Find the equation of the line parallel to the line 3x + 4y-12 = 0 and passing through the point a (2, - 3) Solve the linear equation of the intersection of two lines X + Y-3 = 0 and 2x-y = 0 and perpendicular to the line 3x-4y-12 = 0

Find the equation of the line parallel to the line 3x + 4y-12 = 0 and passing through the point a (2, - 3) Solve the linear equation of the intersection of two lines X + Y-3 = 0 and 2x-y = 0 and perpendicular to the line 3x-4y-12 = 0

3x+4y-12=0
=》4y=-3x+12
=》y=-3/4x+3
Let the equation of line passing through point a be y = KX + B
Because parallel 3x + 4y-12 = 0, k = - 3 / 4
Bring (2, - 3) into
-3=-3/4*2+b
-6=-3+2b
2b=-3
b=-3/2
The equation passing through point a is y = - 3 / 4x-3 / 2

The number of zeros of function f (x) = - x ^ 2 + x-lgx?

There are two. Draw a graph (let y = x ^ 2 + X; Z = lgx) to find the intersection of two functions. You can find that there are two

Known parabola y = MX ^ 2 + (3-2m) x + m-2 (M is not equal to 0)
Judge whether P (1,1) is on the parabola

Substituting x = 1, y = 1
Right = m + 3-2m + m-2 = 1
So x = 1, y = 1
So p is on a parabola

As shown in the figure, the parabola y = - x2 + BX + C passes through point a (2,0), the axis of symmetry is y-axis, and the vertex is p
(1) Find the expression of the parabola, write out the coordinates of its vertex P, and draw its general image;
2) Translate the parabola m units to the right and then M units to the down (M > 0). Note that the vertex of the new parabola is B and the intersection of the new parabola and the Y axis is C
(3) The coordinates of point B and point C are expressed by the algebraic expression of M. if ∠ OBC = 45 °, try to find the value of M
Question 1.2 may not be answered. Main question 3

Point B (m, 4-m)
∵ point C is on the Y axis
∴X=0
Ψ set point C (0, y)
Let the analytic expression of the function be y = - (x-m) ² + 4-m
Point C is on this parabola
∴-m²+4-m=y
｜ point C (0, - M & # 178; + 4-m)

How much is 1 times half plus 2 times one-third plus 3 times one-quarter all the way to 49 times 50?

-One by half plus two by one-third plus three by one-quarter all the way up to 49 by 50/
=1-1/2+1/2-1/3+1/3-1/4+…… +1/50
=1-1/50
=49/50

P = the quadratic power of a + the quadratic power of 5B - 4AB + 2B + 100, find the minimum value of P

p=a²＋5b²－4ab＋2b＋100
=(a²－4ab＋4b²)＋(b²＋2b＋1)＋99
=(a－2b)²＋(b＋1)²＋99
Then the minimum value is 99

Find the solution of quadratic equation with one variable: (9 - 4 √ 5) x & # 178; - (2 - √ 5) x - 2 = 0,

The original equation can be reduced to: (2 - √ 5) &# 178; X & # 178; - (2 - √ 5) x - 2 = 0
The factorization is as follows
[(2-√5)x -2]*[(2-√5)x +1]=0
The solution is: x = 2 / (2 - √ 5) = - 2 (√ 5 + 2) or x = - 1 / (2 - √ 5) = 2 + √ 5

If the polynomials about X & nbsp; X4 + (A-1) X3 + 5x2 - (B + 3) X-1 do not contain X3 and X terms, the values of a and B are obtained

∵ the polynomial X4 + (A-1) X3 + 5x2 - (B + 3) X-1 of X does not contain X3 term and X term, ∵ A-1 = 0, B + 3 = 0, ∵ a = 1, B = - 3. So the value of a is 1, and the value of B is - 3

Who has a question about one variable linear equation in Mathematics in Volume 1 of junior high school
Five questions altogether

In 2011, the total amount of water for production and operation and domestic use in Beijing was 5.8 B cubic meters, and 3 of which were more than 0.6 B cubic meters. How many B cubic meters of water for domestic use and production and operation

It is proved that if an orthogonal matrix is a positive definite matrix, then it must be an identity matrix

Realize that the eigenvalue of an orthogonal matrix is 1 or - 1
Then the matrix is positive definite and the eigenvalues are all 1
If AX = ax, a is the eigenvalue and X is the eigenvector, then transpose x'a '= ax' on both sides. Then x'a'ax = ax'ax
Because a is orthogonal, the left side is x'x, and the right side is AAX'x, so a = 1, and the characteristic root is 1 or - 1
Because a is symmetric and positive definite, there exist orthogonal matrices B and b'ab which are upper triangles (in fact, it can be further known from the orthogonality of a that they are diagonal. We only need to consider AA '= a'a), and they are eigenvalues on the diagonal. It can be seen that the sufficient and necessary condition for positive definite is that the eigenvalues are all positive numbers, and this problem is all 1
So a orthogonal is similar to the unit matrix, that is, the upper b'ab = E. so a = BB '= e (Note b orthogonal)