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Anonymous users2024-02-10This argument of yours is not true.
First of all, you should try to avoid using the concept of the square of the vector, not to say that it is not possible, and it is easy to make mistakes.
It is also not easily distinguishable from the operation of real numbers. The so-called square of a vector is just a name, and it actually represents.
The inner product of the vector and itself.
arithmetic, e.g. a 2 = a · a = |a|^2，b^2=b·b=|b|^2
And: (a·b) 2=(|a|*|b|*cos)^2=|a|^2*|b|2*cos 2, only when cos=1 or -1.
Only: (a·b) 2=(|a|*|b|*cos)^2=|a|^2*|b|^2
That is, the equation is only true if a and b are parallel, and not in all other cases.
For example: = 3, then: a·b=|a|*|b|/2，(a·b)^2=|a|^2*|b|^2/4
2, then: a·b=0, (a·b) 2=0
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Anonymous users2024-02-09You can understand it this way, vector (a*b) 2 = vector a 2 times vector b 2, because the result of the two vectors multiplying is a numeric value, so vector a 2 = |a|2, vector b 2 = |b|^2。So it was established.
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Anonymous users2024-02-08This doesn't seem to look like you, a=(-1,1),b=(1,1) (a*b) 2=0 but a 2*b 2=2*2=4 so it can't be like this Hopefully.
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Anonymous users2024-02-07a2b2 + a2 + b2 + 1 = 4ab shift term (a2b2 - 2ab + 1) +a2 + b2 - 2ab) =0 can be factored into: (ab-1)2 + a-b)2 = 0, both squares should be non-cave negative orange or so there must be ab = 1 and a = b so: circular tremor [solution 1] a = 1, b = 1 [solution 2] a = 1, b = 1....
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Anonymous users2024-02-06Summary. The quadratic of 5a multiplied by the quadratic (ab) of b.
Hello, is the title like this?
Can I do the question specifically?
OK. Did you get it?
Treat ab as the same base, divide by the power of the same base, and subtract the exponent.
Do you understand? That's not quite the right title.
Then you send me the question** to show me.
I'm asking you what kind of topic it is.
Failed to send? Oh.
The quadratic of 5a is not in parentheses.
ab).
Thank you, teacher.
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Anonymous users2024-02-05The quadratic of a + the quadratic of b =, the value of ab =
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Anonymous users2024-02-04Solution: From the known (a+b)?Gizzard?
Armpits? ab=7 (a-b)?Gizzard?
Armpits? ab=3 so a?Explain?
7+3) 2=5 2ab=(7-3) 2=2 i.e. ab=1
Any unary quadratic equation ax 2 bx c 0(a≠0) can be configured as (x+(b 2a)) 2=b 2-4ac, because a≠0, from the meaning of the square root, the sign of b 2 4ac can determine the case of the root of the unary quadratic equation b 2 4ac is called the discriminant of the root of the unary quadratic equation ax 22 bx c 0(a≠0), which is denoted by " " (read delta), that is, b 2 4ac >>>More
private sub command1_click()dim a, b, c, x1, x2, d as singlea = val( >>>More
Solution: In the form of an +bn + c, it can be matched into a(n+b 2a) +4ac-b ) 4a, and the square term in front can determine n, such as a<0, a(n+b 2a) has a maximum value of 0, (if and only if n=-b 2a, etc.), so that the n value can be determined, and then the whole can be determined.
There are x teams.
Then each team will play against the other (x-1) team. >>>More
vb wants to control oh add it yourself.
**：private sub command1_click()dim a, b, c, delta, x1, x2, x as double >>>More