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2 years ago

Solve: 8+2x-12=58 i need help please asap.

Mathematics

2 answers:

Nimfa-mama [501]2 years ago

7 0

Answer: x=31

Step-by-step explanation:

-4 + 2x=58

2x-4=58

2x-4+4=58+4

2x=62

2x/2=62/2

x=31

cricket20 [7]2 years ago

3 0

Answer:

x = 31

Step-by-step explanation:

- Combine all like terms;

8+2x-12=58

2x -4 = 58

- Add 4 to both sides.

2x = 62

- Divide both sides by 2.

x = 31

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Answer:

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Step-by-step explanation:

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2 years ago

-6x + 3y = 9<br>10x + 4y = -6<br>

Stells [14]

Step-by-step explanation:

(-6x +3y=9 )×4

( 10x+4y)×3

-24x +12y =36

30x+12y= -18

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2 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

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$y'=(t+y)^2-1$

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$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

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2 years ago