Answer:
So interval notation is with ( and [ where ( is exclusive and [ is inclusive.
Like (1,2) is between 1 and 2 exclusive. [1,2] is between 1 and 2 inclusive. (1,2] is between 1 and 2, 1 exclusive 2 inclusive.
at the point (6,0) you see that the graph goes from above 0 to below 0 (from positive to negative)
The values are positive when x is less than 6 and negative when x is greater than 6.
so the positive interval is
(-infinity, 6)
and with inifinity you always use exclusive
It's that because everything from all the way to the left (-infinity) to 6, is above the x-axis, which means it's positive
using this logic can you do the negative interval?
Step-by-step explanation:
1.
2.
this may help you(:
Answer:
- 3
Step-by-step explanation:
Using the law of logarithms
x = n ⇔ x = 
let ( 
) = n
( 
) = n, then
=
Since bases are the same on both sides then equate the exponents
⇒ n = - 3 → (4)
Answer:
(2,9)
Step-by-step explanation:
To do this you have to solve one of the equations for a variable.
-x+y=7
2x+3y=31
Because the first equation doesn't have any numbers attached to the variables I decided to solve for the y in that equation.
-x+y=7
y=x+7
With this now one would have to plug this into the second equations:
2x+3(x+7)=31
Use the distributive property.
2x+3x+21=31
simplify.
5x+21=31
Isolate the x.
-21 -21
5x=10
solve for the x.
x=2
Plug the x into one of the equations and solve for y.
-x+y=7
-2+y=7
isolate the y.
+2 +2
y=9
This is the point of solution.
