As shown in the figure, let p be the moving point on the circle x ^ 2 + y ^ 2 = 25, point d be the projection of P on the x-axis, and m be a point on PD, and | MD | = 4 / 5 | PD |. (1) when P moves on the circle, find the equation of the locus C of point M. (2) find the length of the line segment cut by the line passing through point (3,0) with a slope of 4 / 5

As shown in the figure, let p be the moving point on the circle x ^ 2 + y ^ 2 = 25, point d be the projection of P on the x-axis, and m be a point on PD, and | MD | = 4 / 5 | PD |. (1) when P moves on the circle, find the equation of the locus C of point M. (2) find the length of the line segment cut by the line passing through point (3,0) with a slope of 4 / 5

(I) let the coordinates of M be (x, y) and p be (XP, YP)
It is known that {XP = XYP = 54y}
P is on the circle,
The equation of C is X225 + y216 = 1
(2) The linear equation passing through point (3,0) with slope of 45 is y = 45 (x-3),
Let the intersection of a line and C be a (x1, Y1) B (X2, Y2),
By substituting the linear equation y = 45 (x-3) into the equation of C, we get that X225 + (x-3) 225 = 1, that is, x2-3x-8 = 0 ∧ X1 = 3-412, X2 = 3 + 412,
The length of segment AB is | ab | = (x1-x2) 2 + (y1-y2) 2 = (1 + 1625) 2 (x1-x2) 2
=41•4125=415．

For any positive integer n, please prove whether the value of integral (3N + 1) (3n-1) - (3-N) (3 + n) is a multiple of ten
It's due tomorrow, please!

（3n+1)(3n-1)-(3-n)(3+n)
=9n^2-1-(9-n^2)
=9n^2-1-9+n^2
=10n^2-10
=10 (n ^ 2-1) is a multiple of ten

If the tangent line is drawn from the point P (m, 3) to the circle C: (x + 2) 2 + (y + 2) 2 = 1, then the minimum length of the tangent line is ()
A. 26B. 5C. 26D. 4+2

As shown in the figure, when PA ⊥ X axis, the length of the tangent through P point is the shortest. AQ ⊥ PQ is obtained according to PQ as the tangent of the circle and Q as the tangent point. The center (- 2, - 2) of the circle is obtained from the equation of the circle, and the radius is 1. In the right triangle Apq, AQ = 1, PA = 3 - (- 2) = 5, and PQ = 52 − 1 = 26 according to the Pythagorean theorem

Tolerance D of arithmetic sequence 0, 2, 4, 6 =?

The answer is 2

In the pyramid p-abcd with rectangular bottom, PA ⊥ plane ABCD, PA = AB = 2,. BC = 4. E is the midpoint of PD,
In the pyramid p-abcd with rectangular bottom, PA ⊥ plane ABCD, PA = AB = 2,. BC = 4. E is the midpoint of PD
Finding the sine value of the angle between the straight line CD and the plane AEC

Calculate vd-aec = 1 / 2vc-apd = 1 / 4vp-abcd triangle AEC by volume, AE = 1 / 2PD = radical 5 AC = 2 radical 5 Pa vertical CD ad vertical CD vertical plane APD, so CD vertical pdce ^ 2 = CD ^ 2 + de ^ 2 = 3 make ef vertical AC in triangle AEC, let AF = x CF = 2 radical 5-xef ^ 2 = AE ^ 2-af ^ 2 = CE ^ 2-cf ^ 25 -

When x = 1, the value of the cubic power + QX + 1 of the algebraic expression PX is 2010. When x = - 1, the value of the algebraic expression is obtained

x=1
Then p + Q + 1 = 2010
p+q=2009
So x = - 1
The original formula = - P-Q + 1
=-(p+q)+1
=-2008

Given that the moving points a and B are on the x-axis and y-axis respectively, and satisfy | ab | = 2, the point P is on the line AB, and the vector AP = t times of the vector Pb, t is not 0, find the trajectory of P

Let P (x, y) a (x1,0) B (0, Y1) have X1 & sup2; + Y1 & sup2; = 4
Because the vector AP = t times the vector Pb, so x = X1 / T + 1, y = Ty1 / T + 1
Substitute X1 & sup2; + Y1 & sup2; = 4

Factorization: (square of 1-A) (square of 1-B) - 4AB

﹙1－a²﹚﹙1－b²﹚－4ab
The original formula is 1-B & # 178; - A & # 178; - A & # 178; B & # 178; - 4AB
＝﹙1＋a²b²－2ab﹚－﹙b²＋a²＋2ab﹚
＝﹙1－ab﹚²－﹙a＋b﹚²
＝﹙1－ab＋a＋b﹚﹙1－ab－a－b﹚

The point m.n is the midpoint of the edges BB1 and BC1 of the cube abcd-a1b1c1d1, respectively. The angle of Mn and CD1 is calculated

Because m and N are the midpoint of BB1 and BC1 of cube abcd-a1b1c1d1 respectively, Mn is parallel to c1d1. Because c1d1 is horizontal to a1d1, vertical to dd1, c1d1 intersects dd1 to D1, so a1d1 is vertical to cdc1d1. Because CD1 belongs to cdc1d1, a1d1 is vertical to CD1

1 + 3 = 2 square 3 + 6 = 3 square 6 + 10 = 4 square question the equation containing N expresses the law in n-1 square lattice

1 + 3 = 2 square 3 + 6 = 3 square 6 + 10 = 4 square question the equation containing N expresses the law in n-1 square lattice
1+3=2^2
3+6=3^2
6+10=4^2
10+15=5^2
15+21=6^2
.
1+2+3+...+n,+1+2+3+...+n+1=(n+1)^2
n(n+1)/2+(n+1)(n+2)/2=(n+1)^2
(N-2)(N-1)/2+(N-1)N/2=(N-1)^2