If P (- 3, y) is a point on the terminal edge of angle A and Sina = - 2 / 3, what is the value of Y

If P (- 3, y) is a point on the terminal edge of angle A and Sina = - 2 / 3, what is the value of Y

When the terminal edge of angle a is in the third or fourth quadrant, the value of sina is negative; the abscissa of point P is - 3, so point P must be in the third quadrant, that is, y < 0. You can think so, because Sina = - 2 / 3, that is ab = 2 / 3 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; cosa = radical 5 / 3 & nbsp;; OA = radical 5 / 3ob = 1, then
If LIM (an / BN) = a (a is not 0) LIM (an) = 0, prove LIM (BN) = 0
It can be proved by the definition of sequence limit
To the contrary, suppose LIM (BN) is not equal to 0, LIM (an / BN) = LIM (an) / LIM (BN) = 0, which is inconsistent with the meaning of the title, so LIM (BN) = 0
To the contrary, suppose LIM (BN) is not equal to 0, LIM (an / BN) = LIM (an) / LIM (BN) = 0, which is inconsistent with the meaning of the title, so LIM (BN) = 0
The square of half x multiplied by 2x is equal to?
A set of data is known as x1, X2,..., xn, and its variance is the square of S
Verification: the variance of data kx1 + A, kx2 + A,..., kxn + A is the square of K multiplied by the square of S
That's why it's equal to the square of K times the square of S
The average of x1, X2,..., xn is
X average = (x1 + x2 +... + xn) / n
The variance of x1, X2,..., xn is s ^ 2 = [(x1-x average) ^ 2 + (x2-x average) ^ 2 +... + (xn-x average) ^ 2] / n,
The average of kx1 + A, kx2 + A,..., kxn + A is [(kx1 + a) + (kx2 + a) +... + (kxn + a)] / N = K (x1 + x2 +... + xn) / N + a = KX average + a
The variance of kx1 + A, kx2 + A,..., kxn + A is
{[(kx1 + a) - (KX average + a)] ^ 2 + [(kx2 + a) - (KX average + a)] ^ 2 +... + [(kxn + a) - (KX average + a)] ^ 2} / n
={[K (x1-x average)] ^ 2 + [K (x2-x average)] ^ 2 +... + [K (xn-x average)] ^ 2} / n
=(k ^ 2) [(x1-x average) ^ 2 + (x2-x average) ^ 2 +... + (xn-x average) ^ 2] / n
=(K^2)(S^2)
3. To make cosx? SiNx = m-2 meaningful, the value range of real number m is
If cosx * SiNx = m-2, then cosx * SiNx = 1 / 2sin2x ∈ [- 1 / 2,1 / 2], so m-2 ∈ [- 1 / 2,1 / 2], so m ∈ [- 5 / 2,3 / 2]
If cosx ± SiNx = m-2, then cosx ± SiNx = radical 2 * sin (x ± π / 4) ∈ [- radical 2, radical 2], so m-2 ∈ [- radical 2, radical 2], so m ∈ [- 2 - radical 2, - 2 + radical 2]
If LIM (square of an + bn-5) / (2n + 1) = 1, why a = 0, otherwise the limit does not exist
There are two possibilities in this problem. One is that the denominator can be reduced, and the other is that the denominator cannot be reduced. In the first case, if the denominator is reduced, the limit will exist. In the second case, if the denominator is not reduced, the molecule is of second order and the denominator is of first order. The limit can only exist if the order of the molecule is less than or equal to the order of the denominator
If the square of 2x is equal to x, then x is equal to half
Tell me why
Wrong
The square of 2x equals X
Square of 2x - x = 0
x(2x-1)=0
X = 0 or x = 1 / 2
Wrong.
From 2x ^ 2 = x, it is easy to know x = 0 or X '= 1 / 2
X = 1 / 2 is only one of the solutions
Space vector and solid geometry
In the parallelepiped abcd-a'b'c'd ', ab = 2, AA' = 2, ad = 1, and AB.AD.AA The vector AC '* vector BD' =?
AC'=AB+BC+CC'
BD'=BA+AD+DD'
Substitution expansion
*For point multiplication
Vector AC '= vector AA' + vector a'B '+ vector b'c'
=Vector AA '+ vector AB + vector ad
Vector BD '= vector BB' + vector b'a '+ vector a'd'
=Vector AA '- vector AB + vector ad
Vector AC '* vector BD' = (vector AA '+ vector AB + vector AD) * (vector AA' - vector AB + vector AD)
=Vector AA '^ 2... Expansion
*For point multiplication
Vector AC '= vector AA' + vector a'B '+ vector b'c'
=Vector AA '+ vector AB + vector ad
Vector BD '= vector BB' + vector b'a '+ vector a'd'
=Vector AA '- vector AB + vector ad
Vector AC '* vector BD' = (vector AA '+ vector AB + vector AD) * (vector AA' - vector AB + vector AD)
=Vector AA '^ 2 + vector AA' ^ 2 + vector AA '^ 2 + vector AA' * vector ad
=2 ^ 2 + 2 ^ 2 + 1 ^ 2 + 2 by 2 by 1 by 1 / 2
11. Put it away
Function y = (sinx-a) 2 + 1, when SiNx = a, there is a minimum value, when SiNx = 1, there is a maximum value, then the value range of a is ()
A. [-1，0]B. [-1，1]C. （-∞，0]D. [0，1]
∵ function y = (sinx-a & nbsp;) 2 + 1, when & nbsp; SiNx = A & nbsp;, it has the minimum value, ∵ quadratic function y = (T-A) 2 + 1 of t takes the minimum value at symmetry axis t = a, it can get - 1 ≤ a ≤ 1, and ∵ when SiNx = 1 & nbsp;, it has the maximum value, ∵ a ≤ 0, it can get - 1 ≤ a ≤ 0
Does Liman exist, limbn does not exist and lim (an * BN) exists
an=1/n
bn=n
an=1/X
bn=(-1)^n