The math of this problem is solved 30, and the math of this problem is solved

Knowledge points: the application of primary functions.

Solution: The coordinates of point c are (3,60).

The equation for the straight line oc is s=20t

The coordinates of point d are (1,0).

The coordinates of point e are (3,90).

Let the equation for the line de be s=kt+b

k+b=03k+b=90

s=45t-45 ②

Solve the system of equations consisting of t=9 5

Thus, (1) A is 1 hour after B, and the speed is 20 km/h.

2) 9 5 hours after the departure of B, the two met.

1. A is 1 hour after departure than B, and B's speed is 60 3 = 20 km/h.

2. (1) d coordinates (1,0), e coordinates (3,90) then the de equation is set to y=kx+b, and substituting d d and e to obtain.

k+b=0,3k+b=90, solution:k=45, then b=-45de:y=45x-45......（1）

2）o(0，0)，c(3，60)

Then the OC equation: y=20x......（2）

3) Solution (1) and (2): x=9 5, y=36b departure 9 5 hours, the two meet.

1) Solution: 1 hour, 60 3 = 20 (kilometers per hour) Answer... 2) Solution: Straight line oc passes through the point (0,0),(3,60)y=kx+b

60=3k+b

0=bk=20

b=0 gives y=20x

Straight line de pass points (1,0),(3,90).

y=kx+b

90=3k+b 1

0=k+b 2

Get 90 = 2k

k=45b=-45

The solution yields y=45x-45

Make a point F on the PD, and F is the midpoint of the PD, connect EF, and after proving EF AB, you can get EF CD, which can prove that E is the midpoint of PC. It can be proved that the AF vertical plane PCD can launch the AF vertical EF, and the AF vertical AB and AB and EF are in the same plane (this can be briefly said) then the EF ABCD vertical plane pad then CD vertical AF and Pa=AD F is the midpoint, and the AF vertical plane PCDPD is the plane where F, A and B are located, (PD vertical AB, AF) The plane where PD is vertically A, B and E (PD vertical BE, AB), Moreover, these two planes should be parallel and have ab sides, so it should be no problem to say that a, b, e, and f are in the same plane. I'm sorry, I learned this knowledge a long time ago, and I forgot it, so it's more troublesome to answer.

1. First prove that AC is perpendicular to the surface PDB, and DE is on this surface... 2. The dihedral angle e ac p is not easy to find, and it can be made by 180'Subtract the dihedral angle e ac b and subtract the dihedral angle p ac d...

Solution : Let l=cm=co am=a co=b ma ab, ob ab from the Pythagorean theorem:

ac²=l²-a² bc²=l²-b²

ab=ac+bc

The value of ab can be obtained.

Because the angle ocb = 90-angle cob; And the angle OCB = 180-90-angle MCA, so the angle COB = angle MCA

Two right triangles, according to the "corner edges", we can know that these two triangles are congruent.

So ac=bo, am=cb

So ab=ac+bc=bo+am=a+b

Original question: 3(m 2*n + mn) -2 (m 2 * n -mn) - m 2 * n = 3m 2 * n + 3 mn - 2 m 2 * n + 2 mn - m 2 * n = 5 mn

Substituting m=1 5 n=-1

Result: Original = -1

I hope to choose it.

2747 is divided by 67 to 41, and the answer is that the last two digits are 47

Wow, I'm a senior here at Zhejiang University.

I usually write things on the WX number Yiyou.

Take a look at this topic

Equation 1: x-y-1=0

Equation 2: x-y+3=0

These two equations have the same value k, so the two straight lines are parallel;

This needs to be understood;

1.So imagine that the secant values are the same, indicating that the center of the circle is in the middle of the central parallel line of the two lines;

It's like this.

2.So we can make a hanging down from the center of the circle, and we can know the two sides, and use the Pythagorean theorem to calculate that the hypotenuse is the radius.

In this step, you have to draw your own pictures and think about it more

3.Calculate the distance between two parallel lines, and then half of the distance from the center of the circle to one of the lines, right?

Two-line distance formula: |c1-c2|/√a^2+b^2.

It is calculated that it is 4 2 = 2 2

So half a distance is 2

4.The Pythagorean theorem calculates that the radius is 2 2 + 2 2 = 6, and the square of the radius is 6

The area is. 6π5.Finally, the center of the circle is on the horizontal axis, so y is equal to 0, and the center of the circle is in the middle of two parallel lines, and the number is (-1, 0) So, the center of the circle is here.

The radius is equation 6: x 2+(y+1) 2=6