Given the linear function y = 2kx-5k + 1, when k = (), the image must pass through the origin

Given the linear function y = 2kx-5k + 1, when k = (), the image must pass through the origin

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Given the linear function y = (m-2) + M & # 178; - 4, if the image passes through the origin, then M=________ If the first-order function y = (m-1) x-m + 2 image passes through the first, second and third quadrants, then the condition m should satisfy is_______

Given the linear function y = (m-2) x + M & # 178; - 4, if the image passes through the origin, then M=____ -2____ .
If the image of a linear function y = (m-1) x-m + 2 passes through the first, second and third quadrants, then the condition that m should satisfy is U 1

If the image of the positive scale function y = KX is translated down two units and passes through the point (- 2,1), then the image of the original function must pass through the point () a (2,3)
B（－3,－2）C（2,－3）D（－3,

Because after the image of positive scale function y = KX is translated down 2 units, it passes through the point (- 2,1)
So y = kx-2 passes through the point (- 2,1)
That is 1 = - 2k-2
So k = - 3 / 2
So y = - 3x / 2
So the original function must pass through point (2, - 3) and choose C

If the function image of positive scale function y = KX and primary function y = ax + B passes through a (- 1, - 4) B (2, m), then 2A plus B=

Because y = KX over a (- 1, - 4)
So - 4 = - K
k=4
And because k = 4 is substituted by B (2, m)
So y = 4x2
=So m = 8
Therefore, substituting a (- 1, - 4) B (2,8) into y = ax + B, we get the following results:
-4=-a+b
8=2a+b
(the calculation process is omitted)
Finally, a = 4, B = 0
So 2A + B = (2x4) + 0 = 8
I made it myself. It should be right!

Given the first-order function y = (2m + 2) x + (3-m), according to the following conditions, the value range of M is calculated, and the image does not pass through the fourth quadrant?

If x = 0, y = 3-m > = 0, then M

Given the positive proportional function y = - (k-1) x, with the increase of X, the value of y also increases, so what is the value range of K? (to analyze,

With the increase of X, the value of y also increases
So - (k-1) > 0
k＜1

If the function y = (k2-1) x + 3K is a linear function, then the value range of K is ()
A. K ≠ - 1B. K ≠ 1C. K ≠ ± 1D. K is all real numbers

From the meaning of the question: k2-1 ≠ 0, the solution: K ≠ ± 1, so choose: C

Given the square positive proportional function of the absolute value of K-4 of the function y = (square of K + 3K) x, find the value of K

Because the function y = (square of K + 3K) the absolute value of K of X - the square of 4 is a positive proportional function
So we have | K | - 4 & # 178; = 1, K | - 178; + 3K ≠ 0
The solution is k = ± 17

Eighth grade function: known positive proportion function y = (K-2) x ^ k square - 3k-3, y increases with the increase of X, find the value of K
An eighth grade function problem, need complete steps, study bully help to see
The exponent of X is k square - 3k-3
thank you

∵ positive scale function
∴k-2>0
k²-3k-3=1
‖ k = 4 (- 1, rounding off)

Given the function y = (K-2) x + 2K + 1, when k______ It is a positive proportional function when k is a positive proportional function______ It is a function of degree

The function y = (K-2) x + 2K + 1, when 2K + 1 = 0, then k = - 12, it is a positive proportional function; when K-2 ≠ 0, that is, K ≠ 2, it is a linear function. So the answer is: = - 12, ≠ 2