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Because ad bc,ea ad, mbe=45° mbe is an isosceles right triangle, so be = me and because bae= mce, aeb = mec = 90° so abe mce

So ab=cm ( is right).

As you can see from the diagram, the MCE is rotated 90° counterclockwise to coincide with ABE.

i.e. MC and AB at 90° (right).

Because BMC = 180°- mbe - mce = 135°- mce

The title doesn't say what degrees MCE is equal to, so it's not certain that BMC=90° (which is wrong).

Because f and g are the midpoints of ab and cm, respectively.

In a right triangle, the midline length of the hypotenuse is equal to half of the hypotenuse.

i.e. 2ef = ab; 2eg = cm

And because of the abe MCE

So ef = cm ( is right).

Because fb = ef = eg = gm

be = me

So bef emg

So fbe = meg

The same can be seen. aef ≌△ecg

i.e. MEF = MCE

So feg = meg + mef

fbe + mce

fbe + bae

So EFG is an isosceles right triangle ( is correct).

Correct serial numbers 1, 2, 4

Analysis: ABE AEC from AEC= AEB=90°, BAE=MCE, AE=AE

So ab=cm 1 is correct.

Since f g is the midpoint of the hypotenuse of the right triangle, there is ef=(1 2) ab and eg=(1 2) cm

So ef=eg 2 is correct.

Extend the CM line to cross AB to K

Since ecm= bae, bae+ abe=90°,,fem<45, so ecm+ kbc=90°, so ckb=90°

So ab cm 4 is correct.

Since mbe=45°, bme=90°-45°=45°, abe >45°, cme= abe>45°, bmc= bme+ abe >(45+45=90) 90° 3 is wrong.

feb =∠abe>45,，∠fem<45

fg bc,, emf= emg=90, efm emg,fem<45°,= meg<45,°,feg<90° 5 error.

Because. ad bc,ea ad, mbe=45° so mbe is an isosceles right triangle, so. beme again because. bae=∠mce，∠aebmec

So. abe≌△mce

So ab=cm

That's right) as can be seen in the diagram.

A 90° counterclockwise rotation of the MCE coincides with the ABE.

Namely. MC and AB at 90°

That's right) because of BMC

mbemce

MCE is not mentioned in the MCE title

Equal to how many degrees, so.

Not sure. bmc=90°

is wrong) because f and g are the midpoints of ab and cm, respectively.

In a right triangle, the midline length of the hypotenuse is equal to half of the hypotenuse. Namely. 2efab；2eg

cm again.

Abe MCE So. ef

cm is right).

Because fb = ef = eg = gm

beme, so bef

emg, so fbe

The same can be said for meg.

aef△ecg

i.e. MEF = MCE

So. feg

megmef

fbemce

fbebae

So EFG is an isosceles right triangle.

That's right)

Let the bread sold be x, the remaining bread be y, and the remaining biscuits be 2yx+y+2y=8+9+16+20+22+27=102x+3y=102

According to the option, y is known to be an integer.

3y=102-x

102-x must be divisible by 3.

102 is divisible by 3.

then x should also be divisible by 3.

x=9 or 27

Let x=9 y=52

orx=27 y=40

Check x=9y=52=8+22+9+16

Y=40 cannot be composed of the above integers.

So the answer is d

Eliminate z and you're good to go. That is, Lianli.

x + y +4z = 1 and x = y + z

From x +y +4z=1 we get 4z=1-x 2-y 2 so x =y +z gives 16z 2 =16x 2 -16y 2 so the squares of both sides are brought in : get (1-x 2-y 2 ) 2 =16x 2 -16y 2

It can be simplified to get x 4 +y 4 +1-18x 2 +14y 2 +2x 2 y 2=0

The cylindrical equation is: 5x 2 - 3y 2 = 1

Parse: x2 + y2 + 4z2 = 1 .Equation 1).

x^2 = y^2 + z^2 ..Equation 2).

Because the busbar is parallel to the z-axis, there is no z in the cylindrical equation, that is:

We're going to use (Equation 1) and (Equation 2) to get rid of z.

According to (Eq. 2), z 2 = x 2 - y 2

Substituting (Eq. 1) yields: x 2 + y 2 + 4 (x 2 - y 2) = 1

To simplify: 5x 2 - 3y 2 = 1

Actually, this equation is not a cylinder, but a shape similar to a vertical hyperboloid, as shown in the figure below

It should be known that the distance from the point to the straight line is 1, so use the formula in the figure.

d=丨3 2+4 1-4b丨 (3 +4 )=1, then 丨10-4b丨=5

Go to the absolute value symbol, you can get.

10-4b = 5, when 10-4b = 5, b = (10-5) 4 = 5 4, when 10-4b = -5, b = (10 + 5) 4 = 15 4.

In summary, the b-value is 5 4 or 15 4.

The first step is the distance formula, which substitutes the xy value of the coordinates.

The second step is the absolute value of 10-4b = 5, so 10-4b = 10-5 or 10+5, you can solve it.

The distance that Zhang Qiang continued to walk was the distance that Wang Ping had walked before. =10* = 15km。

Total length = 15 (7 12) = 180 7 Zhang Qiang's previous distance = 180 7 - 15 = 75 7 Zhang Qiang's walking time = 75 7 10 =75 70 = Wang Ping's walking time s=l t=15 (75 70)= 14km h