What quadrant does the image of a linear function y = negative 3x plus 2 pass through?

What quadrant does the image of a linear function y = negative 3x plus 2 pass through?

In the function y = KX + B, the problem k = - 3, when k ＜ 0, it goes through two or four quadrants
When B = 2 and b > 0, add the remaining quadrant above the x-axis, that is, the first quadrant
So the function goes through one two four quadrants

It is known that the image of the first-order function is parallel to the positive proportion y = - x, and the expression of the first-order function is obtained by M (0,4)
If the points (- 8, m) and (n, 5) find the value of M, N on the image of a linear function

Since y = KX + B is parallel to y = - x, so k = - 1, when y = - x + B passes through point (0,4), = 4, so y = - x-4.. if points (- 8, m), and (n, 5) are on the image of function, so m = 4, n = - 9

When P1, P2 When PN, are all positive numbers, N / P1 + P2 +... + PN is called the "mean reciprocal" of P1, P2... PN

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The intersection of the image of parabola y = - x ^ 2 + 1 and the positive half axis of X is a. divide the line OA into N equal parts, and set the points as P1, P2, P3 respectively , PN-1, make the perpendicular of X axis through each minute point, and intersect with the parabola at points Q1, Q2 QN-1, remember right triangle op1q1, p1p2q2, ^ The area of S 1, s 2 So S1 = (n ^ 2-1) / 2n ^ 3, S2 = (n ^ 2-4) / 2n ^ 3 And w = S1 + S2 + S3 + Sn-1, when n is larger and larger, what is the closest constant of W? Why

（1）
F (x) + F (y) = f [(x + y) / (1 + XY)], let x = y = 0
There is f (0) = 0
Let y = - x, x, y belong to (- 1,1)
f(x)+f(-x)=f(0)=0
So f (x) is an odd function
（2）
Order 0

Definition: n / P1 + P2 +... + PN is called the mean reciprocal of n positive numbers P1, P2... PN, and the mean reciprocal of the first n items of known sequence {an} is 1 + an / Sn
The formula of finding an general term and Sn

N = 1
1/a1=2
a1=1/2
1+an/Sn=n/Sn
Multiply both sides by SN at the same time
Sn+an=n
n> 1:00
S(n-1)+a(n-1)=n-1
Subtract to get
an+an-a(n-1)=1
2an=a(n-1)+1
2(an-1)=a(n-1)-1
an-1=1/2(a(n-1)-1)
a1-1=-1/2
So {an-1} is an equal ratio sequence
an-1=-1/2*(1/2)^(n-1)=-1/2^n
an=1-1/2^n
According to Sn + an = n
Sn=n-an=n-1+1/2^n

P / w = P1 / W1 + P2 / W2 + P3 / W3 +.. + pN / wn?
Such as the title

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PS: the formula doesn't seem to hold

In the rectangular coordinate plane, the known points P1 (1,2), P2 (2,2 ^ 2) , PN (n, 2 ^ n), where n is a positive integer. For any point A0 in the plane, A1 is the symmetric point of A0 about point P1, A2 is the symmetric point of A1 about point P2 , an is the symmetric point of an-1 with respect to point PN
1. Find the coordinates of vector a0a2;
2. When the point A0 moves on the curve C, the trajectory of point A2 is the image of the function y = f (x), where f (x) is a periodic function with a period of 3, and when x ∈ (0,3], f (x) = 1gx, find the analytic expression of the function with curve C as the image on (1,4];
3. For any even number n, use n to represent the coordinates of vector a 0An

1) Coordinates of vector a0a2 (2,4)
2) Let A0 coordinate (x, y), because A0 and A1 are symmetrical about P1, so A1 = (2-x, 4-y), and because A1 and A2 are symmetrical about P2, so A2 = (x + 2, y + 4)
If x ∈ (0,3], f (x) = lgx, and f (x) period is 3, so when x ∈ (3,6], f (x) = lgx
f（x）=lg（x-3）,x∈（1,4】,x+2∈（3,6】
So when C is x ∈ (1,4], y + 4 = LG (x + 2-3), that is y = LG (x-1) - 4)
3) In △ a0a1a2, P1 is the midpoint of a0a1, P2 is the midpoint of A1A2, so a0a2 = 2p1p2, and so on, a2a4 = 2p3p4 An-2an = pn-1pn, according to the vector addition, a0an = 2 (p1p2 + p3p4 +...) Pn-1Pn）,
Its abscissa is 2 [(2-1) + (4-3) + +（n-n+1）】=n,
Its ordinate is 2 (2 + 23 + 25 +) +2n-1）=4（2n-1）/3
So a 0An = (n, 4 (2n-1) / 3), n is even

On the xoy plane, there are a series of points P1 (A1, B1), P2 (A2, B2) ,Pn(an,bn),… For each natural number n, the point PN lies in the function y = 2000 (A / 10) ^ X
（0

First question: an = [n + (n + 1)] / 2 = n + 1 / 2bn = 2000 * [(A / 10) ^ (n + 1 / 2)] second question: if a = 10, BN = 2000, meet the requirements if a > 10: BN = 2000 * [(A / 10) ^ (n + 1 / 2)] is an increasing function BN + B (n + 1) > b (n + 2) 1 + A / 10 > (A / 10) ^ 25 + 5 √ 5 > a > 5-5 √ 5, so 5 + 5 √ 5 > a > 10 if a

Known points P1 (A1, B1), P2 (A2, B2) , PN (an, BN) (n is a positive integer) are all on the function y = a ^ x (a > 0, a ≠ 1)
An is the formula for finding the general term of an and proving the proportional sequence of BN with 1 as the first term and 2-position tolerance arithmetic sequence

an=1+2(n-1)=2n-1
bn=a^an=a^(2n-1)
b1=a^a1=a
bn=a×a^2(n-1)
=a×(a²)^(n-1)
The first term is a, and the common ratio is a & #

Solve the following equation (process) 5 / 4x-0.7 = 3 / 4x + 0.510x + 7 = 14x-5-3x
Er. Process means that you don't need x = () directly below the formula, and there are many processes in the middle. It's better to explain them and say thank you first,

5/4X-3/4X=0.5+ 0.7
1/2X=1.2
X=2.4
14X-3X-10X=-5-7
X=-12