Answer:
a.
<u>Increasing:</u>
x < 0
x > 2
<u>Decreasing:</u>
0 < x < 2
b.
-1 < x < 2
x > 2
c.
x < -1
Step-by-step explanation:
a.
Function is increasing when it is going up as we go rightward
Function is decreasing when it is going down as we go rightward
We can see that as we move up (from negative infinity) until x = 0, the function is increasing. Also, as we go right from x = 2 towards positive infinity, the function is going up (increasing).
So,
<u>Increasing:</u>
x < 0
x > 2
The function is going down, or decreasing, at the in-between points of increasing, that is from 0 to 2, so that would be:
<u>Decreasing:</u>
0 < x < 2
b.
When we want where the function is greater than 0, we basically want the intervals at which the function is ABOVE the x-axis [ f(x) > 0 ].
Looking at the graph, it is
from -1 to 2 (x axis)
and 2 to positive infinity
We can write:
-1 < x < 2
x > 2
c.
Now we want when the function is less than 0, that is basically saying when the function is BELOW the x-axis.
This will be the other intervals than the ones we mentioned above in part (b).
Looking at the graph, we see that the graph is below the x-axis when it is less than -1, so we can write:
x < -1
The sum of integers p and q is modeled on a number line, where |p| < |q|. In each box write p,q, or p + q. Then write an addition equation
Plug y=2x+1 into <span>x+2y=17, we have:
x+ 2(2x+1)= 17
</span>⇒ x+ 2*2x+ 2*1= 17 (distributive property)
⇒ x+ 4x+ 2= 17
⇒ 5x= 17-2
⇒ 5x= 15
⇒ x= 15/5
⇒ x= 3
<span>
y= 2x+1= 2*3+1= 7
Final answer: x= 3, y= 7~</span>
The simplified version of arccot sqroot 3/3 is
Answer:
1. 1/2, rise/run and since you rise one and run 2 it'll be 1/2
2. 7/0 = undefined automatically, there is no value(undefined) if there is no run
3. rise: 28 - 18 10 5
_______ = _____ = _____ = 5
run: 4 - 2 2 1
4. 1, it rises 1 and runs 1
