Answer:
Dy/Dx=x
(
2
ln
(
x
)
−
1
)/
ln
^2
(
x)
Step-by-step explanation:
We have this function and let's derive it in terms of x.
y =x^2/In x
Dy/Dx=(x^2/In x)'=2/lnx *(x^2)'-(x^2/In x)'=> 2
x
*ln
(
x
)
−
x
*ln
^2
(
x
)
=x
(
2
ln
(
x
)
−
1
)/
ln
^2
(
x)
Answer:
<h3>Graph 3</h3>
Line starting at x = -2
- <u>Domain</u>: x ≥ -2
- <u>Range</u>: y ≥ 0
<h3>Graph 4</h3>
Vertical line
- <u>Domain</u>: x = 3
- <u>Range</u>: y = any real number
<h3>Graph 5</h3>
Quadratic function with negative leading coefficient and max value of 3
- <u>Domain</u>: x = any real number
- <u>Range</u>: y ≤ 3
<h3>Graph 6</h3>
Curve with non-negative domain and min value of -2
- <u>Domain</u>: x ≥ 0
- <u>Range</u>: y ≥ -2
<h3>Graph 7</h3>
Line with no restriction
- <u>Domain</u>: x = any real number
- <u>Range</u>: y = any real number
<h3>Graph 8</h3>
Quadratic function with positive leading coefficient and min value of 4
- <u>Domain</u>: x = any real number
- <u>Range</u>: y ≥ 4
<h3>Graph 9</h3>
Parabola with restriction at x = -4
- <u>Domain</u>: x = any real number except -4
- <u>Range</u>: y = any real number
<h3>Graph 10</h3>
Square root function with star point (2, 0)
- <u>Domain</u>: x ≥ 2
- <u>Range</u>: y ≥ 0
I got 11,304. I don't know if that's right or if it's wrong. Let me check again
15 16 and then 17 to be an equilateral
Answer:
For the first box, the answer is 3.
For the second box, the answer is -1.
For the third box, the answer is 0.
I hope this helped!
