Sup guys, got a couple questions I couldn't figure out.

1. Solve and express the solution in interval notation

5x^2 + 3x ≥ 3z^2 + 2


2. Find an equation of a line that satisfies the given conditions

Slope 2/5 ; y-intercept 4


3. y-intercept 6 ; parallel to the line 2x + 3y + 4 = 0


4. Through (2,6) ; perpendicular to the line y=1


5. Find the domain of the function

f(x) = x^2 + 1, 0 ≤ x ≤ 5


Any help would be appreciated!

are you sure you have stated the question correctly?

what does "z" in question 1 mean?

and in question 5, are you looking for the range? the domain is given as [0,5]

5x^2 + 3x ≥ 3x^2 + 2
2x^2 +3x>= +2
2x^2 +3x -2 >= 0
(2x-1)(x+2)>=0




http://www.purplemath.com/modules/strtlneq2.htm

This post was edited by saber_x3 on Feb 5 2014 05:44pm

yes, it was supposed to be x instead of z but saber figured it out already.

as for question 5, no, it says the domain. so [0,5] is the domain?

Sweet. Thanks a bunch!

still cannot figure out #4

please help if you can

the line y=1 is a horizontal line where the y-value is 1 for every x-value.
You should know that a line perpendicular to a horizontal line is a vertical line.
A vertical line will have the same x-value for every y-value.
Now use your given point (2,6) and take the x-value to figure out your line.
So from the given point and the fact you know it's a vertical line, you get: x=2

E: if you're having trouble following my wording, try sketching what i say line-by-line. It might help you understand it better

This post was edited by TritonV8 on Feb 6 2014 10:12pm

#4) When you need to find a line perpendicular to another line, you find the reciprocal of the
slope. So since y=1 has no slope, your answer will have no slope.

And then you plug in the numbers to the formula: (y-y1)=M(x-x1).

However, in this case you don't need to do any of that. y=1 is a straight horizontal line
at (0,2). So to find the line that is perpendicular to that, you just take the x coordinate
and that's your answer! x=2.