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d [x] here is rounded, so [x] has nothing to do with it.

d Draw an image, a symmetrical figure, continuous at x=0, not a break point.

When x>0, y'=1 3x 2

b c horizontal asymptote: when x tends to infinity, y tends to constant.

Lead straight asymptote: When x tends to be constant, y tends to infinity.

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Select D for the first question

1、a2、b

3、b c b&c

Analysis: 2. Option a, the image with a function can know that the function is continuous at x=0, not a discontinuity.

doption, the inflection point is that the second derivative of the function is equal to 0, and the first derivative is differently signed on the left and right sides of the store. The first derivative of the easily obtained function does not hold the different signs on the left and right sides of x=0.

b c option, there is an image easy to get b option correct.

Last time, I didn't read the title clearly, and I apologize for the misunderstanding.

I wish you a good progress in your university life and study & a good harvest in your love and career.

I thank you here.

1) d The title doesn't say that the function value is 0 at x=xo, are there any other requirements at x=xo? It doesn't seem to be any, so d

2）d3）b

2.From the meaning of the title, it can be seen:

Set: each cow eats x per day, and the growth rate per mu per day is y, and the third piece of land can be used for z cows to eat for 80 days;

Then: 10*30*x = 5 + 30*5*y =>60*x = 1 + 30*y

The first meadow can feed 10 cows for 30 days).

28*45*x = 15 + 45*15*y =>84*x = 1 + 45*y

The second meadow can feed 28 cows for 45 days).

Solution: x = 1 12, y = 2 15z*80*x = 24 + 80*24*y =>z = 42Answer: The third plot of land can be eaten by 42 cows for 80 days.

3.From the meaning of the title: wax ride.

Team B has the least cost of contracting alone;

It takes 6 days and costs 1770 RMB.

400,000 percent.

The percentage of 15 grams to milligrams is: 400,000%; Because 15 (.

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The mathematical model of this problem is very simple, each time point p moves to the left or right, it is equivalent to tossing a coin, throwing heads to the left and throwing tails to the right, this problem is equivalent to: tossing a coin several times, finding the probability that the number of heads and tails is the same.

First the number must be even, otherwise the positive and negative numbers cannot be equal, we assume that it is 2n times and its total probability is 2 (2n), where the number of heads n times is c(2n, n), so the probability of the same positive and negative (i.e., point p back to point q) is: c(2n, n) 2 (2n).

e = root number 3 divided by 2

c=√3/2*a,b^2=a^2-c^2=a^2-3a^2/4=a^2/4

The major axis is on the x-axis, so the elliptic equation can be set as: x 2 a 2 + 4y 2 a 2 = 1

The square of the distance from the point (asinr, acosr 2) on the ellipse to p.

A 2sin 2R+(acosr-3) 2 4-1 4*(3A 2cos 2R+6Acosr-9-4A 2)-[3(acosr+1) 2-12-4A 2] 4So, when acosr+1=0, the farthest distance from the square = (12+4A 2) 4=3+A 2

3+a^2=7

A 2=4 elliptic equation is: x 2 4 + y 2 = 1

Formula 1 = 2sin (2x + 4 = 2sin2 (x + 8) Formula 2 = - 2con (2x + 4) = 2sin (x + 3 4) = 2sin 2 (x + 3 8).

So it's a 2 8 unit to the right.

Pan 4 to the right. Pick D

Using the formula of the sum and difference product of trigonometric functions, it can be obtained.

Both of them have periods of 2, and the translations 1 and 4 periods coincide.

You can see this by substituting the formula specifically.

Solution:1Cylindrical volume: (6 2) square.

into a maximum cone, should be cut off wood 2 3, cut off.

2.The area of the bottom of the conic is (4 2) square.

High 3The volume of the conic is 3 square.

There will be water overflowing.

Good luck with your progress.

1. The volume of the cylinder is 6 (6 2) = 54

The volume of the cone is 1 3 of the cylinder, then.

The wood should be cut off as 2 3 54 = 36 = cubic centimeters 2, meters.

3. 3 6 = cubic centimeters.

1. Because x1 and x2 are the two roots of the equation x 2 2 (m 1) x m 2 7 0.

So x1 x2 2(m 1) , x1*x2 m 2 7 and (x1) 2 (x2) 2 10

So m 2 is substituted into x 2 2 (m 1) x m 2 7 0 to get x 2 2x 3 0

Solve x1 1, x2 3

Therefore, the coordinates of a and b are: a( 1,0), b(3,0)2, substituting a and b coordinates into y ax 2 bx c, a b c 0

9a＋3b＋c＝0

Because the ordinate of the parabola y ax 2 bx c vertex m is 4, (4ac b 2) (4a) 4

The above three equations form a system of equations, which can be solved.

a 1, b 2, c 3 (a 0 does not fit, has been discarded) so the analytic formula of the parabola is.

y＝x^2－2x－3

When x 0, y 3

So the coordinates of point C are (0, 3).

3. The vertices of the parabola y x 2 2 x 3 are m(1, 4), ab 3 (1) 4

The coordinates of the point p are (x,y).

s△pab＝ab*|y|/2＝4*|y|/2＝2|y|Passing m is the mn x-axis, and the intersection of the x-axis is at the n point, then.

S quadrilateral, acmb, saoc, s, bnm, strapoid, mnoc 1*3 2 (3 1)*4 2 (3 4)*1 2 9 if s pab 2s pab

then there are 2|y|＝2*9＝18

So |y|9>4, so p is above the x-axis.

So y 9, so 9 x 2 x 3

i.e. x 2 x 2x 12 0

Solution x 1 13

So the presence of the point P makes the area of the triangle PAB equal to 2 times the area of the quadrilateral ACMB, and the coordinates are: P1[(1 13),9], P2[(1 13),9].

<> for the dismantling of the tape code to participate in the test. Yes.

I don't know anything about high school for a long time.