Answer:
The last two functions are decreasing!
Step-by-step explanation:
You can easily tell the functions are decreasing if the function is going down from L to R. If it is going up, from L to R, then it is increasing!
Times the x’a together and x*10 then 2*x and 2*10
Answer:
(2)
Step-by-step explanation:
Our logarithmic expression is:
.
Remember the logarithmic property that ln(a/b) = lna - lnb. So, we can write this as:

Also, we can write square roots as powers of one-half, so √e =
. There's another log property that:
. We can apply that here for both the √e and the y³:

Finally, note that ln(e) is just 1, so we have:

The answer is thus (2).
<em>~ an aesthetics lover</em>
We know that the Pythagorean Theorem is a² + b² = c² and that the area of a square is l x w.
Firstly, we'll have to find the two measures of the triangle that correspond to the areas.
Since the figures are squares, we know that the length and width values must be the same.
We could square the numbers to find the side lengths, however we would have to square them again when substituting for the Pythagorean Theorem, so we can leave them as-is and adjust the equation accordingly.
(33) + b² = (44)
Next, we'll subtract our smaller value from our larger.
b² = (11)
Once again, we could find the square root of this number, but we'd just have to square it again to find the area of the square, so we can just simply write our answer as 11 units.
Therefore, the area of the square is 11 units!
<em>Hope this helped! :)</em>
No because root(3)(x+8+1) is equal to x^3 -9 and (x-1)^3+8 is equal to root (3)(√x−8−1)