Given that the center of the circle X & sup2; + Y & sup2; + DX + ey-4 = 0 is the point C (1, - 2), find the radius r of the circle

Given that the center of the circle X & sup2; + Y & sup2; + DX + ey-4 = 0 is the point C (1, - 2), find the radius r of the circle

x²+y²+Dx+Ey-4=0
=>(x+D/2)²+(y+E/2)²=4+D²/4+E²/4
The center of the circle is (- D / 2, - E / 2), that is, 1 = - D / 2, - 2 = - E / 2
=>D=-2,E=4
The radius R & # 178; = 4 + D & # 178; / 4 + E & # 178; / 4 = 4 + 1 + 4 = 9
=>r=3

If the equation x2 + Y2 + DX + ey + F = 0 represents a circle with (2, - 4) as its center and 4 as its radius, then f=______ ．

From the general equation of the circle, combined with the conditions in the problem, we can get - D2 = 2, - E2 = - 4, 12d2 + E2 − 4f = 4, the solution is d = - 4, e = 8, f = 4, so the answer is 4

Find the derivative of y = | ln (x + 2) |

Let a = {cos n π / 3, n ∈ Z}, B = {sin (2k-1) π / 6, K ∈ Z}. Then a = B, the answer is a = B, why a = B?

π / 3 = 60 ° and N ∈ Z, indicating that n is an integer, that is to say, n π / 3 is always a multiple of 60 degrees
Cos π / 3 = cos 60 ° = 1 / 2, so cos n π / 3 is a multiple of 1 / 2
π / 6 = 30 ° and K ∈ Z, indicating that K is an integer and 2k-1 is an odd number starting from 1, that is, (2k-1) π / 6 is a multiple of 30 °
So sin (2k-1) π / 6 is a multiple of 1 / 2
So a = B

Let P = {cos n π / 3, n ∈ Z}, q = {sin (2k-3) π / 6, K ∈ Z}
Please analyze!

The P period is 2 π / (n / 3) = 6 π / n
Then n = 1 to 6
cos=1/2,-1/2,-1,-1/2,1/2,1
So p = {1 / 2, - 1 / 2, - 1,1}
In the same way
sin(2k-3)π/6
=cos[π/2-(2k-3)π/6]
=cos(π-kπ/3)
=-cos(kπ/3)
Then q is also {1 / 2, - 1 / 2, - 1,1}
So p = Q

When 1. X → 0, LIM (x ^ 3 * (sin1 / x)) =? 2. When x → 1, LIM (x ^ 2-1) cos1 / (x-1) =? 2?
According to the limit algorithm, only when limf (x) and limg (x) exist, can limf (x) * g (x) = limf (x) * limg (x). Please write the specific steps

1. ∵ sin (1 / x) is a bounded function, that is, ∵ sin (1 / x) │≤ m (M > 0 constant) ∵ 0 ≤∵ x ^ 3 * sin (1 / x) │≤ m ∵ x ^ 3 ∵ LIM (x - > 0) (x ^ 3) = 0 ∵ LIM (x - > 0) [M │ x ^ 3] = 0, so LIM (x - > 0) (x ^ 3 * (sin1 / x)) = 0.2. Similarly, LIM (x - > 1) (x ^ 2-1) cos1 / (x-1) = 0

Lim n approaches infinity n ^ 2 * (2-N * sin 2 / N) =?

Let t = 1 / N, T - > 0+
The original formula = LIM (T - > 0) [2 - sin (2t) / T] / T & # 178;
= lim(t->0) [ 2t - sin(2t) ] / t³
=LIM (T - > 0) [2-2 cos (2t)] / (3T & # 178;) lobita rule
= lim(t->0) 4sin(2t) / (6t)
= 8/6 = 4/3

Y = e ^ - cos square x-sin square x, find the second derivative. Lim n →∞ (2 / N ^ 2 + 4 / N ^ 2 + 6 / N ^ 2 +... + 2n / N ^ 2) =?

-cos^2x-sin^2x=-(cos^2x+sin^2x)=-1
So the original formula is y = e ^ - 1 is a constant, the first derivative is 0, and the second derivative is naturally 0
2/n^2 +4/n^2 +6/n^2 + ...+ 2n/n^2
=Sum of the first n terms of 2 (1 + 2 + 3 +... + n) / N ^ 2 arithmetic sequence
=2/n^2 * n(1+n)/2
=n^2+n/n^2
=1+1/n
lim n→∞ (1+1/n)=1

lim(x→∞)[1/n-2/n+3/n-4/n+...+(2n-1)/n-(2n)/n]=?
The answer in the book is - 1
It seems that your answer is wrong
I figured it out to be - 2
On the second floor, I haven't heard of the pinch rule

1. Is n →∞ not (x →∞)
Simplify the original form
=(-1/N)*N=-1

Infinity: Lim ∑ (1-cos (A / N)) n = 1

lim ∑ （1-cos(a/n)）＝0
Wrong