If x + X Λ 2 + X Λ 3 + 1 = 0, find x + X Λ 2 + X Λ 3 The value of X Λ 2007 + X Λ 2008

If x + X Λ 2 + X Λ 3 + 1 = 0, find x + X Λ 2 + X Λ 3 The value of X Λ 2007 + X Λ 2008

∵X +X∧2+X ∧3+1=0∴x＋x²＋x³＋x^4＝x﹙1＋X +X∧2+X ∧3﹚＝0x^5＋x^6＋x^7＋x^8＝x^5﹙1＋X +X∧2+X ∧3﹚＝0…… x^2005＋x^2006＋x^2007＋x^2008＝x^2005﹙1＋X +X∧2+X ∧3﹚＝0∴X +X∧2+X ∧3… ...

Define a × B = |a ·|b |sin θ, where θ is the angle between vector a and B. If |a | = 2, B = 5, a · B = - 6, then |a × B |is equal to

From a * b = | a | * | B | * cos, 2 * 5 * cos = - 6,
So cos = - 3 / 5,
So sin = 4 / 5,
Then | a × B | = | a | * | B | * sin = 2 * 5 * 4 / 5 = 8
If you are satisfied, please click [satisfied] in the upper right corner~

Given ab ⊥ BC, FA ⊥ AC, BC = 3cm, ab = 4cm, AF = 12cm, find the area of square cdef

≁ ab ⊥ BC, FA ⊥ AC ≁ AC ≁ AC ≁ 178; = ab ≁ 178; + BC ≁ FC ≁ 178; = AF ≁ 178; + AC ≁ 178; ≁ BC = 3cm, ab = 4cm, AF = 12cm ≁ FC = 13cm ≁ square cdef area = 13 ≁ 178; = 169cm ≁ 178; if you agree with my answer, please click "adopt as a satisfactory answer", thank you

If the minimum value of the objective function z = ax + by is 2, then the maximum value of AB is 2
How to calculate? I calculated 1 / 2. I brought (1,2) in. The answer is 1 / 6

A & gt; 0, B & gt; 0! Constraint {x ≥ 2 & nbsp; 3x-y ≥ 1 & nbsp; y ≥ x + 1 & nbsp;, the (x, y) feasible region is an open region, the objective function z = ax + by, let z = 0, the initial value line L; ax + by = 0, cross the origin & nbsp; in quadrant 2 and 4, you can see that no matter how l rotates, it is the boundary point a (2

Is the sum of two vectors equal to 0 or equal to 0 vector

After adding vectors, they are still vectors, so of course they are zero vectors!

As shown in the figure, in △ ABC, ∠ a = 36 °, ABC = 40 °, be bisection ∠ ABC, ∠ e = 18 °, CE bisection ∠ ACD? Why?

The reasons are as follows: ∵ a = 36 °, ABC = 40 °, ∵ BCA = 104 °, ∵ ACD = 76 °. ∵ be = ABC, ∵ CBE = 20 °, ∵ e = 18 °, ∵ BCE = 142 °, ∵ ECA = 38 °, ∵ ECD = 38 °,