What is the distributive property for 8 x 71?

2 answers:

7 0

Distributive property is a(b+c)=ab+ac
it could be
8 is 'a' or
71 is 'a'

I prefer 71=a so
basically b+c=8
probably prefer positive so
some possible combls are
0 and 8
1 and 7
2 and 6
3 and 5
4 and 4
and the rest is just reverse

71(0+8)=71(0)+71(8)=0+568=568
71(1+7)=71(1)+71(7)=71+497=568
71(2+6)=71(2)+71(6)=142+426=568
71(3+5)=71(3)+71(5)=213+355=568
71(4+4)=71(4)+71(4)=284+284=568

VARVARA [1.3K]3 years ago

5 0

8 * 71

8(71 - 0) could be a possibility.

Answer: 8(71 - 0)

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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

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