Which of the following situations can be described by the equation 2x – 4 = 10?

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

8 0

Step-by-step explanation:

Add 44 to both sides.

2x=10+4

2 Simplify 10+410+4 to 1414.

2x=14

3 Divide both sides by 22.

x=\frac{14}{2}

4 Simplify \frac{14}{2}

to 77.

x=7

Done

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6. Blake went to the grocery store at 6:30PM, He returned to his house around 9:14PM after running

zhenek [66]

If you subtract the times periods form each you will be able to get your answer. So, 9:14-6:30 it will be 2 hours and 44 mins because from 6:30 to 7 it 30 mins, and from 7 to 9 it’s 2 hours. So far, add the two and you get 2h and 30mins and from 9 to 9:14 it’s 14 mins. So, 2h 30 mins plus 14 mins is 2h and 44mins.

5 0

2 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

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