Junior high school math problems, seeking the process, urgent need urgent need urgent need urgent ne

The title is not clear, is it 4A?

Then it can be divided into.

4a²+4a+1+b²-4b+4=0

2a+1)²+b-2)²=0

Well, the equation is to be true.

2a+1=0

b-2=0, so a=-1 2

b = 2 plus points, haha.

Solution: 1).Let x=0 get y=1, so the coordinates of point A are (0,1); When x=3, y=-45 4+51 4+1=10 4=5 2, so the coordinates of point b.

for (3, 5, 2); kab = (5 2-1) (3-0) = 1 2, so the equation of the straight line where ab is located is y=(1 2)x+1

2).s=-(5/4)t²+(17/4)t+1-[(1/2)a+1]=-(5/4)t²+(15/4)t=(5/4)(3-t)t，0≦t≦3.

3).To make BCMN a parallelogram, only cm bn is required, i.e. only kcm=kbn; kcm=(t/2+1)/(t-3)；

kbn=(-5t²/4+17t/4+1-5/2)/(t-3)=(-5t²/4+17t/4-3/2)/(t-3)；Hence the equation:

t 2 + 1 = -5t 4 + 17t 4-3 2, 5t 4-15t 4 + 5 2 = 0, that is, t -3t + 2 = (t-2) (t-1) = 0, so t = 1;t₂=2；

i.e. when t = 1 second or 2 seconds, BCMN is a parallelogram.

t=1, mn =-5 4+17 4+1-(1 2+1)=5 2, cm = [(3-1) +3 2) ]= (25 4)=5 2;

Hence mn = cm , so bcmn is diamond-shaped.

t=2, mn =-5+17 2+1-3 2=3, cm = [(3-2) +2 ]= 5;

Therefore MN ≠ cm, so BCMN is not a rhombus.

Because of AD BC

So dac= acb

And because b= acd

So ABC is similar to ADC

So bac= d

Because ABC is similar to ADC

So ac:bc=ad:ac

So ad=4

The trapezoidal median line is equal to 1 2 (4 + 9) =

The triangle ABC is similar to the triangle DCA

So. bc/ca=ca/ad

9/6=6/ad

ad=4median=(4+9)2=

Because there are two angles that are equal ......You've proven it yourself that so ABC is similar to ACD

AC is more than AD=BC than AC and AD=4 so the median is (4+9) 2=

This question is similar to b= acd cad= acb dac abc da ac=ac bc da 6=6 9 da=4 The median line of the trapezoid is half of the sum of the two bases Median line=(4+9) 2=

Kneel down and I'll tell you tomorrow.

A is on the straight line and the x-axis, is it wrong?

This question examines the circumferential angles of the same arc that are equal, or that the inner angles of the arc AB are greater than a, the circumferential angles are equal to a, and the outer angles of the circle are less than a, so if the ship wants to go around the reef area, it must pass through any point q on the route.

Proof that since the intersection of the line l and the x-axis is (1,0) and ab is the diameter of the circle, a is on the circle p, so the intersection is on the top.

? It is impossible to tangent.

60° [The picture above is not right, it does not meet cd=1.] 】

Because cd=1, the connection od, oc, and odc are equilateral triangles with angles of 60°

Then all that remains is the conversion between the inner angle of the triangle and 180°, without going into details.

After the O point, OM is perpendicular to M and ON is perpendicular to Cd to N, and OM = ON can be seen from the properties of the angular horizontal

cd=fe arc cd=arc ef

According to the meaning of the title: EF=50, GF=BF

bae=45, so ae=be

gae = 60°, so ge = 3ae = 3begf = ge-ef = 3be-50

bf=be+ef=be+50=gf= 3be-50 solution: bf=50 3+100

That is, the distance of B from the plane mirror Mn bf = 50 3 + 100

Solution: Let bf=x

From the question EF=50CM, GAE=60°, BAE=45°, so we get Tan60°= 3=(X+50) (X-50) (1)Tan45°=1 =(X-50) (X-50) (2)(2)(1) (2).

3=（x+50）/（x-50）

The solution is x=100+50 3

So the distance B leaves the plane mirror MN is 100+50 3 cm

You yourself look at the diagram and listen to me, let eb=x, then the distance of B from the plane mirror mn=50+x, then gb 2=50+x, ge=100+x, bae=45, so, ae=x, in the right triangle of age, gae=60° so, ge=2ae, that is, 100+x=root3*x, calculate x. Get answers.

If you really count it as the question says, it will be 50 3-100 centimeters or -50 3-100 centimeters.

But they are all less than 0, which is not in line with reality, so there is no solution.

100-50 3 I guess it doesn't feel right.

(1) Connect CE and extend, cross the line AB to G, C to F to connect the line Y=-3 4X+3 to the X axis to A, to the Y axis to B AO=4, Bo=3

c and x-axis tangent to e, and straight line ab tangent to f, c(m,n) in the second quadrant af=ae=4-m, cf=ce=n

cfg∽△aeg∽△aob

ag=5/4ae，ag=gf+af=3/4cf+ae∴5/4(4-m)=3/4n+4-m

4-m=3n

m=4-3n

obce is a rectangle, bo=ce=n=3

m=-5c(-5,3)

2) C and x-axis are tangent to e, and y-axis is tangent to d

n=m=4-3n

n=1 c radius r

The distance from r=n=1 point (x1,y1) to the straight line ax+by+c=0 is d=|ax1+by1+c|a 2 + b 2,1)y=-3 4x+3 i.e. 3x+4y-12=0, c(m,n) to the straight-line distance is |3m+4n-12|3 2+4 2=n, we get: 3m-n-12 = 0 (rounded) or m+3n-4=0, i.e., m=-3n+4

ob=ce=n=3,m=-5

c(-5,3)

2) If c is tangent to the y-axis, then m=n, m=-3m+4, m=n=1, i.e. r=1.