Answer:
r = √13
Step-by-step explanation:
Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms: x^2 + 6x + y^2 -2y = 3. Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."
We need to "complete the square" of x^2 + 6x. I'll just jump in and do it: Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x: x^2 + 6x + 3^2 - 3^2. Then do the same for y^2 - 2y: y^2 - 2y + 1^2 - 1^2.
Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2. Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1. Making these replacements:
(x + 3)^2 - 9 + (y - 1)^2 -1 = 3. Move the constants -9 and -1 to the other side of the equation: (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13
Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r: r^2 = 13, so that the radius is r = √13.
Step-by-step explanation:
Follow my steps. Sorry im bit untidy .
Follow these 1°,2°,3°
Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).
Okay, here we have this:
It says that: "Seven more than four times a number (x) is less than or equal to 6"
So, let's take the number as x and from the statement we obtain the following inequality:
7+4x≤6
And as it contains a variable the inequality is open by definition
Your greatest common factor is 9.
9 * 2 = 18
9 * 4 = 36
9 * 5 = 45
