Hurry!! A few questions about middle school math!

In the first question, I didn't understand it,,, I read your answer, I know,,, you said that it must be worth the two straight lines you give, are straight lines, right??! Then the point at x=1 is both too ,,, otherwise the problem is meaningless,,, because if the point p is arbitrary, then the two straight lines coincide!! What do you think??

I didn't understand question 2, but anyone above question 3 will play very well, and the answer is correct, so I should be able to understand it??? If you don't understand, just say, I'll explain it to you.

Question 4: S1 + S2 = Quarter, Eleven, AC + Quarter, BC = Quarter, (AC + BC) I won't talk about the rest of it.

Question 5, the second one is very good, very correct, you can take a look

Question 6 is not easy to say,,, let me talk about the method, because I can't draw the number axis, first draw a number line and mark the two points of a and b, and then point c must be on the straight line of x=1, make the straight line of x=1, and do point b about x=

1 of the symmetry point d (or as a point a), connect AD, find the analytic formula of the straight line AD, and then find its intersection with x=1,,, you can think about why this point is found??? Think about it

Note: This kind of question can be used in this way,,, it is very important, remember the !!

Question 7: Extend the AD and do the extension line perpendicular to the AD with EF, and then do the DG perpendicular to the BC. (You let me draw the figure first).

Recertification of DGC DFE; Let's go down there

These questions have some difficult ,,, I hope you can learn from them to do the ideas of the questions, and I hope it will be helpful to you!!

There is no solution to the seventh question, and one condition is missing.

7。Extend the AD and do the extension of the EF perpendicular to the AD, and then do the DG perpendicular to the BC

Proof of DGC DFE

Ad*ef=1 2*ad*cg=1 2*2*2*1=1 with area of 1 2

It's 8800, so the last 8 is in the hundreds, so it's c

There's a problem upstairs!

In scientific notation, significant figures only look at the first half, do not look at the power of the next 10, when the first half of the count, keep the integer part greater than or equal to 1 less than 10, and the significant figure refers to the remaining part and integer part from the first digit is not zero from the decimal point, so it is two significant digits! Choose C! Because keep it up to 100

Wow, you have to answer so many questions just 10 points.

I will only answer if the reward is raised.

1.Elapse. You can bring in the numbers and do the math, and then you'll know the answer.

2.Because if it's 3, the solution set is also x>3

3.Solution: The number of days that volunteers plan to complete this work is x2 x+2(x-5) x=1

2+2x-10=x

x=84.Solution: S1+S2= R1 + R2 = (R1 +R2) = (AC +BC)16

5.Are you typing the data wrong, you can't figure it out at all.

If you see that the length of the middle line is 2, then the length of the hypotenuse is 4, so the sum of the length of the two right-angled sides is 4 times the root number three, and the sum of the squares must be 16, and the list of equations is unsolvable.

1.Not necessarily, let point p be (a,b), from point p through the line y=mx+n, then b=ma+n, if point p passes through the line y=nx+m, then b=na+m, when m=n, then both points of point p pass, when m≠n then these two equations cannot be true.

2.Choose C! (Draw a haha on the number line).

3.Choose A. (Three days in advance, the two have the same efficiency, which means that the rest is for A to complete the six-day task alone, plus the two days completed before, for a total of eight days.) ）

By ab=4, angle acb=90 degrees, then bc2+ac2=ab2 (Pythagorean theorem), the area formula is r2,r are half respectively, and s1+s2=1 2 (ac 2)2+1 2 (bc 2)2=1 2 (2)2=2

5.Due to the hypotenuse midline theorem of a right-angled triangle, the hypotenuse is 4 in length, and the right angle side is x and the other right-angled side is y, and the area is 1 2 (x*y), and 1 2xy = 8

6.From the question, we can obtain: ac+bc=[(1-3)2+(n+2)2]+[1-4)2+(n-2)2].

4+n2+4n+4)+(9+n2-4n+4)

2n2+21

When n=0, the minimum value is obtained, and the minimum value is 21

7.I didn't expect it for the time being

Question 1, not necessarily going through, use the reasoning method.

Question 2, c Because x>3 is equal to 4 or more, and because x>a, a cannot be equal to 4, but can only be equal to 3 or less.

Question 3, 2 x+2(x-5) x=1

2+2x-10=x

x=8 Question 4, s1+s2=2s1 or 2s2

In question 5, if the right-angled side of the triangle is twice the middle line of the hypotenuse, then it is 4, and if the height is one of the right-angled sides 4, the area is 8

Question 6, c(1,0).

Question 7, I can't see the picture. Can you put letters on the table?

1. Not necessarily. [p(2,1) can be substituted].

2. C [Draw the number axis, find the common part, A and 3 can coincide].

As shown in Figure 3, a pair of right-angled triangles satisfies ab = bc, ac = de, abc = def = 90 °, edf = 30 °Operation: Place the right-angled vertex E of the triangle def on the hypotenuse AC of the triangle abc, and then rotate the triangle def around the point E, so that de and ab intersect at the point p, and ef and bc at the point q

ep=eq to be

Because ABC is an isosceles right triangle, and point E is the midpoint of hypotenuse AC.

So, be=ce.........1）

pbe=∠qce=45°……2）

and peb+ beq= qec+ beq=90°

So, peb= qec.........3）

From (1)(2)(3), PBE QCE(AAS).

So, ep=eq

3) According to your ** results for (1) and (2), try to write out that when CE ea=m, the quantitative relationship between EP and EQ is satisfied, where the value range of M isWrite out the conclusion, and prove it)

Because peq=90°, pbq=90°

So, p, b, q, e are four points in a circle (the blue circle in the figure).

So, ape= eqb

Because m and q are symmetrical with respect to en, that is, en is the perpendicular bisector of mq.

So, em=eq, and eqb=emc

So, ape = emc.........1）

ABC is known to be an isosceles right triangle.

So, a= c=45°......2）

Known by (1)(2), APE CME

So, ep em = ae ce

Instead, em=eq

So, ep eq = ae ce = 1 m

So: Medium, when CE AE=2, EP=(1 2)EQ

, ep eq = ae ce = 1 m

Let the isosceles right triangle ABC, ab=bc=a, then the hypotenuse ac= 2a

So, in rt def de=ac=2a

EDF=30° is known

So, ef=(6 3)a

When the point E is on AC, make sure that EF and BC intersect.

Then, eq ef=(6 3)a

When taking the equal sign, ef(q) is exactly perpendicular to bc

In this case, the EFC is an isosceles right triangle.

Then, CE= 2*EQ= 2*(6 3)A=(2 3 3)A

i.e., CE (2 3 3)a

then, ae=ac-ce 2a-(2 3 3)a=[(3 2-2 3) 3]a

So: when the point e is infinitely close to the point c, ce is close to 0, then :ce ae=m is close to 0

When the point e reaches its maximum value (i.e., the case discussed above), CE ae = m = [(2 3 3) a] [(3 2-2 3)a 3] = 6+2

So, 0 m 6+2

One. Fill-in-the-blank questions:

Right angle = 60 °, 45 ° = 1 4 flat angle = 1 8 circumferential angle.

3.Expressed in degrees: 56°25 12 =: expressed in degrees, minutes, seconds:

4.If AOB=45° and BOC=30° are known, then the degree of AOC is 75° or 15°

2. Answer the question.

After class, Satoshi and Ming were arguing, Satoshi said, "It's as big as 36°25", and Ming said, "It's not as big as 36°25", what is your opinion?

36°25, so Xiao Ming right.

3. Calculation questions.

2 3 Right angle = 60 °, 45 ° = 1 4 flat angle = 1 8 circumferential angle.

3.Expressed in degrees: 56°25 12 =: expressed in degrees, minutes, seconds:

4.Knowing that AOB=45° and BOC=30°, the degree of AOC is 75° (or 15°).

2. Answer the question.

After class, Satoshi and Ming were arguing, Satoshi said, "It's as big as 36°25", and Ming said, "It's not as big as 36°25", what is your opinion?

So "no 36°25 large".

3. Calculation questions.

One. Fill-in-the-blank questions:

Right angle = (60)°, 45° = (1 4) flat angle = (1 16) circumferential angle.

3.Expressed in degrees: 56°25 12 = (degrees): expressed in degrees, minutes, seconds: degrees 20 minutes and 24 seconds).

4.If AOB=45° and BOC=30°, then the degree of AOC is (285°).

2. Answer the question.

After class, Satoshi and Ming were arguing, Satoshi said, "It's as big as 36°25", and Ming said, "It's not as big as 36°25", what is your opinion?

So 36°25 large.

3. Calculation questions.

The empty fill is greater than it... The power sum of two numbers is greater than the power of the product of these two numbers. (The premise is that the power is not equal to these two numbers...)

I don't know if you're satisfied...

3m-4 and 7-4m are the two square roots of n, which means (3m-4) 2=(7-4m) 2

So there are two cases: 1: 3m-4=7-4m gives m=11 72: 3m-4=-(7-4m) gives m=3 and then returns the value of m to the original formula and the result is obtained.

From the known (3m-4) 2=(7-4m) 2, i.e., 7m2-32m+33=0

The solution yields m=3 or m=22 7

So n = 25 or 1144 49

The square roots of a number are opposite to each other, so 3m-4+7-4m=0, and the solution is m=3

3m-4=5

So n=25

Solution: 3m-4 and 7-4m are the two square roots of n.

3m-4+7-4m=0

m=33m-4=3×3-4=5

n=5²=25

These two numbers are the two square roots of the same number, and if they are the same square roots, then 3m-4=7-4m

7m=11m=11/7

3m-4=5/7

n=25/49

If they are different square roots, then the two numbers must be opposites, then.

3m-4=4m-7

m=33m-4=5

n=25

From the meaning of the title, it can be seen that (3m-4) + (7-4m) = 0, so m = 3

3m-4=5,7-4m=-5

So there is: n = ( 5) squared = 25

First find the sales price of the first month: 625 multiplied by (1 20) to get 500, and then find the sales price of the second month: 500 multiplied by (1 6) to get 530 yuan.

Solution: Let the percentage of the cost price be reduced by x per month.

Since the percentage reduction per month is equal, the square of (1 x) should be reduced in two months from the meaning: 530 (1 x) squared 625 500, solve this equation to get x1 is not in place, round off) x2 0 1

So it should be reduced by 10 per month

The length of the two diagonals of the diamond is 6 and 8 respectively, and the two diagonals of the diamond are perpendicular to each other, so the length of each side = root number (3 2 + 4 2) = 5

Perimeter = 5x4 = 20

20.Because the diagonals of the rhombus are bisected with each other, so 1 2*6=3, 1 2*8=4, and because the diagonals of the rhombus are perpendicular to each other, the Pythagorean theorem 3 2+4 2=25 can be used, and the root number is 25, so the length of the rhombus side is 5, and because the four sides of the rhombus are equal, the circumference is equal to 4*5=20