Given the first-order function y = KX + B, when x = 9, y = 0; when x = 24, y = 20, find the first-order function relation and draw the function image

Given the first-order function y = KX + B, when x = 9, y = 0; when x = 24, y = 20, find the first-order function relation and draw the function image

When x = 9, y = 0; when x = 24, y = 20
∴0=9k+b
20=24k+b
∴k=4/3
b=-12
∴y=4x/3-12
Let y = KX + B be substituted into two sets of data to obtain the system of quadratic equations. Remove y = 4 / 3x-12 image and y-axis intersection at - 12 and X-axis intersection at 9
According to the question:
9k＋b＝0
　　（1）
24k＋b＝20　　　（2）
From (1) - (2), it is concluded that:
-15k＝-20
k＝5/3
Substituting k = 5 / 3 into (1) yields:
9×5/3＋b＝0
15＋b＝0
b＝-15
Then: y = 5 / 3x-15
Don't understand can ask! When x = 9, y = 0; when x = 24, y = 20  0 = 9K + B20 =... Expansion
According to the question:
9k＋b＝0
　　（1）
24k＋b＝20　　　（2）
From (1) - (2), it is concluded that:
-15k＝-20
k＝5/3
Substituting k = 5 / 3 into (1) yields:
9×5/3＋b＝0
15＋b＝0
b＝-15
Then: y = 5 / 3x-15
Don't understand can ask! Question: when x = 9, y = 0; when x = 24, y = 20 ﹣ 0 = 9K + B 20 = 24K + B ﹣ k = 4 / 3, B = - 12 ﹣ y = 4x / 3-12, right?
If the real numbers x and y satisfy the radical X-2 + (y + 1) &# =, then X-Y is equal to
Given that the real numbers x and y satisfy the radical X-2 + (y + 1) &# 178; =. 0, then X-Y is equal to
∵ radical X-2 + (y + 1) & ∵ 178; =. 0
∴x-2=0 y+1=0
x=2 y=-1
∴x-y=2-(-1)=3
Given the image of the first-order function y = KX + B as shown in the figure, find the function relation
The coordinates of these two points are (2,0) and (0,3)
The coordinates are brought into the function 2K + B = 0, B = 3
It is concluded that k = - 3 / 2, B = 3
We get y = - 3 / 2K + 3
If x and y are real numbers and (x + 2) &# 178; + radical Y-2 = 0, then what is x + y equal to
x+2=0,x=-2
y-2=0,y=2
x+y=-2+2=0
（x+2）²+√（y-2）=0
Then x + 2 = 0 and Y-2 = 0
X = - 2, y = 2
x+y=0
If the image of a function y = KX + B passes through (- 3,2) and (1,6), can you determine the expression of the function
Solution: substituting the point (- 3,2) (1,6) into the first-order function respectively has - 3K + B = 2 Formula 1 K + B = 6 formula 2 subtracts formula 2 from Formula 1 to get - 3k-k = 2-6 solution to get k = 1, substituting k = 1 into formula 2 to get 1 + B = 6 solution to get b = 5, so the expression of the first-order function is y = x + 5
Substituting two points in oblique form
y-2=((6-2)/(1+3))(x+3)
Well organized
y=x+5
If the point is brought in, then x + 5 = y
Substituting (- 3,2) and (1,6) into the analytic expressions,
We can get 6 = K + B, 2 = - 3K + B
The solution is k = 1, B = 5
∴y=x+5
y=x+5
Two point coordinates are brought into the equation respectively, and two binary linear equations about K and B are obtained, which can be solved.
It is concluded that b-3k = 2
k+b=6
So the solution is k = 1, B = 5
The function expression is y = x + 5
I = {0,1,2,3,4}, a = {0,1,2,3}, B = {2,3,4}, what is the (CIA) intersection (CIB)
CiA={4}
CIB={0,1}
(CIA) intersection (CIB) is an empty set
Given the image of the first-order function y = KX + 4 and the area enclosed by the two coordinate axes is 8, the expression of the first-order function is obtained
The abscissa of the point of intersection with the X axis is
kx+4=0
∴x=-4/k
The ordinate of the intersection with the Y axis is 4
∴s=|-4/k ×4|/2=8
|k|=1
‖ k = 1 or - 1
Ψ y = x + 4 or y = - x + 4
Let f (x) = |x + 1 | + |x + 2 | - A under the root sign. If the domain of F (x) is r, the value range of a is obtained
Because the minimum value of | x + 1 | + | x + 2 | is 1 (any point between - 1 and - 2 of X can be taken)
In the root sign | x + 1 | + | x + 2 | - a > = 0
It holds for any x, that is, it holds for the minimum value of | x + 1 | + | x + 2 |,
therefore
1-a>=0
A
|X + 1 | + | x + 2 | > = 1, so a = 0
That is to find the range of y = |x + 1| + |x + 2|
Divide into x
Given that the ordinate of the intersection of the image of a function of degree and the Y axis is - 2 and passes through the point (5,3), the expression of the function is______ .
Let the analytic formula of a function be y = KX + B. according to the meaning of the question, the graph of a function passes through two points (0, - 2), (5, 3), then B = - 25K + B = 3, and the solution is k = 1b = - 2. The analytic formula of a function is y = X-2
Lgx + LG (x + 3) = 1 to solve x
lg[x(x＋3)]＝1＝lg10
∴x^2＋3x＝10
The solution is: x = - 5 or x = 2
Because x > 0
∴x＝2
X (x + 3) = 10 to get x = 2 and x = - 5, but lg-5 is meaningless, so x = 2
lgx(x+3)=1
x(x+3)=10
x^2+3x-10=0
x=2(x>0)
lgx+lg(x+3)=lgx(x+3)=1
It is concluded that x (x + 3) = E,
Just solve the root of x > 0 (because lgx itself contains the condition x > 0)
Sorry, error x (x + 3) = 10