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3 years ago

Solve for x 2x+16 X+16

Mathematics

2 answers:

Sati [7]3 years ago

7 0

JL = x+16

JK+KL = x+16

2x+16+7 = x+16

2x+23 = x+16

2x-x = 16-23

x = 7

nydimaria [60]3 years ago

3 0

Answer:

2x(16)+16(16)

32x+256

32x/32=256/32

x=8

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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

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3 years ago

A manufacturer finds that the revenue generated by selling x units of a certain commodity is given by the function R(x) = 60x −

Hitman42 [59]

To find the x value of the max of
f(x)=ax^2+bx+c
when a is negative (if a is positive, we find the minimum)
we do
-b/2a is the x value
to find the y value, we just sub that x value back into the function

so

R(x)=-0.2x^2+60x+0
-b/2a=-60/(2*0.2)=-60/-0.4=150
x value is 150
make 150 units

sub back to find revenue
R(150)=-0.2(150)^2+60(150)
R(150)=-0.2(22500)+9000
R(150)=-4500+9000
R(150)=4500

max revenue is achieved when 150 units are produced yeilding $4500 in revenue

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3 years ago