If x and Y in the system of equations 2x − 3Y = k2x + 3Y = 5 are opposite to each other, the value of K is obtained

Add the equation x + y = 0 according to the meaning of the question, then x = - y, substitute this into the second equation to get y = 5, x = - 5, substitute the value of X, y into the first equation to get k = - 25, so k = - 25

It is known that the system of equations 3x-2y = 15, x + 2Y = n and the system of equations x + y = m, 2x-3y = 5 have a common solution to find the value of M and n

3x-2y=15 ,2x-3y=5
9x-6y=45 4x-6y=10
5x=35
x=7
y=3
7+6=n
n=13
7+3=m
m=10

On the problem of finding the intersection point of a straight line and an ellipse. Find the coordinates of the intersection point of a straight line 3x + 10y-25 = 0 and x ^ 2 / 25 + y ^ / 12 = 1
I think the reference book says this:
(1) 3x+10y-25=0,x^2/25+y^2/4=1,
Let x = 5cosa, y = 2sina,
The results show that 15cosa + 20sina-25 = 0,
That is 3cosa + 4sina-5 = 0, cosa = 3 / 5, Sina = 4 / 5,
So the intersection point is (3,8 / 5);
But I don't know how to get cosa = 3 / 5, Sina = 4 / 5 when 3cosa + 4sina-5 = 0

But I don't know how to get cosa = 3 / 5, Sina = 4 / 5 when 3cosa + 4sina-5 = 0
You understand that's parameterizing the actual equation, right? But you don't understand that?
3cosa + 4sina-5 = 0 and the sum of the square of the sine function of the same angle plus the square of the cosine function is equal to 1
Under these two conditions, cosa = 3 / 5, Sina = 4 / 5 (and in parameterization, the default a angle is acute angle)

The sum of formulas for finding the base length of right triangle
The length of hypotenuse of right triangle is 6 meters and 5 meters
How many meters is it? I don't understand.

The Pythagorean theorem says that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the two right angles. So 6 ^ 2-2.5 ^ 2 = bottom ^ 2. Use a calculator to calculate it, and you will get 5.5

As shown in the figure, two objects are stacked on a horizontal table, M1 = 5kg, M2 = 4kg, and the maximum static friction between M1 and M2 FM = 20n,
The dynamic friction coefficient between M2 and horizontal table is μ = 0.1. Try to find: (1) when M2 is subjected to the horizontal tension of F = 36N, what is the acceleration of the two objects? (2) the value range of the horizontal tension acting on M2 should be such that M1 and M2 do not slide relative to each other

(1) Suppose that m1m2 does not slide, that is, it remains relatively stationary and has the same acceleration a
A = f / (M1 + m2) - UG = 3m / S ^ 2 is obtained by integral method
According to Newton's law, the equation is f-u (M1 + m2) G-F = M2A
Substituting into the data solution, f = 15N

Seventh grade mathematics (Part 2) Chapter 1: the operation of integers
I can't find it on the Internet. Or help. Questions in Baidu Library

Volume two

When the car runs along a straight road at a constant speed of 20 m / s, it suddenly brakes. After braking, the car moves at an acceleration of a = - 5 m / S 2. The displacement of the car after 5 s from braking is calculated

Suppose that the time from starting to stopping is t0, then t0 = V − v0a = 0 − 20 − 5S = 4S from v = V0 + at, then the car stops after 4S, and the displacement within 5S is equal to the displacement within 4S of braking, that is to say, there is & nbsp; x = v02t0 = 202 × 4m = 40m A: the displacement of the car from braking to 5S is 40m

If M > 1, am-1 + am + 1-am2 = 0, s2m-1 = 38, then M is equal to ()
A. 38B. 20C. 10D. 9

According to the property of arithmetic sequence, we can get: am-1 + am + 1 = 2am, then am-1 + am + 1-am2 = am (2-AM) = 0, the solution is: am = 0 or am = 2, if am is equal to 0, it is obvious that s2m-1 = (2m − 1) (a1 + A2M − 1) 2 = (2m-1) am = 38 does not hold, so am = 2, s2m-1 = (2m-1) am = 4M-2 = 38, the solution is m = 10

The initial speed VO of the car is 20m / s. after braking, the car moves in a straight line with uniform deceleration, and the acceleration is a = 5m / S2
What is the movement time when the displacement is 15m

The time from deceleration to stop is t = V / 4S. The displacement to stop is s = V ^ 2 / 2A = 400 / 10 = 40m
When the displacement is 15m, s = v0t-0.5at ^ 2,15 = 20t-2.5t ^ 2
t=4-sqrt(10)s

Xiao Ming and Xiao Gang set out from school to a certain place. Xiao Ming's speed is 6 km / h and Xiao Gang's speed is 9 km / h. xiao gang stops for 3 hours, one hour later than Xiao Ming

Let Xiaoming time be x, and the distance be 6x
6X = 9 * (x + 1-3)
X=6
6X=36

If x and Y in the system of equations 2x − 3Y = k2x + 3Y = 5 are opposite to each other, the value of K is obtained