Desperate!! Math problems for junior high school! To be explained!! ）

,y=k/x

kx=k/x

x=1/xx^2=1

x=1,x=-1

So the abscissa of a and b is 1 and -1

Now the abscissa of a is 1

So the abscissa of b is -1

b(-1,-3)

Substitute y=kx, -3=-k

k=3y=3x,x=1,y=3

So a(1,3),b(-1,-3).

2.The probability is about: 1-30 * 30 40 * 40 = 7 16, solution idea: if the center of the disc with a radius of 5cm is within the range of the center of the square floor tile (40cm 40cm) (30cm * 30cm), it does not intersect with the gap between the floor tiles.

Otherwise, if the center of the disc is outside the range of the center of the square (40cm 40cm) (30cm*30cm), it will intersect with the gap between the floor tiles.

It's unfair to number the 6 cards from 1 to 6 and assume that the 5 and 6 cards are the two old kings. Now, let's list all the different combinations that take two out of 6 cards, for a total of 15 such combinations:

Of these 15 pairs, 9 of them contain old kings (5 and 6). Since every pair has the same likelihood, this means that in the long run, you'll have at least one old king turned over 9 out of every 15 attempts. In other words, the probability of turning out at least one old king is 9 15, which can be reduced to 3 5.

This is of course better than 1 2, so the answer is: you are more likely to have at least an old king than you are to not be able to make an old king. So it's not fair.

4.Suppose none of the six people have the same zodiac sign as event a

Each person has 12 options for the base event of 12 6

The number of events A is 12*11*10*9*8*7 For every one person who chooses well, the next candidate will be less than one.

then p(a) = (12*11*10*9*8*7) (12 6).

The probability is 1-p(a) = 1-385 1728 = 1343 1728

Question 1: We grew up looking at pictures.

(1) One side is x, and the other side is 4-x, so y=x(4-x)=-x +4x

2) s=(ab) 2=[a(16-a)] 2=-a 2+8aWhen a=8, s=32

3) Are you sure that there are no fewer conditions or typos, only two points cannot find the quadratic function relation, unless one of them is a vertex.

4) A is made perpendicular to BC

sinb=ae ab=(root number 3) 2

So ae = (root number 3) 2 times x

y = (60-2x) multiplied by ae divided by 2 = negative (root number 3) 2 times x + 30 times (root number 3) 2 times x

When x = 15, y = 225 multiplied by (root number 3) 2(5)s positive = (4-x)4 2+(4-x)4 2+x 2+y=16y=-x 2+4x

Because AEF is a regular triangle, AE=EF

4²+(4-x)²=x²+x²

x = 4, root number 3-4

y=-48+32 root number 3

Connect od, from the question it can be seen that hb is 1. Let the radius be r

According to ODH is the columnable equation for a right triangle: the square of root 2 + the square of (r-1) = the square of r.

Solution r=3 2

ab=3

hd = 2 so bh=1 have you learned the equation (r-bh) 2+hd 2=r 2 triangle odh pythagorean theorem.

Another theorem is that bh*ah=hd*ch proves bh=ob-oh ah=ao+oh ao=bo==>ob 2-oh 2=hd 2 (Pythagorean theorem).

So the answer is 3

1.Bottom surface radius r, r = 33, r = (33 ), bottom surface circumference = 2 r = 2 (33 ), cylindrical side area = 2 r * h = 16 (33 ) conical bus length l = (r 2 + 2 ) = 1 * 33 + 4 ) conical side area = 1 2 * bottom perimeter * bus length = 1 2 * 2 (33 ) * 1 * 33 + 4 )

A yurt needs canvas.

Square metre. 2.Arc length = 118 360 * 2 * * 24 = 236 15 * cone bottom radius = arc length 2 = 118 15 , cone height = (24

In other words, don't you think this yurt is too high, 10m, even the top of 12m, equivalent to the height of a 4-story building, the radius of the bottom is only more than 3 meters, not to mention the standing and not standing, a little wind will fall, I hope that the person who came out of this question is sober.

pi*r^2=33，r=

Cylindrical area = 2*pi*r*h= square meters.

Cone part bus = root number (2 * 2 + r 2) =

Cone part area = busbar * bottom circumference 2=

So each yurt area =

15 = ---2) arc length = 2 * pi * r * 118 360 = cm bottom radius = cm

Height = root number (r cm.)

Solution: x -|x-1|-1=0

When x 1, the equation can be reduced to .

x²-x=0

The solution is x=0 (rounded) or x=1

When x<1.

The equation can be turned into .

x²+x-2=0

The solution yields x=-2 or x=1 (rounded).

In summary, the solution of the equation is x=1 or x=-2

(1) When x 1 is solved as x 2-x=0 to get x1 = 1, x2 = 0 (not suitable for the topic, discarded) when x=1 (2) x < 1 is solved to x 2 + x-2 = 0 to get x1 = -2, x2 = 1 (not suitable for the topic, rounded) to take x=-2, so the root of the original equation is.

x1=1，x2=-2

x^2-1|x-1|-1=0

Solution:1When x>=0, the original equation can be reduced to x 2-x+1-1=0, i.e., x 2-x=0

x(x-1)=0

x-1=0x=12.When x>=0, the original equation can be reduced to x 2-x+1-1=0, i.e., x 2-x=0

x(x+1)=0

x+1=0x=-1

When x>=0, the original equation is x2-x+1-1=0, and the solution is: x1=0, x2=1

When x<0, the original equation is x2+x-1-1=0, and the solution is: x3=-2, x4=1 (not in accordance with the topic, discarded).

So the solution of the original equation is: x1=0, x2=1, x3=-2

When x 1, the original equation is x -x=0, and x1 = 1 or x2 = 0 (not in accordance with the topic, rounded).

When x 1, the original equation is x + x-2 = 0, resulting in x1 = -2 and x2 = 1

Therefore, the root of the original equation is x1=1, x2=-2

1.When x-1>=0, i.e., x>=1, x2-x=0 is solved to get x=1, x=0 (rounded);

2.When x-1<0, that is, x<1, x2+x-2=0 is solved to obtain x=-2, x=1 (rounded);

x=1，x=-2。

The absolute value (x-1) in the problem is equivalent to the absolute value x in the example, which is x-1 0

Just do the rest.

No, I can't see clearly, you still say it.

Let the missing line f(x)=x2-(2m+1)x+(m-8) open to the volt.

by the title. f(-1)<0

f(1) "File Sun 0

Namely. 1+2m+1+m-8<0

1-2m-1+m-8<0

m<2m>-8

Hence -8

I don't know if you have learned the derivative with the erection, but it is estimated that you haven't learned the cavity shed bend, and Wu Bo uses the derivative to do it very simply.

The derivative equation is 2x-(2m+1), when x=1, the result is greater than 0, when x=-1, the result is less than 0, and the two equations are synthetically mgreater than -3 2 less than 1 2

Let f(x)=x2-(2m+1)x+(m-8) be set with extreme values x1=1 x2=-1

Bring x1 x2 into the Lookout Mega Balance Piece The opening is trembling upwards and one is greater than 1 and the other is less than -1

i.e. f(-1)<0

f(1)<0

1+2m+1+m-8<0

1-2m-1+m-8<0

m<2m>-8

Hence -8

-x
+10x

2. (80-x-40)*(500+10x)=8000 solution x=30 or -40 (rounded).

That is, each piece should be bought 80-30 = 50 yuan.