If we assume the hypotenuse is of side length 5:
a^2+b^2=c^2
3^2+b^2=5^2
9+b^2=25
b^2=16
b=4
If the hypotenuse is the unknown:
a^2+b^2=c^2
3^2+5^2=c^2
9+25=c^2
c^2=34
c=(34)^(1/2)
16.875=ab^3 (divided by what follows)
7.5=ab^1 gives you:
2.25=b^2
b=1.5 now we can solve for a...
7.5=a(1.5^1), a=5 so the function is:
y=5(1.5^x)
Answer:
True :p
Step-by-step explanation:
We'll need to find the 1st and 2nd derivatives of F(x) to answer that question.
F '(x) = -4x^3 - 27x^2 - 48x - 16 You must set this = to 0 and solve for the
roots (which we call "critical values).
F "(x) = -12x^2 - 54x - 48
Now suppose you've found the 3 critical values. We use the 2nd derivative to determine which of these is associated with a max or min of the function F(x).
Just supposing that 4 were a critical value, we ask whether or not we have a max or min of F(x) there:
F "(x) = -12x^2 - 54x - 48 becomes F "(4) = -12(4)^2 - 54(4)
= -192 - 216
Because F "(4) is negative, the graph of the given
function opens down at x=4, and so we have a
relative max there. (Remember that "4" is only
an example, and that you must find all three
critical values and then test each one in F "(x).
