Refer to the image attached.
Given: 
and 
are congruent.
To Prove: 
ABC is an isosceles triangle.
Construction: Construct a perpendicular bisector from point B to Line segment AC.
Consider triangle ABD and BDC,
(given)
(By the definition of a perpendicular bisector)
(By the definition of a perpendicular bisector)
Therefore, 
by Angle Side Angle(ASA) Postulate.
Line segment AB is congruent to Line segment BC because corresponding parts of congruent triangles are congruent.(CPCTC)
This is quite a complex problem. I wrote out a really nice solution but I can't work out how to put it on the website as the app is very poorly made. Still, I'll just have to type it all in...
Okay so you need to use a technique called logarithmic differentiation. It seems quite unnatural to start with but the result is very impressive.
Let y = (x+8)^(3x)
Take the natural log of both sides:
ln(y) = ln((x+8)^(3x))
By laws of logarithms, this can be rearranged:
ln(y) = 3xln(x+8)
Next, differentiate both sides. By implicit differentiation:
d/dx(ln(y)) = 1/y dy/dx
The right hand side is harder to differentiate. Using the substitution u = 3x and v = ln(x+8):
d/dx(3xln(x+8)) = d/dx(uv)
du/dx = 3
Finding dv/dx is harder, and involves the chain rule. Let a = x+ 8:
v = ln(a)
da/dx = 1
dv/da = 1/a
By chain rule:
dv/dx = dv/da * da/dx = 1/a = 1/(x+8)
Finally, use the product rule:
d/dx(uv) = u * dv/dx + v * du/dx = 3x/(x+8) + 3ln(x+8)
This overall produces the equation:
1/y * dy/dx = 3x/(x+8) + 3ln(x+8)
We want to solve for dy/dx, achievable by multiplying both sides by y:
dy/dx = y(3x/(x+8) + 3ln(x+8))
Since we know y = (x+8)^(3x):
dy/dx = ((x+8)^(3x))(3x/(x+8) + 3ln(x+8))
Neatening this up a bit, we factorise out 3/(x+8):
dy/dx = (3(x+8)^(3x-1))(x + (x+8)ln(x+8))
Well wasn't that a marathon? It's a nightmare typing that in, I hope you can follow all the steps.
I hope this helped you :)
Answer:
1536 and 6144
Step-by-step explanation:
you are multiplying by 4
I hope this helps you to understands better
Value of 3 is in the solution set
A) from 2 to infinity
B) from -infinity to -3
C) -3 to 2
D) -infinity to -3 and 2 to infinity
E) -infinity to infinity
F) no because the slope is linnear in 2 stops but has a different slope in both spots meaning it can’t be a linnear function or an exponential funtion
