Sina / 2 = 4 / 5, cosa / 2 = - 3 / 5, which quadrant is a

Sina / 2 = 4 / 5, cosa / 2 = - 3 / 5, which quadrant is a

Sin is positive, cos is negative, which means that a / 2 is in the second quadrant, a / 2 is between 90 and 180 degrees, then a is 180 to 360 degrees, that is, a is the third or fourth quadrant, and then judge cosa is negative, so it can only be the third quadrant
Lingo11 solve the problem of linear programming code min z = 3 * x2 + X4 s.t.2 * x1-x3 + X4 = 2 - X1 + x2 + X3 = 4 X1 + X3 + X5 = 6
Lingo 11 code is as follows
model:
sets:
SI/1..5/:c,x;
SJ/1..3/:xl;
SIJ(SI,SJ):cij;
endsets
data:
c=0 3 0 4 0;
xl=2 4 6;
cij=2 0 -1 1 0
-1 1 1 0 0
1 0 1 0 1;
enddata
[obj] min=@sum (SI:x*c)
[eq1]@sum (SIJ(i,j):x(i)*cij(i,j)=xl(j));
Why can't I compile it?
Those three equations. I think it's expressed by matrix multiplication. Please ask how to write it.
There should be a semicolon at the end of the sentence
The last constraint you wrote is wrong. I don't know what you want to express. You'd better write what you want to express
Inequality f (x) of known function f (x) = KX ^ 2 + (K + 1) x solution about X
B ^ 2-4ac = (K + 1) ^ 2-4k = (k-1) ^ 2 > = 0
Calculate LIM (1 / N2 + 1 + 2 / N2 + 1 + 3 / N2 + 1 +... + n / N2 + 1)
The above formula = LIM (1 + 2 +... + n) / (n ^ 2 + 1) = Lim [n (n + 1) / 2] / (n ^ 2 + 1) = 1 / 2 Lim [(n ^ 2 + n) / (n ^ 2 + 1)] = 1 / 2 * 1 = 1 / 2. It is noted that n is of low order relative to n ^ 2
Solve the following equation: (1) x2 + 4x + 2 = 0 (collocation method) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2) 3x (x-1) = 2 (1-x)
(1) The deformation of the equation is: x2 + 4x = - 2, the coordination is: x2 + 4x + 4 = 2, that is, (x + 2) 2 = 2, the square root is: x + 2 = ± 2, the solution is: X1 = - 2 + 2, X2 = - 2-2; (2) the deformation is: 3x (x-1) + 2 (x-1) = 0, the decomposition factor is: (x-1) (3x + 2) = 0, the solution is: X1 = 1, X2 = - 23
Given that the average of x1, X2, X3, x4, X5 is a, X6, X7, and the average of x15 is B, then the average of x1, x2.x15 is a
Refer to your level, great Xia, think about it
（5a+15b）÷（5+15）
=5（a+3b）÷20
=（a+3b）/4
(a1+a2+a3+a4+a5)/5=a
(a6+a7+a8+a9+a10+a11+a12+a13+a14+a15)/10=b
(a1+a2+...+a15)/15 = (a1+a2+a3+a4+a5)/15+(a6+a7+a8+a9+a10+a11+a12+a13+a14+a15)/15
=(a1+a2+a3+a4+a5)/5*（1/3）+ (a6+a7+a8+a9+a10+a11+a12+a13+a14+a15)/10*(2/3)
=a/3+2b/3.
You can understand it. There should be no mistake
a/3+2b/3
(5a+10b)/15
Given the function f (x) = x ^ 2 / (2-x), let k > 0, solve the inequality f (x) < [(K + 1) x-k] / (2-x) about X, hoping to explain the process in detail,
When 2-x > 0, that is X
The sum of the first n terms of sequence an is Sn, Sn = 4an-3. It is proved that an is an equal ratio sequence. The sequence BN satisfies B1 = 2, BN + 1 = an + BN
The key is the second question~
One
an=Sn-Sn-1=4an-4an-1
4an-1=3an
an/an-1=4/3
a1=4a1-3,a1=1
an=1*(4/3)^(n-1)
Two
b1=2
b2=a1+b1=3
b3=b2+a2=2+1+(4/3)
b4=2+1+4/3+(4/3)^2
bn=2+(1-(4/3)^(n-1))/(1-4/3)=2+3((4/3)^(n-1)-1)
First question:
When n > 1,
an=sn-s(n-1)=4an-4a(n-1),
Then, an / a (n-1) = 4 / 3;
It is concluded that an is an equal ratio sequence with the first term of 1 and the common ratio of 4 / 3.
Second question:
B2-b1 = A1, then B (n + 1) - BN = an
Add: bn-b1 = a1 + A2 +.. + a (n-1) = 3 [(4 / 3) ^ (n-1) - 1]
The result is: BN = B1 + 3 [(4 / 3) ^ (n-2) - 1] = 3 (4 / 3) ^ (n-1) - 1
After the first question is solved, the second question is also an equal ratio sequence
1. Sn=4an-3
S(n-1)=4a(n-1)-3
an=4an-4a(n-1)
an=(4/3)a(n-1)
So the equal ratio sequence whose common ratio of {an} is 4 / 3
2.S1=4a1-3=a1 a1=1
an=a1*(4/3)^(n-1)=(4/3)^(n-1)
b(n+1)-bn=an=(4/3)^(n-1)
Bn-b (n-1) = (4 / 3... Expansion)
1. Sn=4an-3
S(n-1)=4a(n-1)-3
an=4an-4a(n-1)
an=(4/3)a(n-1)
So the equal ratio sequence whose common ratio of {an} is 4 / 3
2.S1=4a1-3=a1 a1=1
an=a1*(4/3)^(n-1)=(4/3)^(n-1)
b(n+1)-bn=an=(4/3)^(n-1)
bn-b(n-1)=(4/3)^(n-2)
...
b2-b1=(4/3)^1
Superposition bn-b1 = (4 / 3) * [1 - (4 / 3) ^ (n-1)] / (1-4 / 3)
=4*[(4/3)^(n-1)-1]
BN = 4 * (4 / 3) ^ (n-1) - 4 + B1 = 4 * (4 / 3) ^ (n-1) - 2
Univariate quadratic equation: (2x-1) square = 2x-1, 6x square + 7x-3 = 0 (formula method) 2x square + 5x-1 = 0 (collocation method)
A:
1）
(2x-1)^2=2x-1
(2x-1)^2-(2x-1)=0
(2x-1)(2x-1-1)=0
x1=1/2,x2=1
2）
6x^2+7x-3=0
x=[-7±√(49+4*6*3)]/(2*6)
x=(-7±11)/12
x1=1/3,x2=-3/2
)
2x^2+5x-1=0
2(x^2+5x/2)=1
2(x^2+5x/2+25/16)=1+25/8
2(x+5/4)^2=33/8
(x+5/4)^2=33/16
x+5/4=±√33/4
x=(-5±√33)/3
What is x + x2 + X3 + X4 + X5 + X6 + X7 + X8 + x9 = 17.39
Please give the calculation formula process
Please give the process and results of the calculation formula, specifically X=
The sum of the first n terms
Formula: SN = A1 * (1-Q ^ n) / 1-Q
a1=x q=x n=9
Just put it in