Known: nonzero vector [vector a], use Pythagorean theorem to correctly make [vector a] 5 times the root

Known: nonzero vector [vector a], use Pythagorean theorem to correctly make [vector a] 5 times the root

As shown in the figure, OA = a (vector). Take ab ⊥ OA, and | ab | = 2 | OA |, take o as the center of the circle and | ob | as the radius to make an arc
Intersection of OA extension line at C, vector OC = √ 5A. & nbsp; [OC & amp; sup2; = ob & amp; sup2; = OA & amp; sup2; + AB & amp; sup2; = 5A & amp; sup2;]

In the isosceles right triangle, ∠ BAC = 90 °, ad = AE, AF ⊥ BC, FG ⊥ CD through F, BG = AF + FG is proved

In △ BAF, △ BHF, AF = FH, BF is the common side, ∠ BFA = ∠ BFH (easy to prove) ≠ BAF ≌ △ BHF ≠ BAF = ∠ BHF, ∠ ABF = ∠ HbF = 45 ≌ BAF = ∠ AEB

If f (x) = AF (x) + BG (x) + 3, the maximum value on the interval (0, positive infinity) is 8,
Find the minimum value of F (x) in the interval (negative infinity, 0)

Because f (x) and G (x) are all odd functions whose domain is r
So f (x) = AF (x) + BG (x) + 3 is centrosymmetric about point (0,3)
F (positive infinity) = AF (positive infinity) + BG (positive infinity) + 3 = 8
F (negative infinity) = AF (negative infinity) + BG (negative infinity) + 3
=-AF (positive infinity) - BG (positive infinity) + 3
=-2

Mathematical problem vector
Let a = (1,2) B = (- 2, y). If a is parallel to B, then 3A + B is equal to
How to add vector? How to do this problem

If a is parallel to B
Then 1 * Y-2 * (- 2) = 0 = = > y = - 4
ab=1*(-2)+2*(-4)=-10
a^2=1+2^2=5
b^2=(-2)^2+(-4)^2=20
|3a+b|=√(3a+b)^2
=√(9a^2+6ab+b^2)
=√(45-60+20)
=√5
A ^ 2 is the square of A

As shown in the figure, ⊙ o is equal to the chord length on the ABC side of the triangle. Let ∠ AOC = x °, BOC = y ° and find the direct function of Y and X
As shown in the figure, ⊙ o is cut on the three sides of the triangle ABC so that the chord length is equal, ⊙ a = x °, ⊙ B = y °, find the direct function of Y and X

It can only be proved that the point O is the center of the inscribed circle of the triangle, and the direct function of Y and X can not be obtained

The unit price of a kind of candy in a store is 20 yuan per kilogram, and the unit price of B kind of candy is 16 yuan per kilogram. In order to promote sales, now 10 kg of B kind of candy is mixed with a package of a kind of candy. If the unit price of the mixed candy is set at 17.5 yuan per kilogram, then the sales volume of mixed sales is the same as that of separate sales. How many kilograms of a kind of candy are there?

Let's suppose that the box of A-type candy has x kg, the sales amount of non mixed sales is (20x + 16 × 10) yuan, and the sales amount of mixed sales is 17.5 (x + 10) yuan. From the title, we know that the sales amount of mixed sales is the same as that of separate sales. The formula is 20x + 16 × 10 = 17.5 (x + 10), and the solution is x = 6

If O is the outer center of the triangle ABC and 3oa + 4ob + 5oC = 0, then the vector OC multiplied by the vector AB = the answer is - 1 / 5

Make a graph, considering that | OA | = | ob | = | OC |, because 3oa + 4ob + 5oC = 0, that is, 3oa + 4ob = - 5oC, there is OA ⊥ ob, that is, OA * ob = 0. In addition, OC = - 3 / 5 * oa-4 / 5 * ob
OC*AB=(-3/5*OA-4/5*OB)*(OB-OA)=1/5*(OA-OB)(3OA+4OB)=1/5*[3OA^2+OA*OB-4OB^2]
=-1/5*OB^2=-1/5*R^2
Where R is the circumscribed circle radius of △ ABC. Unless there is a circumscribed circle radius of 1, the answer is - 1 / 5

If BM = AC, find the degree of ABC

Consider two cases: when ∠ ABC is an acute angle, as shown in Figure 1, ∵ ad ⊥ dB, be ⊥ AC, ∵ MDB = ∠ AEM = 90 °, ∵ ame = ∠ BMD, ≌ CAD = ∠ MBD, in △ BMD and △ ACD, ≌ BDM = ∠ ADC = 90 °∠ dBm = ∠ dacbm = AC, ≌ BMD ≌ ACD (AAS), ≌ ad = BD

If the solution of the equation | x | = 2x + 1 about X is negative, then the value of X is ()
A. −14B. −13C. −12D. -1

① When x ≥ 0, go to the absolute value, x = 2x + 1, get x = - 1, not in line with the preset x ≥ 0, rounding. ② when x < 0, go to the absolute value, - x = 2x + 1, get x = - 13

It is known that a and B are the left and right vertices of the major axis x2 / 36 + Y2 / 20 = 1 respectively, point F is the right focus of the ellipse, point P is on the ellipse and above the X axis, and PA is perpendicular to Pb
(1) Find the coordinates of point P (2) let m be a point on the major axis ab of the ellipse, and the distance from m to the straight line AP is MB (absolute value). Find the minimum distance D of point m on the ellipse (⊙ o ⊙) (⊙ o ⊙) ╮ (╯﹏) ╭

First of all, we can get a = 6, B = 2 √ 5 from the elliptic equation, so C = 4, C / a = 2 / 3, right guide line x = a ^ 2 / C = 9 (1) let P coordinate be (x, y), then the distance from P to right guide line is 9-x, and the distance from P to f is 2 (9-x) / 3, passing P