HOw do i start this problem? let h(x)=12-3x. Determine all values of x so that h(2x)>24
K kkiceman14 New member Joined Oct 30, 2005 Messages 22 Nov 2, 2005 #1 HOw do i start this problem? let h(x)=12-3x. Determine all values of x so that h(2x)>24
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Nov 2, 2005 #2 Write down what h(2x) is by substituting 2x for x in h(x). Make it bigger than 24.
K kkiceman14 New member Joined Oct 30, 2005 Messages 22 Nov 2, 2005 #3 make what bigger than 24? SO should it look like this h(x)=12-3(2x)
K kkiceman14 New member Joined Oct 30, 2005 Messages 22 Nov 2, 2005 #4 make what bigger than 24? SO should it look like this h(2x)=12-3(2x)
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Nov 2, 2005 #5 Absolutely. Expand the parentheses and you have h(2x) = 12 - 6x h(2x) is to be greater than 24. 12 - 6x > 24 Can you take it from there?
Absolutely. Expand the parentheses and you have h(2x) = 12 - 6x h(2x) is to be greater than 24. 12 - 6x > 24 Can you take it from there?
K kkiceman14 New member Joined Oct 30, 2005 Messages 22 Nov 2, 2005 #6 i have another question it is let f(x,y)=x+2y+6. Find the value(s) of a so that f(6,a) - f(a,6)=f(a,a)
i have another question it is let f(x,y)=x+2y+6. Find the value(s) of a so that f(6,a) - f(a,6)=f(a,a)
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Nov 2, 2005 #7 This is not much different than the previous problem. f(6,a) just means substitute x=6 and y=6 into f(x,y): f(6,a) = 6 + 2a + 6 Do the same for the other two. You have an equation to solve for a.
This is not much different than the previous problem. f(6,a) just means substitute x=6 and y=6 into f(x,y): f(6,a) = 6 + 2a + 6 Do the same for the other two. You have an equation to solve for a.