Middle school math questions are more than just answers. I want the process

1.The take-in point (3,0) is found m=3 so the function is y=-x*2+2x+3

is the intersection of the function and the x-axis, so that y=0 finds x to the left of point b.

3.Both triangles have the same base edge ab, that is, the height is the same, then we know that the ordinate of d is the same as c, so that x=0 of c(0,3) Here, y=3 can be brought in to solve the coordinates of the other x value is the point d.

In the first problem, substituting a into a function can solve m=3

The second problem y=0, solve the equation x, x=-1 or 3 b(-1,0).

In the third question, two triangles have the same bottom, so as long as the height is equal, the area is equal, and the symmetry point (0,-3) of c and c about the x-axis is a straight line parallel to the x-axis, and there are three intersection points in the parabola, so d has three,

Solution: (1) If the point d is perpendicular to the x-axis, then the parabolic property can be obtained o,c symmetry with respect to de.

Therefore, the coordinates of point C are (0,2).

The parabola passes (0,0) and (0,2), and the analytic equation (2) trapezoidal area = 3 (root number 3) 2

Since the point P is the moving point on the line segment CD (which does not coincide with the points C and D), then the area of the quadrilateral ADPO = triangle ADO area + triangle DPO area triangle ADO area = (root number 3) 2 = 1 3 trapezoidal area can be obtained from the title The area of the triangle OPC = 2 1 Let the distance from P to the X axis be h (the intersection point of P perpendicular to the X axis is m) and the area of the triangle OPC = H = (root number 3) 2

Then, from the similarity of the triangle PMC and the triangle DHC (H is the intersection point perpendicular to the x-axis), the abscissa of P is 1 2 + 1 = 3 2

Therefore, the coordinates of point p are (3 2, (root number 3) 2).

3) The point q should be at the intersection of x=-(root number 3) 2y straight line and parabola x=- (root number 3) 2y straight line is obtained from the relationship between angles (4) I'm sorry, I have something to do, I can't help think about it, I hope these can help you, come on, you're very smart, you can definitely think of it.

1. 2(x＋6y)－8（x－y）－4－4y=2x+12y-8x+8y-4-4y= -6x+16y-4= -2(3x-8y)-4

The value of 3x 8y is 2010

So: -2(3x-8y)-4= -2*2010-4= -40242At 7:45 p.m., the angle between the hour and minute hands of the clock is.

Minute hand = 270 degrees, clock 19 o'clock integer angle = 360 12 * (19-12) = 30 * 7 = 210, 45 minutes clock angle = 30 * 45 60 = 30 * 3 4 =

At 19:45, the hour hand angle = 210+ degrees.

At 7:45 p.m., the angle between the hour and minute hands of the clock is = degrees.

Answer: 1. The following formula is given by 3x 8y=2010 generations:

2﹙x＋6y﹚－8﹙x－y﹚－4－4y

2x＋12y－8x＋8y－4－4y

6x＋16y－4

2﹙3x－8y﹚－4

2. The speed of the hour hand = 360 12 = 30°.

The speed of the minute hand = 360° 60 = 6° minutes.

19:45 is equivalent to 7:45.

The hour hand travels from 0:00:30° 7 45 60 = the minute hand moves from 0:00:6° 45=270°

Included angle = 270°

Hello, question 1: We can algebraically formulate: -6x+16y-4=(-2) (3x 8y)-4=(-)2010-4=-4024

Question 2: Because there is a 45 minute, the hour hand does not refer to the position at 7 o'clock, but is close to 8 o'clock, 45 accounts for 3 4 of 60, so the difference between the hour hand and 8 o'clock is (1 4) 30=, so the angle is.

Pure hand hitting, looking.

1.The first question is simplified: 2x+12y-8x+8y-4-4y-6x+16y-4

2(3x+8y)-4

2.A clock can be seen as a circle, the sum of the inner angles is 360 degrees, there are a total of 12 scales, 1 scale is 30 degrees, and the hour hand is close to 8 o'clock, the angle is (1-45 60)*30 degrees = degrees, so it is 30 degrees + degrees = degrees.

1。Simplify: 2x+12y-8x+8y-4-4y6x+16y-4

Because 3x-8y=2010. So (3x-8y)*-2=2010*(-2) i.e.: -6x+16y=-4020

6x+16y-4=-4020-4

So algebraic equation 2(x 6y) 8(x y) 4 4y=-4024

The first question will not be the second question.

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The intersection of the diagonal lines of the rectangular o, connecting the PO, the S triangle OCD = S triangle POC + PCD, equal area method.

Obviously, the triangle ABC and the triangle BCD are congruent, and the triangle DEP and the triangle CFP are similar to the triangle ABC, then there is EP DP=3 5, FP PC=3 5, so PE+PF=3 5(DP+PC)=12 5

We're not in a hurry, you drink tea and wait.

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