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3 years ago

7

What’s the least common multiple of the denominators of 2/3x+1/6=-3/4x+1

2 answers:

Dmitrij [34]3 years ago

6 0

Answer:

x = 30/51

Step-by-step explanation:

$\frac{2}{3} x + \frac{1}{6} = - \frac{3}{4} x + 1 \\ \frac{2}{3} x + \frac{3}{4} x = 1 - \frac{1}{6} \\ \frac{8}{12} x + \frac{9}{12}x = \frac{6}{6} - \frac{1}{6} \\ \frac{17}{12} x = \frac{5}{6} \\ x = \frac{5}{6} \times \frac{12}{17} \\ x = \frac{60}{102} \\ x = \frac{30}{51}$

GenaCL600 [577]3 years ago

4 0

30/51 .................

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shannon walked 4 and 7/8 miles and ran 3 1/2 miles during the week. how much further did she walk than run

love history [14]

she walked 1 3/8 miles more than she ran, Hope this helps :D

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4 years ago

In the diagram below, OPis circumscribed about quadrilateral ABCD. What is<br> the value of x?

trasher [3.6K]

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3 years ago

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

$y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v$

Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

$\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}$

$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

then for $w(x)$ ,

$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

$v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx$

Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

$y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)$

$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

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4 years ago

Can somebody help me please

White raven [17]

The answer to this question is
D.

4 0

3 years ago

Read 2 more answers

Mattie Evans drove 350 miles in the same amount of time that it took a turbopropeller plane to travel 1350 miles. The speed of t

Murljashka [212]

let

d1 = 250 mi the distance that Mattie Evans drove

v1 = the speed of Mattie Evans

d2 = 1300 mi the distance the plane traveled

v2 = the speed of the plane

The speed of the plane was 190 mph faster than the speed of the car:

v2 = v1 + 190

since time = distance/speed and they both traveled the same time we have

d1/v1 = d2/v2

250/v1 = 1300/v2 cross multiply

250v2 = 1300v1 divid eboth sides by 50

5v2 = 26v1

by solving the system of equations:

v2 = v1 + 190

5v2 = 26v1

we find

v1 = 45.24 mph

v2 = 235.24 mph

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2 years ago

Read 2 more answers