How do I do this math problem in the second year of junior high school?

p to x-axis distance = 3, then p has an ordinate of 3 or -3

Therefore, let y=3 or -3, and solve the corresponding x value respectively to find the coordinates of p.

3x+2=3

3x=1x=-1/3

3x+2=-3

3x=-5x=5/3

So the coordinates of p are (-1 3,3) or (5 3,-3).

Analysis: If the distance from the point p to the x-axis is 3, then the ordinate of the point p is +3 or -3.

When y=+3, substituting the linear equation gives x=-1 3, so the point p is (-1 3,3).

When y=-3, substitute the linear equation to obtain x=5 3, so the point p coordinate is (5 3, -3).

Let y=-3x+2 in y=3 or -3

Bring in the function. Solving.

Solution p(-1 3,3) or (5 3,-3).

There are two scenarios.

1. When y=3, find the coordinates.

2. When y=-3, find the coordinates.

The distance from point p to x = 3, i.e. y = 3, substitution, x = -1 3

p（-1/3，3）

The straight line mn:y=x-1 intersects the x-axis at the point m(1,0) and the y-axis at the point n(0,-1).

Through A(3,2) as the perpendicular line of the X-axis, the perpendicular foot is the point B(3,0), the bisector of AMB mc:y=(2-1)(X-1) crosses BC:Y=X-3 in C(3+ 2, 2), and intersects AB in D(3,2 2-2).

Find a little p(p,0) on the x-axis so that cdp is an isosceles triangle with cp as the waist, then.

pc=pd or pc=cd, so (3+ 2-p) 2+2=(3-p) 2+(2 2-2) 2, or (3+ 2-p) 2+2=2+( 2-2) 2,2 2(3-p)+4=12-8 2, or 3+ 2-p=soil(2-2), the solution is p1=7-2 2, p2=1+2 2, p3=5

So p(7-2 2,0), (1+2 2,0), (5,0).

The difference in height between the two trees is 20-4=16 meters.

The distance is 12 meters.

By the Pythagorean theorem.

The distance between the two birds (12 +20 )=20 meters.

A: It takes 5 seconds.

The distance between the bird and its partner is [12 2 + (20-4) 2] square root = 20 m, which takes 20 4 = 5 seconds.

No. I really don't understand that this kind of topic has a bird effect on the future life practice.

I haven't done math problems for a long time, and I can't write the process.

Well, this ......I'm only in sixth grade.

That's right, what kind of question are you talking about?

If you are satisfied, please click Accept as a satisfactory answer.

x+88>400 (number of students).

y (price per piece) column: xy=1936

y=1936/x

x=352y=There are 352 students in the ninth grade of the school.

Verification (352+88).

There are x ninth-grade students, and the unit price of stationery bags is y yuan:

x*y=1936

x+88)*(

The second equation is , that is.

The solution is 88*8*y=1936*2 , y=, x=1936

Pass p as line pqLet have: triangle apq is equal to apm (asa) aq=am

Again: The triangle cpn is equal to cpq (asa)cn=cq

So: am+cn=aq+cq=ab