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3 years ago

If 1/6x+2/3y=8 what is the value of 2x+8y

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Answer: 96

Step-by-step explanation:

Simply multiply the first question by 12 to get 2x+8y=96

Hope it helps <3

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Find the coordinates of the other endpoint of the segment, given its midpoint and one endpoint. (Hint: Let (x,y) be the unkno

makkiz [27]

Step-by-step explanation:

Hey, there!!

Here, one point is A(10,8) and P(8,5) is the midpoint.

Let B(x,y) be the another end point.

Now,

Using midpoint formulae,

$p(x.y) = \frac{x1 + x2}{2} . \frac{y1 + y2}{2}$

$p(8.5) = ( \frac{10 + x}{2} . \frac{8 + y}{2} )$

Since they are equal,equating with their corresponding elements we get,

$8 = \frac{10 + x}{2}$

or, 16 = 10 + x

or, x=16-10

Therefore, x = 6

Now,

$5 = \frac{8 + y}{2}$

or, 10 = 8 + y

or, y = 2

Therefore, The coordinates of another point are B(6,2)

Hope it helps .....

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3 years ago

Zeke had 10 marbles in the beginning of the game and he had 7 at the end of the game. What was the percent loss?

zloy xaker [14]

The answer is 30%. It would be like 70% is what you have left of the 100.

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3 years ago

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

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