Simple math in the first year of high school, 10 minutes, to the process, good points are given

1, sin508=sin(508-360)=sin148 sine is a subtraction function in the second quadrant, so sin144 is larger.

2, cos760=cos(760-720)=cos40cos(-770)=cos(720-770)=cos-50 cosine is negative in the first quadrant and positive in the fourth quadrant.

So cos(-770) is big.

sin508＜sin144

Because sin508=sin148 sin144cos760 is greater than cos(-770).

Because cos(-770)=cos770=cos50cos760=cos40

COs40 is greater than COs50

sin508=sin(360+148)=sin148Because sinx is a subtraction function at (90,180), sin144>sin148

cos760=cos(360*2+40)=cos40cos(-770)=cos(-360*2-50)=cos(-50)=cos50

Since cosx is a subtraction function at (0,90), cos40 > cos50

The meaning of the title is unclear. Is the ratio a root number 3?

If so, the solution is as follows:

The square of the distance from the point m to the straight line y=3 is: d1 2=(y-3) The square of the distance from the point m to the point f is: d2 2=x 2+(y-1) 2 is known by the hailstorm: d1 d2=root number 3

So. d1^2/

d2^2=3

sin508=sin(360+148)=sin148 is located in the second quadrant. So sin148 is less than sin144

So sin508 is less than sin144

cos760=cos（720+40）=cos40cos(-770)=cos（-720-50）=cos（-50）=cos310

Because cos40 is in the first quadrant.

COS310 is located in quadrant 4.

So cos310 is smaller than cos40 so.

COS760 is greater than COS (-770).

As shown in the figure, take the midpoint E of BC, connect De, make the projection O of point A on the plane BCD, connect AO and CO, cross the point Q for QP AO, and QP intersect the plane BCD at the point P, and connect CP.

Because the edges of tetrahedra a-bcd are equal in length, the tetrahedron is a regular triangular pyramid, and each face is an equilateral triangle, because the point o is the projection of point a on the plane bcd, so ao plane bcd, point o is on de and is the center of the equilateral bcd, and because qp ao, so qp plane bcd, and the point p is on de, qcp is the plane angle formed by cq and plane bcd, because point e is the midpoint of bc, so de bc, let the length of each edge be x, then it is equilateral In BCD, BE=CE=X 2, DE=(3)X 2, EO=(3)X 6, then CO=(3)X 3, because AO plane BCD, CO is on the plane BCD, so AO Co, then AO=(6)X 3 is calculated in the right angle AOC, because the point Q is the midpoint of AD, QP AO, so CQ=(3)X 2, QP is the median line of AOD, we can know QP=AO2=(6)X 6, so SIN QCP=QP CQ=[(6)X 6] [(3) x 2]=( 2) 3, that is, the sine value of the angle between CQ and the plane BCD is ( 2 ) 3.

The key is to prove that PQ is the line of the vertical bottom that we want to do.

Proof of thirds.

Proof of how the coordinates of the last equinox came about.

Points a, b, and c are not collinear to form a triangle.

That is, the case where the vectors ab and ac are parallel are excluded.

The absolute value of the difference between the two sides is the sum of the two sides.

The vector coordinates of the vectors ab, ac, and bc are calculated by vector calculation, and then the length of ab, ac, and bc is obtained.

According to the absolute value of the difference between the two sides, the sum of the two sides of a side gives x,y satisfies the condition.

The three sides are not parallel and not common points The sum of any two sides is greater than the third side.

5。If we know that the solution set of a= is, then a b=?

Solution: ax +bx+c=a(x+1)(x-2)=a(x-x-2)=ax -ax-2a>0, so a<0, and b=-a, c=-2a;

Substituting the expression of b yields b====

Hence a b =

b = with and only one element, then a b = ?

Solution: b==b= has and only one element;

Its discriminant formula =a -4=0, i.e., a = 2;That is, b = and a=, that is, for any x there is ax + (a-1)x+a-1 0, so there must be a<0;

And its discriminant formula =(a-1) -4a(a-1)=-3a +2a+1=-(3a -2a-1)=-(3a+1)(a-1) 0, i.e., there is (3a+1)(a-1) 0

Therefore a=== then a b= =.

Question 1: As can be seen from the solution set of a, a is definitely less than 0, and -1 and 2 are the two roots of a=0, bringing -1 and 2 into a=0 gives that b=-a, c=-2a. Then bring a, b, and c into b, and divide both sides by a at the same time (since a < 0, greater than 0 must become less than 0) to solve the solution set of b is -2, so anb=-1 The second problem: from b has and only one solution, it can be concluded that b=(x+a 2) 2-a 2 4+1=0 has only one solution, then a 2 4-1 = 0, so a = plus or minus 2. If a is true to a real number, it can be concluded that a must be less than 0

So a=-2i.e. anb = -2 (note that a and b are a set of a).Hope it helps.

In question 6, there is a set of b to know that a is equal to 2

Substituting A, verifying, can get the answer,