The midpoint o (0,0) and a (2,0) B of the plane rectangular coordinate system are the midpoint of the line OA. Rotate OA clockwise about point O for 30 degrees. Mark the corresponding point of point B as C, and find the location of point C

The midpoint o (0,0) and a (2,0) B of the plane rectangular coordinate system are the midpoint of the line OA. Rotate OA clockwise about point O for 30 degrees. Mark the corresponding point of point B as C, and find the location of point C

First calculate the coordinates of a 'and draw it as (2, - 1), because the square of a plus the square of B = the square of C, the square of OA plus the square of AA' = the square of OA ', so the length of OA' is 5, because the corresponding point of B is C, B is the midpoint of line OA, so C is the midpoint of OA ', so the root of OC is 2 / 5, because the square of OB plus the square of CB = the square of OC (Pythagorean theorem, ditto), So the ordinate of C is (- 0.. 5), and the abscissa is (1) with B, so the coordinate of C is (1, - 0.. 5)
Solving linear equations (1) 2x1-x2 + x3-2x4 = 7 (2) X1 + 2x2-3x3 = - 4 (3) - x1-x2 + X3 + 4x4 = 4 (4) 3x1 + x2-x3-6x4 = 0
The answer is X1 = 3, X2 = - 2, X3 = 1, X4 = 1
The augmented matrix is 2 - 11 - 2712 - 30 - 4 - 1 - 11431 - 1 - 60r1 + 2r3, R2 + R3, R4 + 3r3, R3 * (- 1) 0 - 3361501 - 24011 - 1 - 4 - 40 - 22612r1 + 3R2, R3 - R2, R4 + 2r200 - 3181501 - 24101 - 8 - 400 - 21412r1 * (- 1 / 3), R
As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connecting an and cm intersects at P, try to find the degree of ∠ APM
As shown in the figure, △ Kam ≌ MBC, KM = cm, ∠ AMK = ∠ MCB, because ∠ CMB + ∠ MCB = 90 °, so ∠ CMB + ∠ AMK = 90 °, so △ KMC is isosceles straight
In the binomial (1 + x) ^ m + (1 + 2x) ^ n expansion, the coefficient of X is 11, and the minimum value of x ^ 2 coefficient is obtained
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In the (1 + x) ^ m + (1 + 2x) ^ n expansion, the coefficient of X is:
Cm(1)+2Cn(1)=11
m+2n=11
m/2+n=11/2
m=11-2n
The coefficient of x ^ 2 is:
Cm(2)+2Cn(2)
=m(m-1)/2+n(n-1)
=m^2/2-m/2+n^2-n
=(11-2n)^2/2+n^2-11/2
=3n^2-22n+55
When n = 11 / 3, the above formula is the minimum, but n needs to be an integer, so when n = 4, it is the minimum
Minimum = 3 * 16-88 + 55 = 15
11=m+n*2^(n-1),
If n > = 3, then M + n * 2 ^ (n-1) > = m + 3 * 2 ^ 2 = m + 12 > = 12 > 11.
Therefore,
N = 3, then M + n * 2 ^ (n-1) > = m + 3 * 2 ^ 2 = m + 12 > = 12 > 11.
Therefore,
N
As shown in the figure, in the plane rectangular coordinate system, a (4,2), B (3,0), rotate △ ABO 90 ° counterclockwise around the midpoint C of OA to get △ a ′ B ′ o ', then the coordinate of a ′ is______ .
As shown in the figure, the vertical line of o'b 'intersects Y-axis through a' at point n, the distance from a to ob is 2, and the distance from a 'to o'b' is a'm = 2, so a'n = mn-a'm = ob-a'm = 3-2 = 1. OA = 25, a'c = OC = 5, and OA '= 10 are obtained by Pythagorean theorem. In RT △ oa'n, on = 3, a' (1, 3) are obtained by Pythagorean theorem
Finding the general solution of the system of non homogeneous linear equations 2x1 + 7X2 + 3x3 + X4 = 6 3x1 + 5x2 + 2x3 + 2x4 = 4 9x1 + 4x2 + X3 + 7x4 = 2
Augmented matrix = 2731 63522 249 417 2r3-3r2, r2-r12731 61 - 2 - 1 1 - 20 - 11 - 51 - 10r1-2r20 11 5 - 1101 - 2 - 11 - 20 - 11 - 51 - 10r3 + R1, R1 * (1 / 11), R2 + 2r10 15 / 11 - 1 / 11 10 / 111 0 - 1 / 11 9 / 11 - 2 / 110 000
Write augmented matrix, dissolve into the upper triangular matrix on OK
As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connecting an and cm intersects at P, try to find the degree of ∠ APM
As shown in the figure, △ Kam ≌ MBC, KM = cm, ∠ AMK = ∠ MCB, because ∠ CMB + ∠ MCB = 90 °, so ∠ CMB + ∠ AMK = 90 °, so △ KMC is isosceles straight
If only the binomial coefficient of the fourth term in the expansion of (x2 − 12x) n is the largest, then the sum of all the coefficients in the expansion is zero______ .
According to the meaning of the question, in (x2 − 12x) n, let x = 1, the sum of all coefficients in the expansion is (12) n, and only the fourth binomial coefficient in the expansion of (x2 − 12x) n is the largest, so n = 6. Then the sum of all coefficients in the expansion is (12) 6 = 164; so the answer is 164
Given that the coordinates of point P0 are (1,0), point P1 is obtained by rotating point P0 around origin o in a counter clockwise direction by 600, extending OP1 to point P2 so that op2 = 2op1, and then point P3 is obtained by rotating point P2 around origin o in a counter clockwise direction by 600, then the coordinates of point P3 are
The coordinates of P1 point are (half, half root sign 3) P2 (1, root sign 3), P2 and P3 are symmetrical about y axis, P3 (- 1, root sign 3)
s1/t1=(v1+v2)/2 ,s2/t2=(v2+v3)/2 ,(s1+s2)/(t1+t2)=(v1+v2)/2 ,
Let's look at the question again. Is it wrong,
From S1 / T1 = (V1 + V2) / 2, S2 / T2 = (V2 + V3) / 2
It is impossible to deduce that: (S1 + S2) / (T1 + T2) = (V1 + V2) / 2