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3 years ago

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Explain the steps on how to convert these repeating decimals to a fraction (a) 0.77 (77 is repeating) (b) 2.3 (3 is repeating (c

) 1.013 (13 is repeating)

2 answers:

Alex787 [66]3 years ago

7 0

You would have to round all the repeating decimals to the nearest hundredths or tenths or thousandths
A.) 0.78 = 78/100 you can reduce on this answer so it would be 39/50
B.) 2.3 = 2 3/10
C.) 1.013 = 1 13/1000

Arturiano [62]3 years ago

5 0

<span>77/100 = 7/10 I think sorry if im wrong

Best of luck :)</span>

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Answer:

Step-by-step explanation:

Here you go mate

Step 1

(2x-2)=8 Equaion/Question

Step 2

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(2x-2)=8

Step 3

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1 year ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

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2 years ago

Tony took a city bus from his dorm to the school library, and then to a gym for a workout

Feliz [49]

Answer:

27 miles.

Step-by-step explanation:

Here I attach the draw of the coordinates.

Tony traveled 3 segments. The first was from (12,6) to (12, 15), where, leting 12 constant, he moved from 6 to 15 in the ordinates axis, which implies 9 units. This is the section 1 in the draw.

Then he moved from point B to C. If you notice, this distance is the hypotenuse on the the triangle DBC. We can find this value using Pitagoras' theorem:

DB^2 + CD^2 = CB^2

With DB=15 and CD=8 (12 minus 4 = 8)

15^2 + 8^2 = 289

So CB^2=289

Applying sqr root:

CB = 17

So, the second section has a measure of 17 units.

Finally, the 3rd section is the hypotenuse of the DAC triangle and we can use Pitagoras to solve it:

CD^2 + AD^2 = CA^2

8^2 + 6^2 = CA^2

64 + 36 = 100

So, CA=10

In the 3r section we traveled 10 units.

So, in total he traveled 10 + 17 + 9 = 36 units

As every unit is 0.75 miles he traveled 36*0.75 miles:

36*0.75 = 27 miles

He traveled in total 27 miles!!

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3 years ago

love history [14]

The particular construction used will make OS the angle bisector of ∠POQ, so
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This is true for selection C.

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