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If M = the arithmetic square root of 2007-n + the arithmetic square root of n-2007
M = arithmetic square root of 2007-n + arithmetic square root of n-2007
So 2007-n ≥ 0 and n-2007 ≥ 0
So only n = 2007
So m = 0
If the arithmetic square root of the square of M + n is m + N, then the following formula holds: a.m > n B.M > n_ n.C.m
B
It is known that the three sides of △ ABC are a, B, C respectively, and satisfy a & sup2; + B & sup2; + C & sup2; + 50 = 6A + 8b + 10C. Judge whether this is a right triangle
a²＋b²＋c²＋50＝6a＋8b＋10c
The results show that (A-3) & sup2; + (B-4) & sup2; + (C-5) & sup2; = 0
So a = 3, B = 4, C = 5
3*3+4*4=25=5*5
So delta ABC is a right triangle
The area of a parallelogram is 8 square centimeters larger than that of a triangle with the same base and height. What is the area of a triangle
The area of a parallelogram is 8 square centimeters larger than that of a triangle with the same base and height. The area of a triangle is equal to half the area of a parallelogram, that is, 4
Given that point P is a moving point on the parabola x2 = 2Y and the focus is f, if the fixed point m (1,2), then when point P moves on the parabola, the minimum value of | PM | + | PF | is equal to ()
A. 52B. 2C. 32D. 3
Let the distance between the point P and the guide line be | PE |. From the definition, we know | PF | = | PE |, so | PM | + | PF | = | PF | + | PM | ≥ | me | ≥ | Mn | = 52. (the perpendicular foot from m to the guide line is set as n). When equal sign is taken, the minimum value of M, P and E is collinear, and the minimum value of | PM | + | PF | = 52
It is explained by the collocation method that no matter what value x takes, the value of the algebraic formula - 3x ^ 2-x + 1 is no more than 13 / 12
﹣3x²－x＋1
＝﹣3﹙x²＋1/3x－1/3﹚
＝﹣3[x²＋1/3x＋﹙1/6﹚²－﹙1/6﹚²－1/3]
＝﹣3[﹙x＋1/6﹚²－13/36]
＝﹣3﹙x＋1/6﹚²＋13/12≤13/12.
Let p be any point on the straight line L: 2x + y + 9 = 0, pass through point P as two tangent lines PA and Pb of circle x2 + y2 = 9, and the tangent points are a and B respectively, then the straight line AB passes through the fixed point___ ．
Because P is any point on the straight line L: 2x + y + 9 = 0, let P (m, - 2m-9), because the two tangent lines PA and Pb of circle x2 + y2 = 9, and the tangent points are a and B respectively, so OA ⊥ PA and ob ⊥ Pb, then points a and B are on the circle with OP as diameter, that is, AB is the common chord of circle O and circle C, then the coordinates of circle center C are (m2, - 2m + 92), and the square of radius is R2 = M2 + (2m + 9) 24, so the equation of circle C is (x-m2) 2 + (y + 2m + 92) 2 = M2 + (2m + 9) 24, ① and X2 + y2 = 9, ②, ② - ① get MX - (2m + 9) Y-9 = 0, that is, the linear equation of the common string AB is: MX - (2m + 9) Y-9 = 0, that is, m (x-2y) - (9y + 9) = 0, from x-2y = 09y + 9 = 0, x = - 2Y = - 1, so the line AB always passes the fixed point (- 2, - 1), so the answer is: (- 2, - 1)
Given that point a (2,0) B (- 2,4) is symmetric with respect to line L, point P (7,3) and point Q (m, n) are symmetric with respect to line L, then the value of M + n is
∵ point a (2,0) B (- 2,4) is symmetric with respect to line L
∴AB⊥l
While, KAB = (4-0) / (- 2-2) = - 1
The slope product of two vertical lines is - 1
∴kL=-[(4-0)/(-2-2)]^(-1)=1
The product of the slope of PQ line and the slope of L is - 1
∴(n-3)/(m-7)*1=-1
n-3=7-m
m+n=10
In rectangular paper ABCD, ab = 3cm, BC = 4cm. Now fold and flatten the paper so that a and C coincide. If the crease is EF, the area of the overlapping part △ AEF is equal to______ ．
Let AE = x, from folding, EC = x, be = 4-x, in RT △ Abe, AB2 + be2 = AE2, that is 32 + (4-x) 2 = X2, the solution is: x = 258, from folding, we can know ∠ AEF = ∠ CEF, ∫ ad ‖ BC, ∫ CEF = ∠ AFE, ∫ AEF = ∠ AFE, that is AE = AF = 258, ∫ s △ AEF = 12 × AF × AB = 12 × 258 × 3 = 7516
The N + 1 power of 5 divided by the 3N + 1 power of 5 and (AX + b) multiplied by (Cx + D)?
Solving with integral
First question:
5^(n+1)/5^(3n+1)=5^(n+1-3n-1)=5^(-2n)=1/5^2n
Second question:
(ax+b)(cx+d)=acx^2+adx+bcx+bd=acx^2+(ad+bc)x+bd
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