae

Because ac=cf=fe=ae

So --- f is a point on the bisector of the CE hammer.

Because d is a point on the edge of bc, the line de is perpendicular to bc to d, and ab is parallel to the point ecf and parallel to ab is the line de to f

EF, CF cannot be the point of the CF perpendicular bisector.

So it can't be a diamond.

Angle acb = 90 degrees, ac = 2, bc = 3, get ab = root number 13

It is easy to get a parallelogram for EACF, so only require that x when any two edges are equal.

cd=x,db=3-x,cos angle cba=3 root number 13,be=(3-x)root number 13 3,ae=root number 13-(3-x) root number 13 3=x root number 13 3,ae=ac=2

x root number 13 3 = 2, x = 6 root number 13 = 6 root number 13 13

It's like 6/13 of 13

Can't see your diagram.

If it is a rhombus, then ac=ae=ef=fc

By proving that the triangle ABC is similar to the triangle FCD (this one is easily equal by 2 angles), so ab bc=cf cd root number 13 2=2 x x = 6/13 root number 13

2 x = (2*2+3*3) 3 squared

Over E as eg parallel to BC to AC to G

If the quadrilateral ACFE is a rhombus, then AC AE=2

The triangle age is similar to ACB

then ae ge=ab bc

If the EACF is a diamond. Obviously it is already a flat quadrilateral, just let ac=af, i.e. af=2, in abc, ae ab=cd cb is known by the pythagorean theorem ab*ab 13So x 3=2 (root number 13).

x=6*(root number 13) 13

Solution: Set the fee per room to increase by X yuan.

It can be obtained according to the title.

100+x)(100-x/20*10)=1125(100+x)(

x²-100x+2500=0

The solution is x1=x2=50

A: The fee will be increased by $50 per room.

Solution: Set the fee per room to increase by X yuan.

Then: (100+x)(100-x 2)=11250 solution: x=50

A: Increase by 50 yuan.

This problem is a quadratic function.

You take a look at the original price, that is, 100 * 100 = 10000 11250 yuan, so the fee has been increased (you don't write these on the exam paper).

Set up to rent Y room, and set up x 20 yuan, (10 x 0) y = 100-10x, then set up income c yuan.

There is c=y*(100+20x)=(100-10x)(100+20x)=11250

x = because x is an integer so x = 3 or x = 2, so it is increased by 40 yuan or 60 yuan, if the title says that it is for the sake of consumers, this is 40 yuan.

Solution: Set each room to increase by X yuan.

100-x 2)*(100+x)=11250 solution: x=50

Solution: Let increase x yuan, according to the meaning of the topic, get:

100+x)(100-10x 20)=11250 to solve this equation yields: x=50

A: Increase by 50 yuan.

Column equation solution. Let x be increased by 20 yuan, then the equation is (100+20x)(100-10x)=11250, and the solution is x=, so it is increased by 50 yuan.

It's useless for me to tell you this, you just put it in the plane Cartesian coordinate system, and that's how I understood this kind of problem, and the plane Cartesian coordinate system can show its relationship very vividly, and I seem to have learned it in the eighth grade, not in the third year of junior high school.

Let the average rate cut be a%, then.

After the first rate cut.

After the second rate cut.

That is, tidying up, got.

1-a%)^2=198/225

1-a% = plus or minus (3 root number 22) 15

Take 1-a% = (3 root number 22) 15a =

Each drop is a%.

Then there is a solution to the equation.

1.I don't understand it, I can't answer it, please forgive me.

2。 6+（x/3）=8-（3/2）x

Solution: 36+2x=48-9x

36+2x-48+9x=0

11x-12=0

11x=12

x=12/11

5x=10x=2Because the solution of the two equations is the same, substitute x=2 into the equation 5x+2m=2

5x2+2m=2

2m=2-10=-8

m=-44。Set x number to end the training.

x+（x-1）+（x-2）+（x-3）+（x-4）+（x-5）+（x-6）+（x-7）=100

8x-28=100

8x=128

x=128÷8

x=16A: Xiaoguang finished training on the 16th.

5.To be determined.

6.Let the ten digits be x, then the hundred digits are 7+x, and the single digit is 3x, 7+x+x+3x=17, x=2

This number is 926

7.Solution: Let the original single digit be x, then the ten digit is 2x+1

x+10（2x+1）=2x+1+10x

x+20x+10=12x+1+36

9x=27x=3Answer: The original number is to meet in x hours, according to the meaning of the question:

48x+60x=162

x=2.Set x hours later, the distance is 270km, according to the title:

48x+60x=270-162

x=13.Let the two cars meet in x hours, and the slow car will go (x+1) hours when they meet, according to the title:

48（x+1）+60x=162

x=19/18

4.Let the fast train run for x hours and meet the slow train, and the slow train will go (x-45 60) hours when they meet, according to the title:

60x+48（x-45/60）=162

x=11/6

5.After x hours, the fast train can catch up with the slow train, according to the title:

60x-48x=162

x=6.Set x hours later, the distance between fast and slow trains is 200km, according to the title:

60x-48x=200-162

Question 5; :

Let the original tens of digits be x, and the single digits be 10-x

The original number is expressed as 10x+ (10+x).

The inverted ten digits are expressed as: 10 (10+x)+x

Subtracting two numbers to get x=1, then the single digit is 10-1=9

The original number was 19

If the ABE area is the smallest, then the AEO area should be the largest, that is, the OE should be the largest.

And OE = OA*tan OAE, i.e. OAE should be maximum.

Either a straight line intersects a circle at point P, for APC, the sine theorem is applied.

sin p:ac=sin pac:r, where ac,r is a fixed value.

sin pac = sin p * ac r ac r if and only if p is at right angles, the above inequality is equal and sin pac has a maximum.

pac is an acute angle, and the sinusoidal function is an increasing function in the first quadrant, and therefore also the maximum angle.

ad = √(ac^2 -cd^2) = √(9 - 1) = 2√2

tan∠oae = cd/ad = 1/2√2 = √2/4oe = oa * tan∠oae = 2√2/4 = √2/2be = ob - oe = 2- √2/2s△abe = be*oa/2 = 2(2-√2/2)/2 =2 - 2/2

In RTΔAOE t

oe/oa=tan∠oae

OE=OA Tan OAE=2TAN OAE So when TAN OAE is maximum. OE is the largest, BE is the smallest.

And when AD is tangent to the circle O, the OAE is the largest, and the Tan OAE is larger.

So when ad is tangent to the circle o, be is the smallest.

If you don't understand, you can ask Helpful, I wish you progress in your studies, thank you.

Crossing point A can make two tangents of a circle, one crossing.

One, two, four quadrants, the other passes.

One, three, four quadrants.

For the first tangent line, the position of e is closest to b, so be is the shortest at this time;

The second tangent, the position of e is farthest from b, so be is the longest at this time.

The problem needs to be the minimum value, so it's the first tangent.

As for why the point e can't go up and get closer to b, it's because the point d must be on the circle, and if e is going up, the straight line ae has no intersection with the circle, which is not in line with the topic.

I don't know if the landlord can understand

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

Since the length of OA is a fixed value, if the area of ABE is the smallest, the length of BE is the shortest, and AD is tangent to . CD can be connected, and in RT ADC, the length of AD is obtained by the Pythagorean theorem, and the area of ADC can be obtained; It is easy to prove AEO ACD, and according to the square of the area ratio of similar triangles equal to the square of the similarity ratio, the area of AOE can be obtained, and then the area difference between AOB and AOE can be obtained, and the solution can be obtained

Let the coordinates of point d be in the second quadrant.

The equation for a circle (x + 1 ) 2 + y 2 = 1 Let the d coordinate be x1 then y gives the linear equation x - 2 y = 2-x1 - ( x1 + 1) 2 ) under the root number

Point e coordinates 0,y

y = 2( 1-( x1+ 1) 2 ) 2-x1 according to this idea The parabola vertex is the largest, and x1 = 1 3 can know that cd is perpendicular to the da line, so we know.

Because when AD is tangent to a circle, the angle between AD and the X-axis is maximum. And because OA is fixed. So, the bigger the angle, the bigger the OE, so the smaller the BE, and if you do want a company, you can consider the TAN angle OAE

Triangle area = so the area is smallest be the smallest.

According to the figure, only ad is smallest when the second quadrant is tangent to the circle.

To make abe minimum, eao is to be maximum.

To make eao maximum, d needs to be on the circle c, then when the d point is the tangent point< EAO is the maximum.

It's a physics problem.

There is a formula 2as=the square of vt, the square of v0 a is the acceleration, vt is the end velocity, and v0 is the initial velocity.

So there is 2a*20=0-100

a=how much the velocity decreases per second is actually to find its acceleration.

Let the time be: t The acceleration is: a The time to roll to 5m is: t(1)at=-10; (2)10t+

Simultaneous solution: t=4 a=

10t + solution t=4 2 3 and at 4 2 3 the ball has stopped, so t be accurate by yourself.

In this problem, you only need to know a few formulas for linear motion with uniform variable speed.

1.The ball moves at a uniform velocity, and the final velocity is 0m s, so the average velocity is 5m s, and the displacement is 20m, so the time is 4s

2.Because it is a uniform speed movement, that is, the decrease in speed per second is the same, as long as the amount of speed change is divided by time.

3.When rolling 5 meters, the acceleration by displacement = (the square of the end velocity minus the square of the initial velocity) 2 times the acceleration, you can find the final velocity, note that the initial velocity here is 10m s, and the acceleration is negative even if the second question is asked. Knowing the end speed, the time is easy to ask, the same as the first question!

There are many ways to do this, but you can take a look at the knowledge related to uniform speed change in physics.

1) 2as=vt 2 v0 2 (formula) (subscript cannot be typed, distinguish by case).

a=The result must have a unit.

vt=v0+at

t=42) The absolute value of v minus = -a multiplied by t = the velocity of the decrease is not a vector 3) s = v0t + at 2 2t =

Because A + B + C = 0

So c = a + b).

So ax +bx - a + b) =0 factoring yields:

x - 1) (ax + a + b) = 0x = 1 or x = 1 - b a

So the equation one root is 1

It's easy to see that x=1 you see a+b+c=0=ax +bx+c is only true when x=1.

So equation one is x=1

If you're doing exercises, you can just say that it's easy to know by the formula... No points will be deducted for the answer routine.