I believe it's b , hopefully this helps
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
Answer:
x= 66
Step-by-step explanation:
if you set up the equation mathematically, it would be x+0.5x=99, because 50% is essentially just 0.5.
so, now that you have the equation, you have to add like terms, so it would be 1x+0.5x=99, and 1+0.5 is 1.5, so youre left with the equation 1.5x=99.
from there, you solve for x by dividing each side by 1.5:
1.5x=99
/1.5 /1.5
x=66
im sorry if this didnt make much sense, but i hope this helped!! :D
Answer:
Add all the numbers together and u will get your answer
Step-by-step explanation:
We are asked to find the area of the base or the side with the dimensions 4 and 2. The length would be the longest side and the width is the shorter side. The area is calculated as the product of length and width.
Area of base = 4 x 2 = 8 square units.
Therefore, the correct answer is B.
