I'm taking "part 3" to mean the 3rd of the posted questions.
If
, then
![f'(x)=(x^3+2x)'\ln x+(x^3+2x)(\ln x)'=(3x^2+2)\ln x+\dfrac{x^3+2x}x](https://tex.z-dn.net/?f=f%27%28x%29%3D%28x%5E3%2B2x%29%27%5Cln%20x%2B%28x%5E3%2B2x%29%28%5Cln%20x%29%27%3D%283x%5E2%2B2%29%5Cln%20x%2B%5Cdfrac%7Bx%5E3%2B2x%7Dx)
![\implies f'(x)=(3x^2+2)\ln x+x^2+2](https://tex.z-dn.net/?f=%5Cimplies%20f%27%28x%29%3D%283x%5E2%2B2%29%5Cln%20x%2Bx%5E2%2B2)
Then when
, we get
.
Let first number is x
second number be y
one number is four less than a second number: x=y-4
Three times the first is 6 more than 6 times the second: 3x=6y+6
substitute x=y-4 in 3x=6y+6
3x =6y+6
3(y-4)=6y+6
3y-12=6y+6
3y-6y=6+12
-3y=18
y=-6
substitute y=-6 in x=y-4
x=-6-4=-10
so first number is -10
second number is -6
Yes what’s the question ?
229-133 = -y
96/-1 = -y/-1
-96 = y
Answer:
Pythagorean Theorem:
+
=
OR
-
= ![b^{2}](https://tex.z-dn.net/?f=b%5E%7B2%7D)
Step-by-step explanation:
You can use it to solve a side of a triangle when you have 2 of the sides of the triangle. I hope this helps! Please mark me brainliest!