I posted this question before:
What is the area enclosed by the graph of the absolute value(3x)+ the absolute value(4y)=12
I tried solving for y and got: y=4-x
when I graphed this, I got a linear equation graph...I am wondering if I am supposed to find the area of triangle by finding the lengths of the sides??? Am I doing this right? Any suggestions would be great...
There is more than that triangle.
consider when y=0 to -3, x ranges as + or - 4 to zero. Did you graph both those plots? I think you will get four triangles in which to find the area. I would find the area an easier method, but computing the inner and outer triangle between the x axis and y, then subtracting. THere are other ways to find the triangle area also.
Are you that stupid that you cannot solve a problem that simple. Here idiot: graph it and then solve for x then y. Duh! I am a engineer that makes millions of dollars, this is just a waste of time.
Yes, graph both plots;then what do you get;Come on genious your the one on the interent looking for help on something so simple. Why don't you just drop out of school and work at McDonalds, I heard that they are hiring idiots. KillK whats up my wigga!
i nedd help working this problem
2 log X = log 2+log ( 3x-4)
To solve the equation 2 log X = log 2 + log (3x-4), you can follow these steps:
Step 1: Apply logarithm properties
Using the logarithm properties, you can rewrite the equation as a single logarithm:
log X^2 = log (2 * (3x-4))
Step 2: Remove the logarithms
Since the logarithm is a one-to-one function, you can remove the logs by setting the arguments equal to each other:
X^2 = 2 * (3x-4)
Step 3: Expand and rearrange the equation
Multiply out the right side of the equation:
X^2 = 6x - 8
Step 4: Rearrange the equation
Move all terms to one side to create a quadratic equation:
X^2 - 6x + 8 = 0
Step 5: Factor or use the quadratic formula
To solve the quadratic equation, you can either try factoring it or use the quadratic formula.
Factoring:
(X - 2)(X - 4) = 0
So X = 2 or X = 4.
Alternatively, you can use the quadratic formula:
X = (-(-6) ± √((-6)^2 - 4(1)(8))) / (2(1))
X = (6 ± √(36 - 32)) / 2
X = (6 ± √(4)) / 2
X = (6 ± 2) / 2
X = 4 or X = 2
So the solutions to the equation are X = 2 and X = 4.