9th Grade Mathematics, Emergency, Junior 3 Mathematics Urgent

Untie; From ab ad=bc de=ac ae, the triangle abc is similar to the triangle ade.

So angular bac=angular dae , so angular bad=angular cae and because angular abc = angular ade

Corner ade = corner abe + corner bad

Angular ABC = Angular ABE+Angular EBC

So the angle bad = the angle ebc = 20 degrees.

Ab ad=bc de=ac ae is similar to ade, so ade= abc

So ebc= abc- abe= abc-(ade-20°)= abc-(abc-20°)=20°

20 degrees. The triangle ADE is similar to the triangle ABC, the angle E = angle C, so the angle Cae = the angle EBC, because it is similar, so the angle BAC = the angle DAE, so the angle EBC=CAE=ABE=20 degrees.

Solution: The triangle ABC obtained from ab ad=bc de=ac ae is similar to the triangle ade, so the angle bac = angle dae and the angle bca = angle dea. Because E is in the triangle ABC, the angle EBC = the angle CAE(1).

Because the angle BAC=angle DAE has been obtained, the angle BAD=ANGLE CAE(2). The angle obtained from (1) and (2) is EBC = angle BAD = 20 degrees. The answer is OK.

k=4 gives the equation for the straight line bc as.

y=3x+1

Therefore, it can be made c(m,3m+1).

o is the midpoint of a,c because a(3,0).

Therefore, o((m+3) 2,(3m+1) 2) substitute o,c into the hyperbolic equation in parallel and bipolar.

m=1,k=4

There are two cases when ABE is an isosceles triangle:

1、ae=be

In this case, AEB=90°, because AEF=45°, so CEF= C= B=45°, CEF is an isosceles right triangle, and CFE is a right angle.

ce=bc-be=bc-[(bc-ad)÷2]=4√2-[(4√2-√2)÷2]=(5√2)/2

According to the Pythagorean theorem: CF2 + EF2 = CE2, that is, 2CF2 = CE2 = 25 2 gives CF=5 2=

2、ab=ae

At this point AEB= B= C=45°, we get that AECD is a parallelogram and CE=AD= 2

Because AEF= AEB=45°, CEF=90° and CFE=180°- CEF- C=45°= C, then CEF is an equilateral right-angled triangle, and CEF is a right angle, and the edge CE=EF is known by the Pythagorean theorem: CF2=CE2+EF2=2+2=4, so CF=2<>

Is 9th grade the third year of junior high school?

(1) ao＝oc

From S oac=ao*oc2 18, ao=oc=6a(-6,0) c(0,6).

Substituting y -1 6x bx c respectively, to obtain.

0=-6-6b+c ..1)6=c ..2) solution (1) (2), b=0, c=6

The analytic formula is: y -1 6x 6

2) Let d(m,0).

Cross the point e as the perpendicular line of the x-axis and cross the x-axis at g

ocd≌δgde

Point E coordinates: xe=m+|oc|=6+m ye=m i.e. e(6+m,m).

From b(6,0), it can be obtained.

kbe=(m-0)/(6+m-6)=1

kbc=-1

kbc*kbe=-1

bc⊥be

<> did tease the rock skin mountain and sent Zaotan.

<> Tu Heng Ming instructed the Huai Huai to hold it.

After <>, Natan Litong continued to defend himself.

The 19 questions are simple. Question 20 1800-1200 is the soil that is less filled, exactly the soil of four cars, divided by four is the soil of one car is 150, and then, 1800 150 = 12 1200 150 = 8The second question, assuming that x car is adjusted, then vehicle A becomes 8+x car B is 12-x, then, A has more 30 is 150+30 is 180, so the equation is 180*(8+x)+150*(12-x)" 3300, and it is easy to calculate that x is 2

Will take the exam so 6? It's just a pity that I can't help you.

If you look at how many of them have a right-angled triangle with an angle of 30 degrees, look at the opposite side equal to half of the hypotenuse, and the line between the midpoint and the right-angled point on the hypotenuse is equal to half of the hypotenuse.