Your function is not defined at x=0 and x=4 (you have vertical asymptotes there). So, the denominator must be zero at x=0 and x=4. This is the case for
So, the function with that denominator is graphed.
Make the equation the same or a multiple of the other to have an infinite number of solutions. See how the -8/2= -4 and 6/2=3 ? then -2/2 = c
Make c = -1 and they are the same linear equation. (same slope, same intercepts, same line) and therefore have infinite solutions.
In order to have no solutions, the lines cannot cross at all and so they must be parallel but not the same line. Parallel lines have the same slope with different intercepts. if you rearrange both equations in slope intercept form:
y = x/3 - 4/3
y = - cx/3 - 4/3
no matter what you make c the lines will always cross at the y-intercept (0, -4/3). this is a solution and therefore there's no value of c that would produce a system with no solution.
The answer should be 3 since its the absolute value and thats just how far a number is from 0 so it would be 8-5=3
The answer is x-5 First collect like term so it will be x-9+4 and then subtract 9-4= 5 so now it will be x-5
The values that make this statement falser are any in which a and b do not have the same sign.
For instance, if a was equal to 3 and b was equal to -3 than see the results.
|a+b|=
|3+-3|=
|0|= 0
Then see the next equation with the same selections
|a|+|b|
|3|+|-3|
3 + 3 = 6
And this would be true no matter which is the negative, as long as there is one negative and one positive.
