When the value range of independent variable x satisfies what conditions, the value of function y = 1.5x + 6 satisfies the following conditions? 1. Y = 0.2. Y0.4. Y

How to find the value range of independent variable in drawing positive scale function image
I just learned positive proportion function, but I don't understand it. For example, how to find the value range when y = - 3x
The original question is as follows:
Take a point P on the image of the function y = - 3x, pass through the point P and make the PA ⊥ X axis. Given that the abscissa of the point P is - 2, calculate the area of △ POA (o is the origin of the coordinate)
But I want to draw an image. I don't understand. I hope someone can help me

The abscissa of point P is - 2, and P is substituted on the image of function y = - 3x
The ordinate of P is 6
OA=2,AP=6
The area of △ POA = 1 / 2 * OA * AP = 6
There is no way to draw the diagram, and for the value range, remember that the area can never be zero or negative

The value range of independent variable of inverse proportion function
Given the inverse scale function y = - 3 / 2x, the value range of the independent variable is_____ .
Thank you! The sooner the better!

y=-3/2x
The denominator is not equal to 0
therefore
X is not equal to 0

6X = 18 solve equation and test

6x÷6=18÷6
x=3
test
x=3
6x=6×3=18
establish
So x = 3 is the solution of the equation

Judge the position relationship between the line 4x-3y = 50 and the circle x2 + y2 = 100. If there is a common point, calculate the coordinates of the common point

The vertical distance d between the center of the circle and the straight line at (0,0) = 10, so the straight line is tangent to the circle

How to use calculator to calculate 10! 10?

10! Is the factorial of 10, which means 10x9x8x7x6x5x4x3x2x1. Use the calculator to calculate the order of keys. First input the number 10, then press the 2ndf key, and finally press the CE key (or find the key with n! Above the key)

Univariate quadratic equation [200X · (1 + x) - 100] · (1 + x) = 108

Hey, it looks like the univariate cubic equation [200X + 200X & # 178; - 100] * (1 + x) = 108. Did you give the wrong question?

If and only if the limit of function f (x) exists when x → X0

F '(x0 +), f' (x0 -), and f '(x0 +) = f' (x0 -)

If x ^ 2 + X + 1 = 0, find (1) x ^ 2005 + x ^ 2003 + x ^ 2001 + 2 (2) x ^ 3 + 2x ^ 2 + 2x + 2006

x^2+x+1=0
x^2+1=-x
square
x^4+2x^2+1=x^2
x^4+x^2+1=0
x^2005+x^2003+x^2001+2
=x^2005+x^2003+x^2001+2
=x^2001((x^4+x^2+1)+2
=x^2001*0+2
=2
x^3+2x^2+2x+2006
=(x^3+x^2+x)+(x^2+x+1)+2004
=x(x^2+x+1)+(x^2+x+1)+2004
=x*0+0+2004
=2004

4x2-8x + 1 = 0 (with matching method)

The equation 4x2-8x + 1 = 0 is divided by 4 to get x2-2x + 14 = 0. Move the constant term of the equation 4x2-8x + 1 = 0 to the right of the equal sign to get x2-2x = - 14. Add the square of one-half of the first term on both sides of the equation to get x2-2x + 1 = 34, ｜ X-1 = ± 32. The solution is X1 = 2 + 32, X2 = 2 − 32

When the value range of independent variable x satisfies what conditions, the value of function y = 1.5x + 6 satisfies the following conditions? 1. Y = 0.2. Y0.4. Y