Matlab solution of a binary quadratic equations x = 2.1*v*cosa*t+0.5*w*2.1*t*t y = 2.1*v*sina*t-0.5*85*t*t V and T are unknowns, and others are known expressions for solving V and t It's better to solve it with MATLAB. I just want the result

Matlab solution of a binary quadratic equations x = 2.1*v*cosa*t+0.5*w*2.1*t*t y = 2.1*v*sina*t-0.5*85*t*t V and T are unknowns, and others are known expressions for solving V and t It's better to solve it with MATLAB. I just want the result

Wrong, it should be CLC, clear; Syms x y a w [T, v] = solve ('x = 2.1 * V * cosa * t + 0.5 * w * 2.1 * t * t ','y = 2.1 * V * Sina * t-0.5 * 85 * t * t','t ','v') > > t = 21.0 * (COSA * y - 1.0 * Sina * x) * (- 0.04535147392290249431065759637188 / ((850.0 * co

Solving binary quadratic equations (x + 2) ^ 2 + (Y-1) ^ 2 = 1 x ^ 2 + (Y-2) ^ 2 = 4

There is too little input. Drawing shows that there is an intersection (- 2,2), that is, X1 = - 2, Y1 = 2
The two intersection points are symmetrical with respect to the straight line passing through the centers of two circles
Let y = KX + B and substitute (- 2,1) (0,2)
y=x/2+2
Let two intersection lines y = - 2x + C
Substituting (- 2,2) gives y = - 2x-2
Solve the equations y = - 2x-2, x ^ 2 + (Y-2) ^ 2 = 4
We get x2 = - 1.2, y2 = 0.4

To solve 100 binary linear equations, the problem should be a little simpler, and the number should not be too large
The number is so big, it should be smaller

Why are there two distance formulas √ [(x2-x1) &# 178; + (y2-y1) &# 178;] and √ [(x1-x2) ^ 2 + (y1-y2) ^ 2] what's the difference between them!

The two formulas are the same, (x2-x1) = (x1-x2) ^ 2, (y2-y1) = (y1-y2) ^ 2

A (x1, Y1), B (2 √ 2,5 / 3), C (X2, Y2) are three points on X & # 178; / 9 + Y & # 178; / 25 = 1. The distance between F (0,4) and a, B, C is an arithmetic sequence
Find the value of Y1 + Y2

This topic is actually the second definition of ellipse
So, for this ellipse, because 25 > 9, its focus is on the Y axis, a = 5, B = 3, C = 4
Therefore, we know that point F (0,4) is a focal point of the ellipse. If it is on the positive half axis of the y-axis, then the Quasilinear equation of the ellipse on the positive half axis of the y-axis is y = a ^ 2 / C = 25 / 4
Then for any point P (XP, YP) | PF | = e (a ^ 2 / c-yp) = A-E * YP on the ellipse
The centrifugal ratio is e = C / a = 4 / 5
So for the point ABC, according to the question: | AF | + | CF | = 2 | BF|
That is: A-E * Y1 + A-E * y2 = 2 (A-E * 5 / 3)
That is: Y1 + y2 = 2 * 5 / 3 = 10 / 3
That's it,

In the space rectangular coordinate system, find the distance between a (- 3,2, - 4) and B (- 4,3,1)

d={[(-3)-(-4)]^2 +(2-3)^2 +(-4-1)^2}^0.5
=27^0.5=3*3^0.5

In the space rectangular coordinate system, the distance between points a (1, - 3,0) and B (2,0,4) is?

√(2-1)^2+3^2+4^2=√26

In space rectangular coordinate system, the distance between point a (- 3,4,0) and point B (2, - 1,5) is______ ．

∵ point a (- 3, 4, 0), point B (2, - 1, 5) | the distance between a and B | ab | = (− 3 − 2) 2 + (4 + 1) 2 + (0 − 5) 2 = 75 = 53, so the answer is: 53

How to calculate the linear distance from any given point to the origin in the space rectangular coordinate system

Let the coordinates of known points be (x, y, z) and the distance to the origin be d
D = [(x-0) ^ 2 + (y-0) ^ 2 + (z-0) ^ 2] open root

If a (1,1,1) and B (- 3, - 3, - 3) are known, the distance between point a and point B is______ ．

In the space rectangular coordinate system, a (1, 1, 1), B (- 3, - 3, - 3), then the distance between point a and point B is: (1 + 3) 2 + (1 + 3) 2 + (1 + 3) 2 = 43