#1
If
v=(6-(r^3/3))h and h/(2-r)=5/2
write V as a function of h .
Simplify the function so that you can express it as a polynomial in h .
#2
If y=ln(x) , write ln(x^x) as a simple function of y
thxs alot!
2 h = 10 - 5 r
5 r = 10-2h
r = 2 -(2/5) h
v = h(6 - (1/3)(2 - (2 h/5))^3 )
Yuck, you can multiply it out
ln x^x = x ln x = x y
but
e^y = x
so
ln x^x = y e^y
#1 To write V as a function of h, we substitute the given value of h/(2-r) into the equation for v.
Given: h/(2-r) = 5/2
Substituting this value into the equation for v, we have:
v = (6-(r^3/3)) * (5/2)
To simplify this expression and express it as a polynomial in h, we need to expand and simplify the expression.
First, let's simplify the expression inside the parentheses:
v = (6 - (r^3/3)) * (5/2)
= (6 - (r^3/3)) * (5/2)
= (6 - (r^3/3)) * 5/2
= 30/2 - (r^3/3) * 5/2
Next, let's multiply the fractions:
v = 30/2 - (r^3/3) * 5/2
= 30/2 - (5r^3/6)
Now, let's simplify the expression further:
v = 30/2 - (5r^3/6)
= 15 - (5r^3/6)
= 15 - (5/6)r^3
Therefore, V as a function of h is expressed as V(h) = 15 - (5/6)r^3. This is a polynomial function in h.
#2 To write ln(x^x) as a simple function of y, we need to use the properties of logarithms.
Given: y = ln(x)
Using the logarithmic property, ln(a^b) = b * ln(a), we can rewrite ln(x^x) as:
ln(x^x) = x * ln(x)
Now, substitute the value of y = ln(x) into the equation:
ln(x^x) = x * ln(x)
= x * y
Therefore, ln(x^x) can be expressed as a simple function of y, which is ln(x^x) = x * y.