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In order to get the image of function GX = 2cos (2x + π / 3), just how to translate the image of FX = sin (2x + π / 3) to get it? Find the specific steps
GX = 2Sin (2x + π / 3 + π / 2) = 2Sin (2x + 5 / 6 π) = 2Sin [2 (x + 5 / 12tt)] FX = sin (2x + π / 3) = sin [2 (x + 1 / 6tt) ---- left 1 / 4tt ------- the ordinate becomes twice the original
The image of function f (x) = - 2cos (x + π / 4) is shifted a (a > 0) units to the left to get the image of function y = g (x). If the function g (x) is even, the minimum value of a is 0
3π/4
When d ^ 2 + e ^ 2-4f > 0, the quadratic equation x ^ 2 + y ^ 2 + DX + ey + F > 0 is called the general equation of circle
Is that right?
When d ^ 2 + e ^ 2-4f > 0, the quadratic equation x ^ 2 + y ^ 2 + DX + ey + F = 0 is called the general equation of circle. (it was I who marked = as >)
Wrong
First of all, it is not a quadratic equation of two variables
In addition, this is an inequality, not an equation
If the circle x2 + Y2 + DX + ey + F = 0 is symmetric with respect to the line y = x, then
A.D+E=0 B.D-E=0 C.D+F=0 D.D-F=0
The center of x2 + Y2 + DX + ey + F = 0 is (- D / 2, - E / 2), because the circle is symmetrical about y = x, then the center of the circle must be on y = x, that is - D / 2 = - E / 2, so D = e, that is, D-E = 0, the answer is B
If the circle x2 + Y2 + DX + ey + F = 0 is symmetric with respect to the line L1: X-Y + 4 = 0 and the line L2: x + 3Y = 0, then d=______ ，E=______ ．
Let the intersection point of lines L1 and L2 be the center of a known circle. From X − y + 4 = 0x + 3Y = 0, we can get x = − 3Y = 1, the coordinates of the center of a circle are (- 3,1), D2 = - 3 and E2 = 1, then D = 6 and E = - 2. So the answer is: 6; - 2
In the sequence {an}, an = 1 / N (n + 1) (n + 2), find the limit of Sn
When an = 1 / N (n + 1) (n + 2) = [1 / N (n + 1) - 1 / (n + 1) (n + 2)] / 2, A1 = 1 / 6, so S1 = A1 = 1 / 6N > = 2, Sn = A1 + A2 +... + an = [1 / 1 * 2-1 / 2 * 3] / 2 + [1 / 2 * 3-1 / 3 * 4] / 2 +... + [1 / N (n + 1) - 1 / (n + 1) (n + 2)] / 2 = [1 / 2-1 / (n + 1) (n + 2)] / 2, so the limit of Sn is 1 / 4
If the sum of the first n terms of the sequence {an} is Sn, Sn = 1-23an, then an=______ ．
When n ≥ 2, an = sn-sn-1 = (1-23an) - (1-23an-1) = 23an-1-23an, we can get Anan − 1 = 25. So the sequence {an} is an equal ratio sequence with the first term of 35 and the common ratio of 25, so an = 35 × (25) n − 1, so the answer is 35 × (25) n − 1
In the sequence {an}, an = 1 / N (n + 1) (n + 2), then the limit of Sn is
Give me some ideas
An=[1/n(n+1)-1/(n+1)(n+2)]/2=[1/n-1/(n+1)-(1/(n+1)-1/(n+2)]/2
If the sum of the first n terms of an is the nth power of Sn = 0.5 + A, then limsn =?
The answer is - 1. How do I do it? I also do a
The answer is a
This is called an infinite recursive sequence
If the absolute value of a sequence has a limit, then the sequence has a limit
No. for example, a_ When n = (- 1) ^ n, the sequence {a}_ N |} is a constant sequence with limit 1, but the sequence {a}_ n} For - 1,1, - 1,1, - 1,1,..., there is no limit
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