Given that the solution of the equation AX + B = 2 (2x + 7) + 1 about X is any rational number, find the value of a and B

Given that the solution of the equation AX + B = 2 (2x + 7) + 1 about X is any rational number, find the value of a and B

ax+b=4x+15
(a-4)x=15-b
∵solution to any rational number
∴a-4=0
15-b=0
∴a=4
b=15

Given that m and N are rational numbers, the equation x2 + MX + n = 0 has a root of 5 − 2, then the value of M + n is___ ．

Substituting 5-2 into the equations are: (5-2) 2 + m (5-2) + n = 0, 9-45 + m5-2m + n = 0, (9-2m + n) + (M-4) 5 = 0, ∵ m, n is a rational number, ∵ 9-2m + n = 0m-4 = 0. Solving the equations, m = 4N = - 1 ∵ m + n = 3

An isosceles right triangle with a waist length of 8. The area of this triangle is () square decimeters
The area of the two triangles A and B is equal. The bottom of a is twice that of B. the height of the bottom edge of a is equal to that of B. the height of the bottom edge of a is equal to that of B. the height of the bottom edge of B is equal to that of a, B and C

1. S △ & # 189; × 8.5 × 8.5 = 36.125 square decimeters
The area of this triangle is (36.125) square decimeters
2. The area of the two triangles is equal, and the bottom of a is twice that of B,
The height on the bottom edge of a is the height on the bottom edge of B (B, half)

It is known that hyperbola x ^ 2-y ^ 2 = 1, a is the right focus, BC is on the left branch, and triangle ABC is an equilateral triangle

Hyperbola x ^ 2-y ^ 2 = 1, a is the right focus, BC is on the left branch, and triangle ABC is an equilateral triangle,
Let B (x1, Y1), then C (x1, - Y1) a (√ 2,0) X10 Y1 = √ (x1 ^ 2-1)
The midpoint D of BC is on the x-axis,
|AD|=√2-x1
|BD|=y1 |AD|=√3*|BD|
√3*√(x1^2-1)=(√2-x1)
2x1^2+2√2x1-5=0 x1=-[√2/2]±√3 x1

The derivative of y = ln (x power of 5 / SiNx)

y=ln(5^x)-ln(sinx)
=xln5-ln(sinx),
y'=1*ln5-cosx/sinx
=ln5-cotx.

Fill in the appropriate reduplication as it is
Quiet (quiet)
Wet ()

Wet, poor me. I'll be broke if I don't give it any more

(1) In the same coordinate system, draw the image of the following two functions, Y1 = 2x + 1, y2 = 1-x (2) mark their intersection coordinates and their intersection coordinates with x-axis and y-axis respectively in the coordinate system. (3) calculate the triangle area between them and x-axis

(3) Point B (1,0), point d (- 0.5,0), point C (0,1)
Then the area of △ DBC is equal to: 1 / 2 × | - 0.5-1 | × 1 = 0.75

Let f (x-m) = the square of X - 2x + 3. If f (x) is a decreasing function from negative infinity to 3, then the value range of M is?

F (x-m) = the square of X - 2x + 3
Then f (x) = (x + m) ^ 2-2 (x + m) + 3
f(x)=x^2+(2m-2)x+m^2-2m+3
The axis of symmetry is x = (2m-2) / 2 = M-1
If f (x) is a decreasing function from negative infinity to 3, then the symmetry axis of the function image is on the right side of x = 3
So M-1 > = 3
m>=4

Given that the two roots of the quadratic equation x2-5x-6 = 0 are X1 and X2, then X12 + X22=______ ．

∵ x1, X2 are the two real roots of the equation x2-5x-6 = 0. Then ∵ X12 + X22 = X12 + X22 + 2x1x2-2x1x2 = (x1 + x2) 2-2x1x2 substitute X1 + x2 = 5, x1 · x2 = - 6 to get X12 + X22 = (x1 + x2) 2-2x1x2 = 52-2 × (- 6) = 37

As shown in the figure, △ a = 50 ° in △ ABC, points E and F are on AB and AC, and △ AEF is folded inward along EF to get △ def, then ∠ 1 + 2 in the figure is equal to ()
A. 130°B. 120°C. 100°D. 65°

∵∠ a = 50 °, ∵ AEF + ∠ AFE = 180 ° - 50 ° = 130 °, ∵ fold △ AEF inward along EF to get △ def, ∵ AED + ∠ AFD = 2 (∵ AEF + ∠ AFE) = 2 × 130 ° = 260 °, ∵ 1 + ∠ 2 = 180 °× 2-260 ° = 360 ° - 260 ° = 100 °. So select C