Please write all subsets of {1,2,3,4} by enumeration, and how many are there

Please write all subsets of {1,2,3,4} by enumeration, and how many are there

There are 2 ^ 4 = 16, as follows
empty set
{1}
{2}
{3}
{4}
{1、2}
{1、3}
{1、4}
{2、3}
{2、4}
{3、4}
{1、2、3}
{1、2、4}
{1、3、4}
{2、3、4}
{1、2、3、4}
I would like to ask the triangle perimeter formula and triangle area formula, with instructions,
Suppose that the three sides of the triangle are a, B and C, respectively. The perimeter of the triangle is l and the area is s
Then the perimeter L = a + B + C
The area formula of triangle
(1) S △ = 1 / 2ah (a is the bottom of the triangle, h is the height of the bottom) (2) s △ = 1 / 2acsinb = 1 / 2bcsina = 1 / 2absinc (three angles are ∠ a ∠ B ∠ C, opposite sides are a, B, C, see trigonometric function) (3) s △ = √ [P (P-A) (P-B) (P-C)] [P = 1 / 2 (a + B + C)] (hailun qinjiushao formula) (4) s △ = ABC / (4R) (R is the radius of circumscribed circle) (5) s △ = 1 / 2 (a + B + C) r (R is the radius of inscribed circle) (6). | a B 1 | s △ = 1 / 2 | C D 1 | E F 1 | [| a B 1 | C D 1 |. | E F 1 | is a third-order determinant. The triangle ABC is in the plane rectangular coordinate system a (a, b), B (C, d), C (E, f). Here, it is best to select ABC from the upper right corner in counterclockwise order, because the results obtained in this way are generally positive. If we do not follow this rule, we may get negative values, but we only need to take absolute values, (7) s △ = C ^ 2sinasinb / 2Sin (a + b)
=(1 / 2) * bottom * height
S = (1 / 2) * a * b * sinc (C is the angle between a and b)
Perimeter of triangle = sum of three sides = a + B + C
Triangle area = (1 / 2) * bottom * height = ah / 2
If the circumference of the triangle af1b is 16, the eccentricity of the ellipse e = √ 3 /
If the circumference of the triangle af1b is 16 and the eccentricity of the ellipse e = √ 3 / 2, the standard equation of the ellipse can be obtained
It is known that F1 and F2 are the two focal points of the ellipse. If the circumference of the triangle af1b is 16 and the eccentricity of the ellipse e = √ 3 / 2, the standard equation of the ellipse is obtained.
From the meaning of the title, AF1 + BF1 + AB = 16, af2 + AF1 = 2A, BF1 + BF2 = 2A, so the perimeter of the triangle is 4a, so 4A = 16, so a = 4. Because e = C / A, so C = 2 √ 3, so B2 = a2-c2 = 4, so the elliptic equation is x2 / 16 + Y2 / 4 = 1
What is the general formula of the logarithm of the power of the base?
The form of power with single item and different base on both sides of the equation: F (x) = BG (x) a the form of power with the same base: - Take logarithm on both sides, change the equation into logarithm on both sides, f integral equation (x) LG a = g (x) LG B
It is known that function f (x) is a positive proportional function and function g (x) is an inverse proportional function, and f (1) = 1, G (1) = 1. (1) find f (x), G (x); (2) prove that function s (x) = XF (x) + G (12) is an increasing function on (0, + ∞)
(1) Let f (x) = ax, ∵ f (x) be a positive proportional function and f (1) = 1 ∵ a = 1, f (x) = x, let g (x) = BX ∵ function g (x) be an inverse proportional function, G (1) = 1 ∵ B = 1, G (x) = 1x (2) s (x) = XF (x) + G (12) = x2 + 2, then s ′ (x) = 2x in (0, + ∞) s ′ (x) = 2x > 0, so s (x) = XF (x) + G (12) is an increasing function in (0, + ∞)
F1F2 is the left and right focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1. The line L passing through the fixed point m (0.2) intersects the ellipse at two different points AB, and the angle AOB is an acute angle
F1F2 is the range where the left and right focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1 passes through the fixed point m (0.2), the line L and the ellipse intersect at two different points AB, and the angle AOB is the acute angle, and O is the origin to find the slope of L
According to cosine theorem: cos ∠ AOB = (OA ^ 2 + ob ^ 2-AB ^ 2) / 2oa * ob ∠ AOB is an acute angle, then cos ∠ AOB > 0, then OA ^ 2 + ob ^ 2-AB ^ 2 > 0, let a (x1, Y1), B (X2, Y2) let the linear equation be y = KX + 2, then X1 + x2 = - 16K / (4K ^ 2 + 1) x1x2 = 12 / (4K ^ 2 + 1) and OA ^ 2
With 1 / 3 as the base, what is the logarithm of 2? In other words, what is the power of 1 / 3 to 2?
With 1 / 3 as the base, what is the logarithm of 2? In other words, what is the power of 1 / 3 to 2
(1/3)^x=2
x=-(Log2/Log3)
It is known that the function f (x) is a positive proportion function, and the function g (x) is an inverse proportion function, and f (1) = 1, G (1) = 2. (1) find the function f (x) and G (x); & nbsp; & nbsp; & nbsp; (2) judge the parity of the function f (x) + G (x)
(1) Let f (x) = K1X, G (x) = k2x, where k1k2 ≠ 0, ∵ f (1) = 1, G (1) = 2, ∵ K1 × 1 = 1, K21 = 2, ∪ K1 = 1, K2 = 2, ∪ f (x) = x, G (x) = 2x; (2) let H (x) = f (x) + G (x), then H (x) = x + 2x, ∪ (0, +)
It is known that the focal point of an ellipse is F1 (- 1,0), F2 (1,0), P is a point on the ellipse, and | F1F2 | is the mean difference term of | Pf1 | and | PF2 |. The equation of an ellipse is obtained
Let the elliptic equation be (X & sup2 / A & sup2;) + (Y & sup2 / B & sup2;) = 1, (a > b > 0). From the problem, we know that | Pf1 | + | PF2 | = 2A = 2 | F1F2 | = 2 × 2. = = = = = > A = 2. C = 1, | B & sup2; = A & sup2; - C & sup2; = 3. The elliptic equation is (X & sup2 / 4) + (Y & sup2 / 3) = 1
It is known that f (x) is equal to the power X of the piecewise function 2 (x is greater than 1) and the logarithm (x + 3) (x is greater than - 1 but less than 1) with a base of a, which satisfies the following conditions: F (x) exists for any X1 not equal to x2
1) If - f (x2) divided by x1-x2 is greater than zero, then the value range of a is?
Satisfy f (x) for any x 1 not equal to x 2
1) - f (x2) divided by x1-x2 is greater than zero
That is, f (x) increases monotonically
So a > 1 and log (a) (1 + 3) < 2 ^ 1 = log (2) 4
∴a＞2
So the value range of a is (2, + ∞)