Item 5

Mathematics

1 answer:

kotykmax [81]2 years ago

6 0

Here’s the link to the answe

:510 around th e cme3 cthulhu.com

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500=100(x-1)<br> Someone tell me step by step

Andre45 [30]

First, distribute the 100 to the (x-1) portion:
500 = 100x - 100

then, add 100 to both sides of your equation to get “x” alone on the right side:
500 + 100 = 100x - 100 + 100
600 = 100x

next, divide 100x by 100 and 600 by 100 to isolate a singular “x”:
600/100 = 100x/100
6 = x

therefore, x=6 is your solution

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2 years ago

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Step-by-step explanation:

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3 years ago

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Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

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A protractor can measure all types of angles except which type?

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C reflex because it goes over 180 degrees!