If loga (1 / 2)

If loga (1 / 2)

0

∫∫ Ye ^ (XY) DXDY, where D is a planar region bounded by the curves xy = 1 and x = 1, x = 2, and y = 2

∫∫D ye^(xy) dσ
= ∫(1→2) dx ∫(1/x→2) ye^(xy) dy
= ∫(1→2) (2x - 1)/x² • e^(2x) dx
= [(1/x) • e^(2x)] |(1→2)
= (1/2)e⁴ - e²
= (1/2)(e² - 2)e²

If LG2 = A and Lg3 = B, then LG12 / LG15 is equal to?
RT

LG12＝LG3＋LG2＋LG2
LG15＝LG3＋1－LG2
So the result is a + 2B / A + 1-b

Can we find the maximum value of LNX / X by using lobita's law,

f(x)=lnx/x
Domain x > 0
f'(x)=(1-lnx)/x^2
Let f '(x) = 0
So LNX = 1
So x = e is the maximum point
Substituting f (x)
Then f (x) max = f (E) = 1 / E

The ellipse with (- 3.0) (3.0) as the focus is tangent to the straight line X-Y + 9 = 0

The equation of ellipse is: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1; the straight line is X-Y + 9 = 0, which can be substituted into the equation to get: x ^ 2 / A ^ 2 + (x + 9) ^ 2 / b ^ 2 = 1, which can be reduced to: (1 / A ^ 2 + 1 / b ^ 2) x ^ 2 + 18x / b ^ 2-1 + 81 / b ^ 2 = 0. Because the relationship between the two is tangent, the equation has a unique solution, so the discriminant = 0, that is: 18 ^ 2A ^ 2-4 (a ^ 2 + B ^ 2) (8

A is a prime, and the sum of a + 8 and a + 14 is prime. What is the minimum of a?

The minimum of a is 3, 3 + 8 = 11, 3 + 14 = 17, all prime numbers

We know that logab = logba, (a > 0, b > 0 and a ≠ 1, B ≠ 1), and prove that a = B or a = 1b

∵ logab = logba ∵ from the bottom changing formula, we can get: lgblga = lgalgb, that is, (LGA) 2 = (LGB) 2, ∵ LGA = LGB or LGA = - LGB, from the operation property of logarithm, we can get: a = B or a = 1b

First list, and then draw the function y = 6 / X in figure 14-3
Answer if you know

(1) By y = 2x-1
When y = 0: x = 1 / 2
∴A（1/2,0）
When x = 0: y = - 1
∴B（0,-1）
The line connecting two points AB is y = 2x-1
(2) By y = 6 / X
x=2,y=3.∴C（2,3）
x=3,y=2,∴D（3,2）
The hyperbola through CD is y = 6 / X
Hi, give me a picture

If the distance between two points a and B representing X and - 1 on the number axis is equal to 2, calculate the value of X

X = 1 or x = - 3

It is known that the probability density of two-dimensional random variables is
Find the distribution function of (1) constant and (2)

First question:
From the normalization of two-dimensional random distribution a = 2,
The function solution of F (x, y) is to integrate the density function of two-dimensional random distribution. The integration regions are (- ∞, x) and (- ∞, y). The results are shown in the picture
Second question:
The solution is the same as the first question
A=1/π
The probability is 1 / 3