Senior 1 math problem to specific step 20

Vector A Vector B

a| *b|*cos pi 3 = 1*2*1 2 = 1 vector a+n vector b and n vector a-vector b are obtuse angles.

So. Vector a+n vector b*n vector a-vector b < 0n(|a|^2 - b|^2 ) n^2 - 1) |a| *b|* cos pi/3 < 0

3n + n^2 - 1) <0

Solve n and you're done! I believe that the next steps will not be difficult for you.

Senior 1 math problems should be specific steps.

Reward Points: 20 - 7 days and 15 hours until the end of the question5Known |a|=1，|b|=2, the angle between vector a and vector b is 31).Find the vector a vector b

2）.The angle between vector a+n vector b and n vector a-vector b is an obtuse angle, and the value range of the real number n is obtained.

Vector A Vector B

a| *b|*cos pi 3 = 1*2*1 2 = 1 vector a+n vector b and n vector a-vector b are obtuse angles.

So. Vector a+n vector b*n vector a-vector b < 0n(|a|^2 - b|^2 ) n^2 - 1) |a| *b|* cos pi/3 < 0

3n + n^2 - 1) <0

Vector A Vector Ba|*|b|cosα

3 So: vector a vector b=1*2*

Solution: It is known that y increases with the increase of x, and the minimum value is obtained when x 3, that is, -3=2*3 +a*3+b

When x 4 is used to obtain the maximum value, that is, 6=2*4 +a*4+b, we can get a=-5 and b=-6 by concatenating the two equations

So the analytic formula is y=x -5x-6

The process of solving the equation is -3=2*3 +a*3+b 6=2*4 +a*4+b

4a+b=-26 ④

Obtained by - , a=-5

Substituting a=-5 into yields: b=-6

Because, on [3,4] y increases as x increases.

So, when x=3, y=-3, when x=4, y=6, substituting into the function, the solution is, a=-5, b=-6

[Analysis: (1) It is easy to know that the straight line l is separated from the circle o, so any point a on the straight line can make two tangent lines am and an, where the points m and n are tangent points. When man 60, oam=(1 2) man 30

At this time, the combination of numbers and shapes can be seen that there must be a point b, satisfying oab=30Let the distance from the line ab to the center of the circle o be d, then 0 d 4 3===>oa=2d≤8/3.

2) Since the point a is on the straight line l, the point a(x,(8-x) 3).∴x²+[8-x)/3]²≤8/3)².===>0≤x≤8/5.】

OAB = 30°, AB should be at least tangent to the circle O in the case of the most extreme case, at this time OA = 2 OB = 8 3

Let point A be the abscissa x and the ordinate (8-x) 3

oa²=x²+(8-x)²/9=64/9

Simplified to 10x -16x=0, x1=0, or 8 5, so the abscissa range is [0,8 5].

oa²=x²+(8-x)²/9=64/9

Simplify to 10x -16x=0, x1=0, or 8 5

So the abscissa range is [0,8 5].

Because m=,n=,so the subset of m is ,,, the subset of n is, .

So p=,n=, so p q is the intersection of p and q——, i.e. p q= ,

2 or empty set (with symbols).

1.If p=1, q=2, then the empty set.

2.If p=2 and q=2, then 2

3.If p=1, q=3, then the empty set.

4.If p=2, q=3, then the empty set.

Look... It's all typed out by yourself.。。

p={nullset, q={emptyset, p q={emptyset,}

Here the elements in p and q are sets.

Empty set or 2 Hee hee give me a share See other people's so detailed shame I'm lazy wow.

Solution: Composed of b={x|4x+p=0} gives x=-p 4, and because b a, a={x|x -1 or x 2}, so -p 4 -1 or -p 4 2, the solution is p -8 or p 4, i.e., p = {p|p -8 or p 4}

Solution: a=, b== when b is included in a:

There is: -p 4 -1

So: p 4

For the problem of very 2b, the left and right real parts and imaginary parts correspond to the same, x-y=0, 2x-3y=1, and the solution is x=y=-1

1. Find the probability that A will get exactly three points on three shots, and only one shot will be made three times c3(1)1 3(1-1 3)(1-1 3)=4 9 2. Suppose A shoots once and B shoots twice, let x be the difference between A's score on this shot minus the sum of B's two shots, and find the distribution column of the random variable x. A has 0 or 3 points (0 points, 2 3 points, 3 points, 1 3) B may get 0 points or 3 points, or 6 points (0 points, 9 16 points, 3 points, 6 16 points, 1 16 points, so the value of x is 0, -3 points, -6 points, 3 points. 0 points, A 0 points, B 0 points, A 3 points, B 3 points, 2 3 * 9 16 + 1 3 * 6 16 = 1 2 -3 points, A 0 points, B 3 points, A 3 points, B 6 points, 2 3 * 6 16 + 1 3 * 1 16 = 13 48 -6 points, A 0 points, B 6 points, 2 3 * 1 16 = 1 24 3 points, A 3 points, B 0 points 1 3 * 9 16 = 3 16