The square of 1. / 2x + Y / + (X-Y + 9) = 0 Note: represents absolute value 2. Equations: x + Y / 2 + X-Y / 3 = 2 and X + Y / 2-x-y / 3 = - 4 Note: represents the fractional line

The square of 1. / 2x + Y / + (X-Y + 9) = 0 Note: represents absolute value 2. Equations: x + Y / 2 + X-Y / 3 = 2 and X + Y / 2-x-y / 3 = - 4 Note: represents the fractional line

In the first problem, because the sum of the two square terms of / 2x + Y /, (X-Y + 9) is equal to 0, and the two terms are non negative, so the two terms are equal to 0, that is, 2x + y = 0, X-Y + 9 = 0

Throwing arc y = - 1 / 10x square - 9 / 2x + C passing through point (2, - 12), C=

Let x = 2, y = - 12 generations equation,
I have to,
-12=-1/(10*4）-9/2*2+c,
c=-389/40

Let Ω be a closed space bounded by a cone z = root (x ^ 2 + y ^ 2) and a hemispherical z = (R ^ 2-x ^ 2-y ^ 2) ^ (1 / 2)
If ∑ is the outside of the whole boundary of Ω, then ∫ ∫ (subscript ∑) xdydz + ydzdx + zdxdy=________ The answer is (2 - (radical 2) / 4) π R ^ 3
Ask for detailed explanation

Just use Gauss theorem directly
Original integral = ∫ (subscript is ∑) xdydz + ydzdx + zdxdy = ∫ ∫ (1 + 1 + 1) DV
=3∫∫∫dxdydz
=3∫(0->2π)dθ ∫(0->π/4)dφ ∫(0->R) rdr
=3π^2R^2/4

What's the speed of 70 yards per hour? Is that 70 kilometers per hour?

Late Nebula level 4
Well, it's different
This building is just a fool
70 times 1.61
About 112 kilometers!

1.135°43′16″-57°18′28″=

135°43′16″-57°18′28″
=135°42'76''-57°18'28''
=78°24'48''

As shown in the figure, the eccentricity of the ellipse is ()
A. 12b. 33c. 32D

∵ suppose the diameter of the bottom surface of the cylinder is D, the section is 30 ° to the bottom surface, the length of the short axis of the ellipse is D, the length of the long axis of the ellipse is 2A = dcos30 ° = according to C = A2 − B2, and the half focal length of the ellipse is C = (33d) 2 − (D2) 2 = 36d, then the eccentricity of the ellipse is e = CA = 12

The length of the rectangle is 2A + B, and the width is A-B smaller than the length. Calculate the area of the rectangle

Length: 2A + B
Width: 2A + B - (a-b), namely a + 2B
Rectangular area: (2a + b) (a + 2b) = 2A & # 178; + 5ab + 2B & # 178;
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Factorization of 2x ^ 3 + x ^ 2-6x-3

2x^3+x^2-6x-3
=x^2×（2x+1）-3×（2x+1）
=（2x+1）（x^2-3）
=（2x+1）（x-√3）（x+√3）

RT triangle ABC area is 8cm square; find the minimum perimeter of the triangle
Well It's a concrete process,

Let the two right sides of RT triangle ABC be a cm and B cm respectively
Then AB = 16
The perimeter of the triangle is: a + B + √ (a ^ 2 + B ^ 2)
》2√(ab)+√(2ab)
=2*√16+√(2*16)
=8+4√2.
So the minimum perimeter of the triangle is 8 + 4 √ 2

Prime numbers, a, B, C satisfy a ^ 4 + B ^ 4 + C ^ 4-3 = P to find all the values of P
P is also a prime number

If P is even and prime, then p = 2. Obviously, it doesn't hold. Because a ^ 4 + B ^ 4 + C ^ 4 must be greater than 5