If the side area of a cylinder with a square cross section is s, then the volume of the cylinder is small Ha ha! Give him the best answer in one minute!

If the side area of a cylinder with a square cross section is s, then the volume of the cylinder is small Ha ha! Give him the best answer in one minute!

The answer is 4 π s * root sign S. because the analysis in the title shows that the side area is a square, so the perimeter and height of the bottom surface are equal, which are root sign S. you can calculate the radius of the bottom surface as root sign S / 2 π, then calculate the area of the bottom surface, and then multiply it by the height

If the cross-sectional area of the cylinder axis is a square of 4, then the volume of the cylinder is

Diameter = height = √ 4 = 2
Radius = 2 △ 2 = 1
Base area = π X1 & # 178; = π
Volume = bottom area x height = π x2 = 2 π

If log (2) 3 = (1-A) / A, then log (3) 12
RT for detailed explanation

log(3)12 =log(3)(4*3) =log(3)4+log(3)3 =log(3)2^2+1 =2log(3)2+1 =2*[a/(1-a)]+1 =(1+a)/(1-a).

How can matlab solve the equation of higher order represented by letters? Because the equation has log function and the bottom is of higher order, the solution function can not be solved. Is there any other way?

There is no analytic solution to the equation, so we can only use fzero function to find a numerical solution

Matlab solves the equation log (x) = Tan (x),

syms x
f = log(x) - tan(x);
X=solve(f,'x');

4X * x + 2x * x-kx + K-4 can be divisible by 2x, where k is a constant
*Representative ride

Divisible by 2x means that the constant term is 0, that is, K-4 = 0
K must be equal to 4 if it exists

log(a-1) (2x-1)-log(a-1)(x-1)>0

log(2x-1)-log(x-1)>0
log(2x-1)>log(x-1)
① When 0

Given the function f (x) = loga (x + 1), G (x) = loga (4-2x) (a > 0, and a ≠ 1); (I) find the domain of definition of function y = f (x) - G (x); (II) find the range of values of X that make the value of function y = f (x) - G (x) positive

(I) we can get x + 1 ＞ 04 − 2x ＞ 0, the solution is - 1 ＜ x ＜ 2, we can get the definition domain of function f (x) is (- 1,2); (II) f (x) = f (x) - G (x) = log a (x + 1) - log a (4-2x) = log a & nbsp; X + 14 − 2x, & nbsp; when a ＞ 1, we can get the solution from x + 14 − 2x ＞ 1 − 1 ＜ x ＜ 2

Sin π / 12 with log base 2 plus cos π / 12 with log base 2
fast

Analysis: log (2, sin π / 12) + log (2, cos π / 12)
=log(2,sinπ/12*cosπ/12)
=log[2,sinπ/6/2]
=log(2,1/4)
=-2

Given 0 〈α 〈π / 2, then log (sin α) (1 / (1-cos ^ 2 α)) =? = = RT

log(sinα)(1/(1-cos^2α))=log(sinα)(1/(sin^2α)=log(sinα)(sinα)^(-2)==-2