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3 years ago

18. Find the value of x that makes the equation true.

Mathematics

1 answer:

IgorC [24]3 years ago

6 0

Step-by-step explanation:

✧ $\large{ \tt{ \: \frac{1}{2} x + 2 = \frac{3}{2} x - 6}}$

We need to get rid of the fractions. Notice that there are 4 terms in the equation. Multiply both sides of the equation by 2 to get rid of the fractions. Multiply by 2 because 2 is the denominator.

⤑ $\large{ \tt{2( \frac{1}{2} x + 2) = 2( \frac{3}{2} x - 6)}}$

⤑ $\large{ \tt{x + 4 = 3x - 12}}$

Subtract 4 from both sides in order to isolate the variable on the left.

⤑ $\large{ \tt{x + 4 - 4 = 3x - 12 - 4}}$

⤑ $\large{ \tt{x = 3x - 16}}$

Move 3x to left hand side and change it's sign

⤑ $\large {\tt{x - 3x = - 16}}$

Subtract 3x from 1x

⤑ $\large{ \tt{ - 2x = - 16}}$

Divide both sides of the equation by -2

⤑ $\large{ \tt{ \frac{ - 2x}{ - 2} = \frac{ - 16}{ - 2}}}$

⤑ $\large{ \tt{x = 8}}$

$\boxed{ \boxed{ \underline{ \tt{Our \: Final \: Answer : \boxed{ \underline{ \tt{x = 8}}}}}}}$

Hope I helped ! ♡

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4 years ago

Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1

storchak [24]

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field

$\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})$

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

$\large W=\int_{C}Fdr$

where C is the segment that goes from A to B.

$\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt$

Integrating by parts the last integral:

$\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1$

and

$\large \boxed{W=\int_{C}Fdr=e+24}$

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

$\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1$

The segment that goes from (1,0,0) to B can be parameterized as

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

$\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}$

Hence, the work required to move the particle from A to B along D is

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

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4 years ago