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3 years ago

What is the solution to the equation e^3x=12 Round your answer to the nearest hundredth

Mathematics

2 answers:

ch4aika [34]3 years ago

7 0

Answer:

.83 A

Step-by-step explanation:

Lunna [17]3 years ago

3 0

Keywords:

<em>equation, variable, clear, round, centesima, neperian logarithm, exponential </em>

For this case we have the following equation $e ^ {3x} = 12$ , from which we must clear the value of the variable "x" and round to the nearest hundredth. To do this, we must apply properties of neperian and exponential logarithms. By definition:

$ln (e ^ x) = x$

So:

$e ^ {3x} = 12$ We apply Neperian logarithm to both sides:

$ln (e^{3x}) = ln (12)\\3x = ln (12)$

We divide between "3" both sides of the equation:

$\frac {3x} {3} = \frac {ln (12)} {3}\\x = \frac {ln (12)} {3}\\x = 0.828302217$

Rounding out the nearest hundredth we have:

$x = 0.83$

Answer:

$x = 0.83$

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butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

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lakkis [162]

Answer:

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svetlana [45]

Answer:

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Step-by-step explanation:

3:5= 1.5:2.5

Each number is divided or multiplyed by the same number.

Ex. 3/2=1.5 and 5/2=2.5

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