Take any two numbers P and Q (P ≠ q) from the four numbers 2, 3, 4 and 5 to form the functions Y1 = px-2 and y2 = x + Q, so that the intersection of the two function images is on the left side of the straight line x = 2, then such ordered array (P, q) has () A. 4 groups B. 5 groups C. 6 groups D. uncertain

Take any two numbers P and Q (P ≠ q) from the four numbers 2, 3, 4 and 5 to form the functions Y1 = px-2 and y2 = x + Q, so that the intersection of the two function images is on the left side of the straight line x = 2, then such ordered array (P, q) has () A. 4 groups B. 5 groups C. 6 groups D. uncertain

Let px-2 = x + Q, the solution is x = q + 2p − 1, because the intersection point is on the left side of the line x = 2, that is, q + 2p − 1 ＜ 2, then q ＜ 2p-4 is sorted out. By substituting P = 2, 3, 4, 5 respectively, the corresponding value of Q can be obtained. The ordinal number pairs are (4, 2), (4, 3), (5, 2), (5, 3), (5, 4), (5, 5), and because P ≠ Q, so (5, 5) is rounded off, and there are 5 pairs that satisfy the condition

It is known that the two intersections of the circle x ^ 2 + y ^ 2 = 1 and the X axis are a and B. if the moving point P in the circle makes PA, Po and Pb into an equal ratio sequence, then the value range of the scalar product of PA vector and Pb vector is obtained

A. The coordinates of B are (- 1,0) and (1,0), let P (x, y), vector PA = (- X-1, - y), vector Pb (- x + 1, - y), PA · Pb = x & # - 178; + Y & # - 178; - 1, ∵ PA, Po, Pb form an equal ratio sequence, then √ (x - 1) &# - 178; + Y & # - 178; × √ (x + 1) &# - 178; + Y & # - 178; = x & # + Y & # - 178;
Ψ X & # 178; = y & # 178; + 0.5 ∵ P in the circle ｜ X & # 178; + Y & # 178; - 1 < 0, X & # 178; + Y & # 178; - 1 = 2Y & # 178; - 0.5 ≥ - 0.5
5 ≤ dot product ＜ 0

Given x ^ 2-3x + 1 = 0, find the value of √ x2 + 1 / x2-2
If we know that the square of x minus 3x plus 1 equals 0, we can find the square of X under the root sign plus 1 / 2 of x minus 2

Divide X by the condition left and right to get x + 1 / x = 3
Square: x2 + 1 / x2 + 2 = 9
So x2 + 1 / x2-2 = 5
The root sign is 5

Given the parallelepiped ABCD-A ′ B ′ C ′ d ', ab = 5, ad = 3, AA ′ = 7, ∠ bad = 60 ° and ∠ baa ′ = ∠ DAA ′ = 45 °, the length of AC ′ can be obtained

All are expressed as vectors
AC '= AB + BC + CC' = radical (AB + BC + CC ') ^ 2
=Radical [AB ^ 2 + BC ^ 2 + CC '^ 2 + ab · BC + BC · CC' + ab · CC ')
=Radical [25 + 9 + 49 + 5 * 3 * cos60 + 3 * 7 * cos45 + 5 * 7 * cos45]
=Radical [83 + 15 / 2 + 28 √ 2]
Check to see if the calculation is wrong. It should be simplified

How to calculate 6 / x + 6 / x + 5 = 1, the square of 7x-x = 30,

1： 6/x+6/（x+5)=1? 6/x+6/(x+5)=16(x+5)/[x(x+5)]+6x/[x(x+5)]=1 [6(x+5)+6x]/x(x+5)=1 (12x+30)/x(x+5)=1 x²-7x=0 ...

Simple math problem 1 in Senior High School
1. Evaluation: Tana / 2tanb / 2 + tanb / 2tanc / 2 + Tanc / Tana / 2 =?
2.(1+tan1°)（1+tan2°）.（1+tan44）=?
3. Tan70 ° cos10 ° + radical (3) sin10 ° tan70 ° - 2cos40 ° =?
Online. Thank you

Tanb / 2 = Tan (π - A-C) / 2 = Tan [π / 2 - (a + C) / 2] = cot (a + C) / 2 = (1-tana / 2 * Tanc / 2) / (Tana / 2 + Tanc / 2) so Tana / 2tanb / 2 + tanb / 2tanc / 2 + Tana / 2tanc / 2 = tanb / 2 (Tana / 2 + Tanc / 2) + Tana / 2tanc / 2 = [(1-tana / 2 * Tanc / 2) / (Tana / 2 + Tanc / 2)] * (ta

One out of 100, nine inches plus one inch equals one foot

Advance by an inch

The bottom of a parallelogram is 4.5cm, the height of the bottom is 4cm, and the area of the parallelogram is () square cm
The base of a parallelogram is 4.5cm, the height of the bottom is 4cm, the area of the parallelogram is () square cm, and the area of the triangle with the same base and height is ()

Area of parallelogram: 4.5 × 4 = 18 square centimeter
Triangle area: 4.5 × 4 △ 2 = 9 square centimeter

40% X-1 / 4 = 7 / 12
40%X-1/4=7/12

40% x = 7 / 12 + 1 / 4 = 10 / 12 gives x = 25 / 12

Find the tangent and normal equation of the curve y = (X & # 178; - 3x + 6) / X & # 178; at x = 3

This point is (3,2 / 3)
y'=[(2x-3)x^2-(x^2-3x+6)*2x]/x^4=(3x-12)/x^3
Substituting x = 3, the tangent slope is - 1 / 9
The tangent is Y-2 / 3 = - 1 / 9 * (x-3)
The results show that x + 9y-9 = 0
The normal slope is the negative reciprocal of the tangent slope, with a value of 9,
The normal is Y-2 / 3 = 9 (x-3)
The result is y = 9x-79 / 3