A question from Senior 1 Compulsory 2 and a review reference question for Senior 1 Mathematics Compu

1. Connect AC, AC is parallel to A1C1, and AC is parallel to MN, then A1C1 is parallel to MN, and A1N= 5A=C2M, Mn is not equal to A1C1, then the quadrilateral is equilateral trapezoidal;

2. In the isosceles trapezoid, the upper bottom is 2a, the lower bottom is 2 2a, the waist is 5a, and the height can be found as (5-1 2) a=3 2 2a then the area is s=1 2(2 2a + 2a)(3 2 2a)=9a 2

Solution 1): From the known square ABCD squares A1B1C1, Mn, A1C1 on the square ABCD and the square A1B1C1D1, respectively. So mn a1c1.

And because m and n are the midpoints of cd and ad, respectively. So aa1n cc1m. So na1=mc1.

So the quadrilateral mna1c1 is an isosceles trapezoid.

Let the radius of the bottom surface of the cylinder be r, then the bottom area of the cylinder is r 2, and the bottom area of the triangular prism is 3 3 4*r 2

Because the two cylinders are equal in height, the volume ratio is equal to the ratio of the base area, i.e., 3 3 4

So the triangular prism volume is 3 3 4 v

Let the radius of the bottom surface of the cylinder be r, then the height is 2r. So I knew.

r 2(2r) = v, i.e. r 3 = v 2.

So the triangular prism volume is [( 3r 3r 2) 2]2r=(3 3r 3) 2=(3 3 )2 v 2 =3 3v 4

Note the case where the intercept is 0 for discussion:

When the intercepts on both the x and y axes are 0, the straight line crosses the origin, and the equation is y=3x

When the intercept on the x and y axes is not 0, let the equation be x a-y a=1, and substitute the points (1,3) to obtain a=-2, and the equation is x-y+2=0

Let a(x1,y1) be on l1:2x-y-2=0, b(x2,y2) be l2:x+y+3=0, and ab at the midpoint of p(3,0).

2x1-y1-2=0,x2+y2+3=0,x1+x2=6,y1+y2=0, the solution: x1=11 3, y1=16 3, is obtained from the two-point equation.

The equation for the straight line l: 8x-y-24=0

Let k(x-3)=y

Find the abscissa of the two intersections of two straight lines, a b.

x1=(3k-2)/k-2 x2=(3k-3)/k+1(x1+x2)/2=3

The solution is k=8 l and y=8x-24

Let the straight line be y=kx+b, then.

Because of the point p, the straight line is y=kx-3k

It intersects with two straight lines a((3k-2) (k-2), 4k (k-2)) and b ((3k-3) (k+1), -6k (k+1)).

ap = bp, we get k = 8

The straight line is y=8x-24

From the original formula: l (bus) = (r 2 + r 2) r + r) = (r 2 + r 2) (r + r).

Let the cuboid be a in length, b in width, and c in height

Then the volume of the rectangular body is a*b*c

The volume of the pyramid is (1 3) * (1 2) * a*b * c = 1 6) abc The remaining volume of the bench is abc-(1 6) abc = 5 6) abc then 1 6abc: 5 6abc

So: 1:5

The ratio of the volume of the triangular pyramid to the remaining volume is 1:5

Isn't there a formula for calculating the first one? I forgot the formula, just solve a system of equations, the side area is the umbrella area, and the other is the bus formula, so will it be solved soon?

The second one I can give you advice is very simple, set up a coordinate, put this rectangle in it, and use the three-dimensional coordinates of space to solve it? The book also said that if you encounter these special and more formal, it is very simple to use the coordinates, and I said that you should understand it. I think it's easier to lead you to disturb Li Siwei, and it will be easy to solve similar problems in the future, which is how my teacher cultivated my thinking before.

You can first convert the general equation into a standard equation, i.e., y=kx+b.

Then we will discuss the relationship between the two straight lines k and b.

If k1=k2 and b1=b2, the two lines coincide.

If k1=k2 and b1≠b2 then the two lines are parallel.

If k1 ≠ k2 then the two straight lines intersect.

If k1 k2=-1, then the two straight lines are perpendicular.

Then you know what to do.

and c≠-2 -4 3

When a takes 3 to be parallel, to make the two lines parallel, their slopes must be equal, that is, a 2 = 6 4, and the solution is a = 3

When a takes -4 3 perpendicular, to make the two lines perpendicular, their slopes must be negative reciprocal to each other, that is, a 2·6 4=-1, and the solution is a = -4 3

As for c, no matter what value is taken, as long as the slopes of the two lines are not equal, they will intersect.

When a and c are valued, the line ax-2y-1=0 intersects with the line 6x-4y+c=01; 2.Parallel; 3.Vertical.

Analysis: If two lines are parallel a1b2-a2b1=0 and b2c1≠b1c2 intersect a1b2-a2b1≠0

If two straight lines are perpendicular a1a2+b1b2=0

The teacher should have talked about this question, as if it was the seventh question & the eleventh question on page P84 of the compulsory second course of the Sujiao version.

h2oco2

C2H4 ethylene.

ch2=ch2 + br2 --ch2br -ch2br1mol 160g 1

The x-column solution yields x= 16g

The addition product CH2BR contains hydrogen atoms and needs to be completely replaced!

The dryer tube is weighted, i.e. water is generated.

Lime water gains weight, i.e. generated.

Therefore, the ratio of hydrocarbons to hydrogen is 1:2, which is (CH2)N

The density of this hydrocarbon (gas) is 14 times that of hydrogen under the same conditions, indicating a relative molecular mass of 28

So it's C2H4 ethylene.

The hydrocarbon energy reacts with the reaction, which is 16 grams of bromine.

The addition product needs to be completely replaced by bromine steam.

c2h4

Analysis: The weight gain of the drying tube corresponds to the quality of the generated water, there are n(h2o)= mol = mol n(h)= mol

The weight gain of lime water corresponds to the mass of carbon dioxide generated, which is n(CO2) = mol = mol

Therefore, the ratio of the number of hydrocarbon atoms in the hydrocarbon = 1:2

In addition, "the density of the hydrocarbon (gas) is 14 times the density of hydrogen under the same condition", there is 1 volume of hydrogen 2g, 1 volume of the hydrocarbon is 28g, and the hydrocarbon is C2H4 in summary

ch2 = ch2 double bond open, with a br atom addition on each side, there is 1molc2h4 1molbr2, so it reacts with bromine (bromine relative atomic mass can't remember == check it yourself).

The reaction to generate 1,4 bromoethane (CH2BR-CH2BR) substitution, 4 H, each corresponding to 1 BR2, therefore, 4 bromoethane and the reaction.

Do you want the idea or the process.

Parallelism can be set to the straight line as 2x+y+b=0

According to the formula for the distance between two parallel lines.

ax+by+c1=0

ax+by+c2=0

d=(|c1-c2|)/a^2+b^2)

d is equal to 7 to go to the absolute value yourself.

(x2)+(y2)-2x+4my+3(m2)=0(x-1) 2+(y+2m) 2=4m 2+1-3m 2=m 2+1 The center coordinate of the circle is (1,-2m), on the line x+y+2=0, then there is:

1-2m+2=0

Get m=3 2

Radius = root number (m 2 + 1) = root number (9 4 + 1) = root number 13 2

First, the equation of the circle is organized into a standard form, i.e., (x-1)2+(y+2m)2=m2+1

In this way, we can see that the center of the circle is (1,-2m), and substituting the coordinates of the center of the circle into the linear equation gives m=3 2

r) 2=m 2+1 solution r=root number 13 2