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2 years ago

How many 4 letter arrangements can be made by using different letters from the word hexagon

Mathematics

1 answer:

anastassius [24]2 years ago

3 0

Answer:

I can find at least 10 4-letter words out of hexagon.

Step-by-step explanation:

1.) Hexa (means 6)

2.) gone

3.) nage (a type of bullion)

4.) gane (to yawn)

5.) hage (a frightening old woman; a witch)

6.) hane (ghostly)

7.) axon (a part of a nerve cell)

8.) hang

9.) oxen

10.) hoax

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3 years ago

X^(2)-4x-17=0<br> Please show work step by step

Dimas [21]

Step-by-step explanation:

It seems here that they are asking us to solve for x

to do this we first need to factor

Since we can't factor this using the normal method we can instead do this

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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

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

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3 years ago

A is 60% of B , B is 30% of C, what percentage of C is A

zaharov [31]

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3 years ago