Bo and Co are angular bisectors of ∠ ABC and ∠ ACB respectively. The intersection point is o. it is proved that ∠ BOC = 90 ° + 1 / 2 ∠ a

Bo and Co are angular bisectors of ∠ ABC and ∠ ACB respectively. The intersection point is o. it is proved that ∠ BOC = 90 ° + 1 / 2 ∠ a

∠BOC=180-0.5∠B-0.5∠C=180-0.5(∠B+∠C)=180-0.5(180-∠A)=90+0.5∠A

If the system x + y = 3, X-my = 2 and the system X-Y = 1, nx-y = 2 have the same solution, we can find the value of M and n

The first solution is x + y = 3
x-y=1
Add
2x=4
x=2,y=3-x=1
Put in the other two
2-m=2
2n-1=2
So m = 0
n=3/2

If the vector a = (3, - 2), B = (- 1,2), then (2a + b) × (a-b) is equal to?
Please make every step in detail with instructions

2a+b=(6,-4)+(-1,2)=(5,-2)
a-b=(3,-2)-(-1,2)=(4,-4)
Should it be on demand?
=5*4+(-2)*(-4)=20+8=28

In the triangle ABC, C = 90 °, BC = a, AC = B, ab = C, the height CD on the side of AB = h. prove: 1 / A & # 178; + 1 / B & # 178; = 1 / H & # 178;

This is an AB / C \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\﹣ 1 / A & ﹣ 178; + 1 / B & ﹣ 178; = A / H & ﹣ 178;

Factorization: (1) a ^ 2-4b ^ 2 + 3a-6b (2) 4m ^ 2-4m + 2n-n ^ 2

a^2-4b^2+3a-6b
=（a^2-4b^2）+（3a-6b）
=(a+2b)*(a-2b)+3(a-2b)
=(a-2b)(a+2b+3)
4m^2-4m+2n-n^2
=(4m^2-n^2)-（4m-2n）
=(2m+n)(2m-n)-2(2m-n)
=(2m-n)(2m+n-2)

The relationship between the circle whose diameter is the line segment connected by any point and the focus on the ellipse and the circle whose diameter is the major axis is ()
A. Separation B. intersection C. inscribe D. indeterminate

As shown in the figure, F1 and F2 are the left and right focal points of the ellipse. Point P is any point on the ellipse, then | Pf1 | + | PF2 | = 2A. The center of the circle with | F2P | as the diameter is C. connect f1p and OC. From the median line theorem of the triangle, we can get: | OC | = 12 | Pf1 | = 12 (2A - | PF2 |) = A-12 | PF2 |, that is, the distance between the centers of two circles is equal to the difference of the radii of two circles The relationship between the position of a circle whose line segment is diameter and a circle whose diameter is major axis is inscribed

It is known that: as shown in the figure, in △ ABC, extend the midline be and CD to N and m respectively, so that en = EB and DM = DC

It is proved that: am and an are connected, ∵ DM = DC, ∵ ADM = BDC, ad = dB, ≌ AMD ≌ BCD. ≌ mad = DBC. Similarly, it can be proved that: ∵ NAE = ∵ ECB, ∵ BAC + ∩ DBC + ∩ ECB = 180 °, and ∩ mad + ≌ BAC + ≌ NAE = 180. The three points m, a and N are on the same straight line

How many grams is equal to one kilogram

1000g = 1kg. I hope my answer will help you

If the image of function f (x) = - 5loga (x-1) + 2 (a is greater than zero and a is not equal to zero) passes through the fixed point P, then the coordinates of point P are?

loga （1）=0
So let X-1 = 1
x=2
y=-5×0+2=2
So p (2,2)

If the length of the rectangle is (a + 2) cm and the width is ACM, the perimeter of the rectangle is () cm?

If the length of the rectangle is (a + 2) cm and the width is ACM, the perimeter of the rectangle is (4 (a + 1)) cm?
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