It is known that F1 and F2 are two focal points of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0). Take the edge of line F1F2 as equilateral triangle mf1f2, if the midpoint of MF1 is in hyperbola How to do? Thank you

You see, hyperbola has a characteristic, that is, the distance difference between a point on the curve and two focal points is 2a, so it leads a line to F2 through the midpoint Q of MF1. This line is a middle line of the triangle. Because the side length of the triangle is 2c, the length of F1q is C, and the length of QF2 is C + 2A. Therefore, according to Pythagorean, 4c2-c2 = (c + 2a) 2, divide both sides of the equation by A2 to get 2e2 = 4E + 4

The mixed second partial derivatives are equal
For a function of two variables, if the first order partial derivative is differentiable (FX (x, y) is differentiable for y and FY (x, y) is differentiable for x), is the second order mixed partial derivative continuous, that is, the mixed partial reciprocal equal?

The first order partial derivative is differentiable, but the second order mixed partial derivative is not continuous
Counter example: piecewise function, when x ^ 2 + y ^ 2 ≠ 0, f (x, y) = XY (x ^ 2-y ^ 2) / (x ^ 2 + y ^ 2); when x = y = 0, f (x, y) = 0
If the second mixed partial derivatives are continuous, the second mixed partial derivatives are equal

One kilometer road is finished in three days on average. How much is it finished the next day?
Be right

One third of the total distance, that is 1000 / 3M, 333.33m

As shown in the figure, points a (m, M + 1) and B (M + 3, m-1) are all on the image of inverse scale function y = KX. (1) find the value of M and K; (2) if M is a point on the x-axis, n is a point on the y-axis, and the quadrilateral with points a, B, m and N as vertices is a parallelogram, try to find the functional expression of straight line Mn

(1) From the meaning of the question, we can see that m (M + 1) = (M + 3) (m-1), the solution is m = 3, (2 points)  a (3,4), B (6,2),  k = 4 × 3 = 12; (3 points) (2) there are two cases, as shown in the figure: ① when m point is on the positive half axis of X axis and N point is on the positive half axis of Y axis, let M1 point coordinate be (x1,0), N1 point coordinate be (0, Y1),  quadrilateral an1m1b be a parallelogram,  Line segment n1m1 can be seen as the result of line segment AB moving 3 units to the left and then 2 units to the down (also can be seen as the result of moving 2 units to the down and then 3 units to the left). From (1), we know that the coordinates of point a are (3,4), the coordinates of point B are (6,2), the coordinates of point N1 are (0,4-2), namely N1 (0,2), and the coordinates of point M1 are (6-3,0), namely M1 (3,0), (4 points) Let the function expression of m1n1 be y = K1X + 2, substituting x = 3, y = 0, the solution is K1 = − 23, the function expression of m1n1 is y = − 23x + 2; (5 points) ② when m point is on the negative half axis of X axis and N point is on the negative half axis of Y axis, let M2 point coordinate be (X2, 0), N2 point coordinate be (0, Y2), ∫ ab ∥ n1m1, ab ∥ m2n2, ab = n1m1, ab = m2n2, ∥ n1m1 ∥ m2n2, n1m1 = m2n2, The quadrilateral n1m2n2m1 is a parallelogram, the points M1 and M2 are centrosymmetric with the line segments N1 and N2 about the origin o, the coordinates of M2 are (- 3,0), the coordinates of N2 are (0, - 2), (6 points). Let the function expression of the straight line m2n2 be y = k2x-2, substitute x = - 3, y = 0, and the solution is K2 = - 23, and the function expression of the straight line m2n2 is y = - 23x − 2. Therefore, the function expression of the straight line Mn is y = 23x + 2 or y = − 23x − 2. (7 points)

The known function f (x) = x square - 2mx + 3
(1) If the range of F (x) on the interval [0,2] is [- 2,3], find the value of M
On the first floor, when 2 > m > 0, isn't it out of range?

I'll make it clear:
f(x)=x^-2mx+m^2-m^2+3
=(x-m)^2-m^2+3
The axis of symmetry is x = m with the opening upward
Substituting x = 0, x = 2 into f (x), we get
F (0) = 3, f (2) = 7-4m, indicating that the maximum value appears at x = 0, so m > 0
Then x = 2 may appear on the left or right side of the axis of symmetry
(1) Let x = 2 be on the left side of the axis of symmetry
m> 2, then x = 2 has the minimum value, 7-4m = - 2 does not meet the meaning of M = 9 / 4 = 2.25
(2) Let x = 2 be on the right side of the axis of symmetry
When 2 > m > 0, the minimum value appears at x = m, that is, 3-m ^ 2 = - 2, M = ± √ 5
Because m > 0, so m = √ 5, contradiction, exclusion
Thank you for your reminding so as not to leave wrong answers

What is the size relationship between A2 + B2 and 2Ab?
a. B is any real number

a2+b2-2ab=(a-b)²>0
a2+b2≥2ab

How to draw circle in CAD axonometric drawing
Use the ellipse command to draw a circle on the x-z plane: (3D view / Southwest isometric)
////////////////////////////////////////////////////////
Command: ellipse / / start ellipse command
Specify the end point of the ellipse axis or [arc (a) / center point]: C / / enter the center point command
Specify the center point of the ellipse:_ From / / start the capture from command
Base point: / / snap a line
: 21 / / move the mouse up and enter the offset distance
Specify the end of the axis: 9 / / enter the end of the axis
Specify another half axis length or [rotate]: 9 / / enter another half axis length
/////////////////////////////////////////////////
The circle I drew turned out to be in X-Y coordinates,
This is from Page 103
Is there any other way to draw a circle in an axonometric drawing

Axis side and 3D coordinate are not the same thing
Axis side refers to drawing a plane diagram on XY plane (only XY plane, 2D plane) with the technique of tilting 30 degrees, so that it looks like a three-dimensional diagram, while 3D creates a three-dimensional entity, so make it clear first
You're not talking about the axis side, but about making a circle on the XZ plane in 3D, right
Enter the UCS carriage return
Enter x enter (rotate axis along X axis)
Enter 90 (specify rotation angle) enter
The XY plane of the newly generated coordinate is the x-z plane of the old coordinate
At this point, you can make a circle (circle is OK, you don't need to make an ellipse), just do it on the XY plane of the current coordinate, of course, that is, the XZ plane before modifying the coordinate has been changed
After drawing, input UCS and press enter twice to restore the coordinates to the default coordinates
About the axis side drawing circle:
1. Start the axis side environment: right click object capture / settings to open the sketch settings toolbar;
The increment angle is set to 30 degrees;
Point capture and grid deletion are set to equal axis side capture;
OK, close
2. Click the ellipse command and select the equiaxed side circle option (I) enter;
Choose the center of the circle. You don't have to teach it
F5 key can switch to simulate XY XZ YZ plane in 3D coordinate, making the drawing on the "plane" you want to draw
You can select the center of the circle (don't input the radius first) and press F5 continuously to figure out the function of F5. It's easy to learn
OK

How much is 3 × 40

40 times root 3

There is a rectangle. If the length remains unchanged and the width is changed to 14 meters, the area will increase by 40 square meters. If the width is changed to 17 meters, the area will increase by 100 square meters

If the length is x, then 14x-40 = 17x-100
We can get x = 20
So the original length is 20 meters, width = (20 * 14-40) / 20 = 12 meters, and the original area is 20 * 12 = 240 square meters

Let vector group A: A1, A2 Am is linearly independent, and vector B1 can be expressed linearly by vector group A,
Vector B2 can not be expressed linearly by vector group A. It is proved that M + 1 vectors A1, A2 Am, Lb1 + B2 must be linearly independent

To the contrary:
hypothesis
a1,a2……… Am, Lb1 + B2 linear correlation
be
There exists x1,..., XM such that
lb1+b2=x1a1+...+xmam.(1)
Also known
Vector B1 can be expressed linearly by vector group A
be
There exists Y1,..., YM such that
b1=y1a1+...+ymam.(2)
(1) - (2) * l
b2=(x1-ly1)a1+...+(xm-lym)am
have to
B2 can be expressed linearly by vector group A
Conflict with the known
So, M + 1 vectors A1, A2 Am, Lb1 + B2 must be linearly independent

It is known that F1 and F2 are two focal points of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0). Take the edge of line F1F2 as equilateral triangle mf1f2, if the midpoint of MF1 is in hyperbola How to do? Thank you