Use the equation to solve: there is a right triangle with an area of 8 square centimeters. The length of one right triangle is 5 meters. What is the length of the other?

The other is x cm
8=500x÷2
x=0.032㎝

The equation of the circle whose center is on the line 2x + y = 0 and tangent to the point (2, - 1) with the line x + Y-1 = 0 is______ ．

Let the center coordinates of the circle be (a, b), then 2A + B = 0|a + B − 1|2 = & nbsp; (a − 2) 2 + (B + 1) 2, the solution is a = 1, B = - 2, so r = 2, so the equation of the circle is (x-1) 2 + (y + 2) 2 = 2

Factorization 1: (1) X3 + 3x2-4; (2) x4-11x2y2 + Y2; (3) X3 + 9x2 + 26x + 24; (4) x4-12x + 323
The second question of factorization: (1) (2x2-3x + 1) 2-22x2 + 33x-1; (2) X4 + 7X3 + 14x2 + 7x + 1; (3) (x + y) 3 + 2XY (1-x-y) - 1; (4) (x + 3) (x2-1) (x + 5) - 20 the above two questions are factorization and require complete process

Love fashion

1990/19901991+1990/19911992+.+1990/1999*2000
Using the simplified method

Because 1990 / 1990 * 1991 = 1990 * (1 / 1990-1 / 1991), others and so on, after extracting 1990, every addend term is decomposed into two terms, and the following addend term can be added and subtracted to offset, so there are some problems
1990*(1/1990-1/1991+1/1991-1/1992+...+1/1999-1/2000)
=1990(1/1990-1/2000)
=1/200

The ellipse equation is the square of X parts of the square of a + the square of Y parts of 9 = 1, and its two focuses are F1 and F2 respectively, if | F1F2 | = 8,
If string AB passes through F1, the circumference of triangle Abf1 is

If the string AB passes through F 1, we should find the circumference of △ ABF 2, or if the string AB passes through F 2, we should find the circumference of △ ABF 1
The focal length of the ellipse | F1F2 | = 2C = 8, | C = 4
∵ the elliptic equation is: X & # 178; / A & # 178; + Y & # 178; / 3 & # 178; = 1
∴ b=3
Because the half focal length C = 4 > b, the focus is on the x-axis
From C & # 178; = A & # 178; - B & # 178;, we get a & # 178; = C & # 178; + B & # 178; = 16 + 9 = 25, # a = 5
According to the definition of ellipse: the sum of the distances between the points on the ellipse and the two focal points = 2A
That is | BF1 | + | BF2 | = | AF1 | + | af2 | = 2A
The circumference of abf2 = | BF1 | + | BF2 | + | AF1 | + | af2 | = 4A = 20

The product of three prime numbers is 1001. What are they?
They = these three numbers!

1001 decomposition quality factor
1001=7*11*13
7+11+13=31
The sum of these three primes is 31

Let a and B be constants and the solution set of the inequality X / A + 1 / b > 0 be x greater than 1 / 5, then what is the solution set of the inequality bx-a > 0 about x?

x/a+1/b>0
x/a>-1/b
A > 0 x > - A / B is obtained
-a/b=1/5
So B0
bx>a
x

Given the function f (x) = 1 / x, what is f (- 1) equal to

Solve the following system of linear equations of three variables: 1. {X: Y: z = 1:2:3,2x + y-3z = 15
2.{2x+y+z=-26,x+2y+z-30,x+y+2z=-28.

x: Y: z = 1:2:3 (1) 2x + y-3z = 15 (2) let x = t, y = 2T, z = 3T, substitute (2) formula to get: 2T + 2t-9t = 15-5t = 15t = - 3, so x = - 3Y = - 6Z = - 9 the second correct topic is: 2x + y + Z = - 26 (1) x + 2Y + Z = 30 (2) x + y + 2Z = - 28. (3) (1) + (2) + (3) formula to get: 3x + 3Y + 3Z =

In MATLAB, how to find the maximum value of the function of two independent variables (urgent! 1)
Please give an example

Requirement: Using MATLAB to solve binary function y = f (x1, x2)
=The maximum value of (339-0.01 * x1-0.003 * x2) * X1 + (399-0.004 * x1-0.01 * x2) * X2 - (400000 + 195 * X1 + 225 * x2)
Steps: 1. Symsx1x2;
2. Y = (339-0.01 * x1-0.003 * x2) * X1 + (399-0.004 * x1-0.01 * x2) * X2 - (400000 + 195 * X1 + 225 * x2) gives y = - 195 * X1 - 225 * X2 - X1 * (x1 / 100 + (3 * x2) / 1000 - 339) - x2 * (x1 / 250 + x2 / 100 - 399) - 400000
3. If y = simple (y), y = - X1 ^ 2 / 100 - (7 * X1 * x2) / 1000 + 144 * X1 - x2 ^ 2 / 100 + 174 * X2 - 400000
4. The partial derivative dydx1 = diff (y, x1) gives dydx1 = 144 - (7 * x2) / 1000 - X1 / 50, dydx2 = diff (y, x2) gives dydx2 = 174 - x2 / 50 - (7 * x1) / 1000
5. Let the partial derivative be equal to 0, and solve the equation s = solve (dydx1, dydx2) to get s = X1: [1x1 sym] x2: [1x1 sym]
6. The ans of s.x1 is 554000 / 117, and the ans of s.x2 is 824000 / 117
7. Substitute the result into the original f (x1, x2) and find the maximum y value
Y = subs (y, x1554000 / 117); y = subs (y, x2824000 / 117), y = 5.5364e + 005
Of course, whether the maximum is the real maximum still needs to be further verified in combination with the actual situation (by drawing a graph and observing the sign of the second derivative, etc.). For example, symsx1x2; y = (339-0.01 * x1-0.003 * x2) * X1 + (399-0.004 * x1-0.01 * x2) * X2 - (400000 + 195 * X1 + 225 * x2); then use ezsurf (y, [0 10000], [0 10000]) to get the three-dimensional graph, as shown in the figure below:
It can be seen that in the region of interest, the function has the maximum value, that is, y = 5.5364e + 005, which is obtained at X1 = 554000 / 117, X2 = 824000 / 117

Use the equation to solve: there is a right triangle with an area of 8 square centimeters. The length of one right triangle is 5 meters. What is the length of the other?