If two circles x2 + y2 = m and X2 + Y2 + 6x-8y-11 = 0 have a common point, then the value range of real number m is () A. （-∞，1）B. （121，+∞）C. [1，121]D. （1，121）

If two circles x2 + y2 = m and X2 + Y2 + 6x-8y-11 = 0 have a common point, then the value range of real number m is () A. （-∞，1）B. （121，+∞）C. [1，121]D. （1，121）

Circle x2 + Y2 + 6x-8y-11 = 0 can be reduced to (x + 3) 2 + (y-4) 2 = 62, circle center O1 (0, 0), circle center O2 (- 3, 4), distance between two circle centers d = 5, ∵ two circles x2 + y2 = m and X2 + Y2 + 6x-8y-11 = 0 have common points, | M-6 | ≤ 5 ≤ m + 6 | 1 ≤ m ≤ 121, so C is selected
When summing up the method of polynomials, Mr. Wang said, "first, arrange the original polynomials in the order of ascending or descending power according to a certain letter, and then add and subtract
Please use the skills mentioned by Mr. Wang to calculate the following problem: a = 5x & # 179;, B = - 2x & # 178; - 7 + 3x & # 179;, find the value of a + 2B
A+2B
=5x³+2(－2x²－7＋3x³)
=5x³－4x²－14＋6x³
=5x³+6x³－4x²－14
=11x³－4x²－14
At the same time, 6x + 6 / 7 > 4x + 7 and 8x + 3 are satisfied
8, 4-11
The first solution equals x > 43 / 14, and the second solution equals X
2x>7-6/7,4x4x>13-5/7
x=4,5,6,7,8,9,10,11
At the same time, 6x + 6 / 7 > 4x + 7 and 8x + 3 are satisfied
The quadratic equation x2 + PX + q = 0 (p2-4q ≥ 0) is solved by formula method
x²+px+q=0
x²+px+p²/4+q=p²/4
(x+p/2)²=p²/4-q
And p2-4q ≥ 0, square root on both sides
x+p/2=±√(p²/4-q)
x=-p/2±√(p²/4-q)
Given that circle x2 + y2 = m is inscribed with circle x2 + Y2 + 6x-8y-11 = 0, then the value of real number m is______ .
Circle x2 + Y2 + 6x-8y-11 = 0, that is, (x + 3) 2 + (y-4) 2 = 36, which means the circle with (- 3, 4) as the center and radius equal to 6. Then according to the inscribed phase of two circles, the distance between the center of two circles is equal to the difference of radius, we can get (− 3 − 0) 2 + (4 − 0) 2 = | 6-m |, the solution is m = 1, or M = 121, so the answer is 1 or 121
The correct way to arrange the polynomial - A2 + a3 + 1-A by the ascending power of the letter A is ()
A. a3-a2-a+1B. -a-a2+a3+1C. 1+a3-a2-aD. 1-a-a2+a3
∵ in the polynomial - A2 + a3 + 1-A, the exponent of - A is 1, the exponent of - A2 is 2, the exponent of A3 is 3, and the ascending order of a is 1-a-a2 + a3
Solution: 6x + 8x = 4x + 6
14x=4x+6
10x=6
x=0.6
If two of the quadratic equations x2 + PX + q = 0 are 2 + 3 and 2-3 respectively, then p=______ ，q=______ .
2 + 3 + 2-3 = - P, (2 + 3) (2-3) = q, so p = 4, q = 1
If two circles x2 + y2 = m and X2 + Y2 + 6x-8y-11 = 0 have a common point, then the value range of real number m is ()
A. （-∞，1）B. （121，+∞）C. [1，121]D. （1，121）
Circle x2 + Y2 + 6x-8y-11 = 0 can be reduced to (x + 3) 2 + (y-4) 2 = 62, circle center O1 (0, 0), circle center O2 (- 3, 4), distance between two circle centers d = 5, ∵ two circles x2 + y2 = m and X2 + Y2 + 6x-8y-11 = 0 have common points, | M-6 | ≤ 5 ≤ m + 6 | 1 ≤ m ≤ 121, so C is selected
What is the arrangement of 5x & sup2; y + Y & sup3; - 3xy & sup2; - X & sup3; by the ascending power of Y?
5x & sup2; y + Y & sup3; - 3xy & sup2; - X & sup3; in ascending order of Y is
-x³+ 5x²y-3xy²+y³
The ascending order of Y is
I'm looking forward to the answer, too
-x³+5x²y-3xy²+y³