Equation x △ 70% = 5 / 6 1.8-18, the difference is 100 times larger. 2 =40% of the number B is 250

Equation x △ 70% = 5 / 6 1.8-18, the difference is 100 times larger. 2 =40% of the number B is 250

(8-18*25%)*100=800-450=350
Four fifths of a = 40% of B. B is 250
A × 4 / 5 = 250 × 40%
A × 4 / 5 = 100
A = 100 × 5 / 4
A = 125

40% of a number is 0.5 more than one third of it. To find this number, we need to solve the equation

Let this number be X
40%X-1/3X=0.5
1/15X=0.5
X=7.5

If the four vertices of the square are on the three sides of the right triangle, and the lengths of the two right sides of the right triangle are 3 and 4 respectively, then the length of the square is?

1. The right vertex of a right triangle is a vertex of a square, and the other three vertices are on the three sides of the triangle. Let the side length of the triangle be x by parallel relation, X / 3 = (4-x) / 4
x=12/7
2. The two vertices of the square are on the hypotenuse, and the other two vertices are on the two right angle sides
Use similar triangles to draw diagrams!
Number: 60 / 37

It is known that F1 and F2 are the two focuses of x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 0 (a > 0, b > 0) on the hyperbola, and P is any point different from the vertex on the right branch of the hyperbola,
O is the origin, the following four propositions
1. The center of the inscribed circle of △ pf1f2 must be on the straight line x = a;
2. The center of the inscribed circle of △ pf1f2 must be on the straight line x = B;
3. The center of the inscribed circle of △ pf1f2 must be on the straight line Op;
4. The center of the inscribed circle of △ pf1f2 must pass through (a, 0)
The number of true proposition is 1,4
Please help me analyze why,

Let the inscribed circle of △ pf1f2 be tangent to points a and B respectively with Pf1 and PF2, and tangent to point m with F1F2, then we can see that | PA | = | Pb |, | F1A | = | f1m |, | f2b | = | F2m |, and point P is on the right branch of the hyperbola, so | Pf1 | - | PF2 | = 2A = 6, so | f1m | - | F2m | = 6, and | f1m | + | F2m | = 213,
Let the coordinates of point m be (x, 0),
Then from | Pf1 | - | PF2 | = 2A = 6, we can get (x + 13) - (13-x) = 6, and the solution is x = 3. Obviously, the line between the center of the inscribed circle and the point m is perpendicular to the X axis,
So the answer is (1) and (4)

How to find the second partial derivative?
I'm stupid,
I don't understand the one in the book

For example, I don't know hi
X ^ 2 * y ^ 2 is the second derivative of X
Take y as a constant, and then find a mediator to get 2 * y ^ 2 * X
Take y as a constant, and then calculate the two mediators to get 2 * y ^ 2

Write the words as they are
It must be in the form of () (),

Suddenly, my heart is like a knife

In the following plane coordinate system, draw the inverse scale function y = 4 / X and y = - 4 / X

Y = 4 / x, y = - 4 / x, take several groups of values of X and y to form a smooth curve
Y = 4 / X over (1,4) (2,2) (3,4 / 3) (4,1)
Y = - 4 / X over (1, - 4) (2, - 2) (3, - 4 / 3) (4, - 1)

The function f (x) = x square - 2 | x | - 3 is known
1. Prove that f (x) is an even function
2. Draw an image of this function
3. Write the monotone interval of this function
fast

1、f(x)=|x|^2-2|x|-3
Because f (- x) = | - x | ^ 2-2 | - x | - 3 = | - x | ^ 2-2 | - 3 = f (x)
So f (x) is even function
2、
3. It can be seen from the figure
The increasing interval of F (x) is [- 1,0], [1, + infinity]
The decreasing interval is [- infinite, - 1], [0,1]

If a = 2008, B = 2009, then the value of the algebraic formula a2-2ab + b2-b + A is equal to

a-b=-1
So A-B + 1 = 0
The original formula = (a-b) & #178; + (a-b)
=(a-b)(a-b+1)
=0

As shown in the figure, point a is inside circle O, point B is outside circle O, and points c and D are on circle O. compare the angle CAD and angle CBD
Tulio

Proof: take any point F in the circle, connect CF, DF,
Let BC intersect at point E and connect De
Because the circular angle of arc CD pair is ∠ F and ∠ CED
Then ∠ f = ∠ CED
Because ∠ CAD > F, ∠ CED > CBD (the outer angle of the triangle is greater than the inner angle not adjacent to it)
So, CAD > CBD