Answer:
x+6
Step-by-step explanation:
to find f(-x) plug in -x for all the values of x in the equation
since the absolute value of -x is x you get x+6
There are 14 possibilities for first place.
There are 13 possibilities for second place, since a dog can't get both first and second place.
There are 12 possibilities for third place, since a dog can't get both first, second, and third place.
There are 11 possibilities for fourth place, since a dog can't get both first, second, third, and fourth place.
Multiply them together and get 24024 possibilities.
Hope this helps!
Answer:
1.18
Step-by-step explanation:
5+12=17 20diviede by 17=1.17647 rounded is 1.18
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 :)