college algebra, HELP!?

Question

h(x)= 2-x ; 8 is less than or equal to x is less than 0
x squared; 0 is less than or equal to x is less than 8

I think it is supposed to be a graph

Answer 1

bonne chance....

you didn't write what the question was asking.

I can only assume you want h(x).


a) h(x) = 2 - x ; where 8 is less than or equal to x is less than 0
( 8 <= x < 0 ??? But, that is what you've typed.)

well, even though the problem says X < 0, let's just stary by saying x = 0 to make it easier to explain. that would mean h(0) = 2.

But, x is less than zero (some negative number),
so, h(x) where x < 0 gives
2 - (some negative number), or
2 + (some positive number)

which makes h(x) more than 2, or greater than 2.

right now we have this:
8 <= x < 0
??? h(x) > 2


This time it says X can equal 8.
So, h(8) = 2 - 8 = -6

but, it really says 8 <= x... so X can be more than 8... like 10, or 100, or 10,000,000 and that would mean that h(x) only gets more and more negative. In other words, h(x) gets smaller and smaller as X gets bigger and bigger. Or, h(x) gets less and less than -6 as X gets bigger.

now we know:
8 <= x
-6 >= h(x)

So, it is written:
┏━━━━━━━━┓
┃ ☞ 8 <= x < 0 ☜ ┃
┃ ✓ -6 >= h(x) > 2 ┃
┗━━━━━━━━┛


by similar methods

b) x squared; 0 is less than or equal to x is less than 8
h(x) = x^2
┏━━━━━━━━┓
┃ ☞ 0 <= x < 8 ☜ ┃
┃✓ 0 <= h(x) < 64 ┃
┗━━━━━━━━┛


================================

scroll down to "compund inequalities" here:
http://www.math10.com/en/algebra/solving-linear-inequalities/solving-linear-inequalities.html

Example 3 about the disjunction best describes the first part of the question. Where I assume what you have written is a disjunction, and not accidentally changed a conjunction.

an alternate way of graphing for the <= and < is to replace the Bracket ] or [, which means <= or >= respectively with filled circles at the point (a solid dot).

The use of Parenthesis ) or (, which means < and > respectively, is to use an open circle around the point.

this place shows how to draw the line graphs:
http://www.purplemath.com/modules/ineqlin.htm

Answer 2

Where's the question? Looking for the range given the restricted domain?

if h(x) = 2-x and 8 ≤ x < 0, the range is as empty as the domain, since there are no real numbers less than 0 and greater than 8.

if f(x) = x² and 0 ≤ x < 8, then 0 ≤ f(x) < 64.

Answer 3

Your placement try does not qualify you to get right into a greater physically powerful lever of Math so which you will would desire to take the pre-algebra previously you may examine in for a Math that's tranferable to a school. See a councelor.

Answer 4

if you are trying to find x, then
x = all real numbers/ infinite number of solutions

8<=x<0<=x<8
<= means less than or equal
so you have x=0 through 7;8 and up;anything less than 0
so your left with everything.

this is what i think anyway, i'm only in high school math... so
sorry if it's wrong

Answer 5

Please explain the problem better. It looks familiar, but doesn't make sense as it is.