Let f (x, y) = sin (x + y), then f (0, XY) = ()

Let f (x, y) = sin (x + y), then f (0, XY) = ()

Let f (x, y) = sin (x + y), then f (0, XY) = (sinxy)
It should be sin0 + sinsy = 0 + sinxy = sinxy

The set of all the discontinuous points of the function z = sin (XY) / (1-x ^ 2-y ^ 2)

1-x^2-y^2=0
x^2+y^2=1
The set of all discontinuities = {(x, y) | x ^ 2 + y ^ 2 = 1}

Quadratic function y = x ^ 2 + X + 1, ∵ B ^ 2-4ac =, ∵ there is an intersection between function image and X axis

b^2-4ac=-3
There are 0 intersections between function image and X axis

In which cases can the multiplier be omitted

Numbers, letters, between letters, as long as omitted does not affect the original meaning on the line

Application of discriminant for roots of quadratic equation with one variable
It is known that the equation x & sup2; - (M + 1) X-6 = 0 has a root which is twice the root of the equation x & sup2; - MX + 3 = 0, and one of the two equations guards the opposite number. To find the value of the real number m (a detailed process is required)

Let two roots of X & sup2; - (M + 1) X-6 = 0 be 2a and B, X & sup2; - MX + 3 = 0, two roots of a and - B be substituted into the equation: 4A ^ 2-2a (M + 1) - 6 = 0A ^ 2-AM + 3 = 0, B ^ 2-B (M + 1) - 6 = 0b ^ 2 + MB + 3 = 0. Solve the first equation: am = a ^ 2 + 3, substitute 4A ^ 2-2am-2a-6 = 04A ^ 2-2a-6-2a-6 = 0A ^ 2-a-6 = 0

Linear regression equation of one variable

Steps: 1. Column calculation table, find ∑ x, ∑ XX, ∑ y, ∑ YY, ∑ XY. 2. Calculate LXX, lyy, Lxy, LXX = ∑ (x-x ˇ) (x-x ˇ) lyy = ∑ (Y-Y ˇ) (Y-Y ˇ) Lxy = ∑ (x-x ˇ) (Y-Y ˇ) 3. Find correlation coefficient and test; r = Lxy / (LXX lyy) 1 / 2

Ask about a definite integral problem
The upper limit is 1 and the lower limit is 0
How can I ask?
thank you

Let x = sin (a) a be the 3 / 2 power of [0, PI / 2] (2 / 3) ∫ (1-x ^ 2), DX = (2 / 3) ∫ cos ^ 3 (a) DSIn (a) = (2 / 3) ∫ cos ^ 4 (a) Da = (2 / 3) ∫ cos ^ 4 (a) Da = (2 / 3) ∫ [(1 + cos2a) / 2] ^ 2da = (2 / 3) ∫ (1 + cos ^ 2 (2a) + 2cos (2a)) / 4da = (2 / 3) ∫ 1 / 4D

A ribbon, the first time to use the full length of 25, the second time to use 14 meters, then the remaining ribbon length is exactly 13. How long is the original ribbon?

14 (31 + 3 − 25) = 14 × 207 = 40 (m) a: this ribbon used to be 40 meters long

If P is a point on the ellipse x2100 + y264 = 1, F1 and F2 are the focus, and ∠ f1pf2 = 60 °, then the area of △ f1pf2 is___ ．

Let | Pf1 | = D1, and | PF2 | = D2, then D1 + D2 = 2A = 20. In the triangle pf1f2, | F1F2 | 2 = D12 + d22-2d1d2cos60 °, that is, 122 = D12 + d22-d1d2 = (D1 + D2) 2-3d1d2c = 400-3d1d2  D1D2 = 2563  s △ f1pf2 = 12d1d22sin60 ° = 6433

The top and bottom of trapezoid are 1.5 times of the bottom, 18 cm high and 540 square cm in area

Solution 1
Area = (upper bottom + lower bottom) × height / 2
Upper bottom + lower bottom = 2 × area / height = 2 × 540 / 18 = 60 (CM)
Bottom: 60 / (1.5 + 1) = 24 (CM)
Upper sole: 24 × 1.5 = 36 (CM)
Solution 2
If the lower bottom is x, the upper bottom is 1.5x
Lie equation
（X+1.5X）×18/2＝540
Solution
X＝24
1.5X＝36