If f(x) = 9^x , them prove that : f(m+n+p) = f(m) × f(n) × f(p) .

Mathematics

1 answer:

blondinia [14]3 years ago

4 0

Answer:

f(m+n+p) = 9^(m+n+p)

while f(m) × f(n) × f(p) = 9^m+9^n+9^p = 9^(m+n+p)

so the second property explains why it's a + instead of ×

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Find the volume of a cone with a base diameter of 12 m and a height of 10 m.

dexar [7]

Answer:

=376.8 $m^{3}$

Step-by-step explanation:

V=n $r^{2}$ $\frac{h}{3}$ Where V is the volume , r is the radius and h is the perpendicular height .

r = $\frac{12}{2}$ m=6

h = 10m

n=3.14

v= 3.14 * 6*6 * $\frac{10}{3}$

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3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$