5 / 12 - 1 / 3 = - 1 / 4 (one variable linear equation) the whole process

5 / 12 - 1 / 3 = - 1 / 4 (one variable linear equation) the whole process

5 out of 12 = 1 out of 3-1 out of 4
5 out of 12 = 4 out of 12-3 out of 12
5 out of 12 = 1 out of 12
X = 1 / 12 * 12 / 5
X = 1 / 5
5x of 12 - 1 of 3 = - 1 of 4
5 out of 12 = 1 out of 3 - 1 out of 4
5 / 12 x = 1 / 12
X = one in five

For example: - x = - 2 / 5x + 1

I don't understand. Do you want to write the process format of solving the problem or the general formula of linear equation with one variable -X = - 2 / 5x + 1-x + 2 / 5x = 1-3 / 5x = 1x = 1 × (- 5 / 3) x = - 5 / 3. As for the general formula, it seems to be a function For example: first order function: y = KX + B, second order function: y = ax & sup2; + BX + C

In the process of the clock from 3:30 to 4:30, when does the hour hand coincide with the minute hand? When is the hour hand in a straight line with the minute hand? When is the hour hand at a right angle with the minute hand?
A series of linear equations with one variable

At 3:30, the minute hand points to 180 degrees, and the hour hand points to 90 + 360 △ 12 △ 2 = 105 degrees. When the coincidence angle is equal or different from 360 degrees, if the hour hand turns x degrees and coincides, there will be 105 + x = 180 + 12x or 105 + X = 1

It is known that the lines 2x-y-3 = 0 and 4x-3y-5 = 0 intersect at point p 1. Find the coordinates of point P 2. Find the equation of the line passing through point P whose distance from the origin is equal to 2

2x-y-3 = 0, i.e. y = 2x-3,4x-3y-5 = 0, i.e. y = 4x / 3-5 / 3,2x-3 = 4x / 3-5 / 3, x = 2, so y = 1, P-point coordinate is (2,1), P-point equation is Y-1 = K (X-2), i.e. kx-y - (2k-1) = 0, distance formula from point to straight line is d = | - (2k-1) | / √ (k ^ 2 + 1) = 2, solution is k = - 3 / 4, linear equation is - 3x / 4-y + 5 / 2 = 0

Given the two points P1 (1,3), P2 (5,5), write the algorithm of the vertical equation in p1p2 and draw the corresponding flow chart

1. Calculate the slope of p1p2
2. Calculate the midpoint coordinates of p1p2
3. Calculate the equation of the middle vertical line by using the point oblique formula

Given the symmetric point P1 (B + 1, A-1) of point P (a, b) with respect to line L, then the equation of circle C; X + y-6x-2y = 0 with respect to circle C1 with respect to line L symmetry

Circle C: (x-3) ^ 2 + (Y-1) ^ 2 = 10
Circle C, center O is (3,1), radius is root sign (10)
So let a = 3, B = 1
So the symmetric point of O is (2,2)
So C1 equation is (X-2) ^ 2 + (Y-2) ^ 2 = 10

It is known that the equation of circle C is x2 + y2-6x-2y + 5 = 0. The moving line L passing through point P (2,0) intersects with circle C at two points P1 and P2. Let P1 and P2 be tangent lines L1 and L2 of circle C respectively. Let the intersection of L1 and L2 be m. prove that point m is on a fixed line, and work out the equation of this fixed line

The center of C: (x-3) 2 + (Y-1) 2 = 5 is (3, 1) Let P1 (x1, Y1), P2 (X2, Y2), m (x0, Y0) Because p1m is tangent to circle C, MP1 ⊥ CP1 (4 points) so (x1-x0) (x1-3) + (y1-y0) (y1-1) = 0

If the distances from points P (a-1,5) and P2 (2, B-1) to X axis are equal and p1p2 / / Y axis, then (a + b) is to the power of 2008

Solution
Because they are equidistant from the x-axis
So the values of | y | are equal
So | B-1 | = 5
Solution B = 6 or - 4
And because p1p2 is two points
So B = - 4
P1p2 is parallel to the Y axis
So A-1 = 2, a = 3
So (a + b) ^ (2008) = (3-4) ^ (2008)
=-1^2008
=1

Given the quadratic function y = a (x + 2) ^ 2 + 1.5m, the intersection of image and X axis is P1, P2, and | p1p2 | = 6?
I've forgotten my knowledge. Thank you

y=a（x+2）^2+1.5m
The axis of symmetry is x = - 2,
P1 and P2 are symmetric with respect to x = - 2, and | p1p2 | = 6, P1 and P2 are on the x-axis. We can see that the coordinates of these two points are (- 5,0), (1,0), and,
Substituting y = a (x + 2) ^ 2 + 1.5m, we get
a=(-1/6)m

Given that the intersection of the image of quadratic function y = a (x + 2) ^ 2 + 3 / 2 and X axis is P1, P2, and | p1p2 | = 6, find the value of A
I'm not very good at finding intersections,
Intersection P1 (- 2 + radical - 3 / 2a, 0) P2 (- 2 - radical - 3 / 2a, 0)
a=-1/6

No need
y=ax²+4ax+4a+3/2
By Weida theorem
x1+x2=-4
x1x2=4a+3/2
|P1P2|²=(x1-x2)²
=(x1+x2)²-4x1x2
=16-16a-6
=6²
16a=-26
a=-13/8