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3 years ago

What % of 60 is 35?

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

Answer:

58.33%

Step-by-step explanation:

is %

-------- = ---------

of 100

Just plug in the numbers based on what it tells you.

What % = goes in the % spot

of 60 = goes in the of spot

is 35 = goes in the is spot

35 x

--------- = ---------

60 100

Cross multiply:

60 * x = 60x 35 * 100 = 3500

60x = 3500

Divide by 60 to get x by itself.

3500 / 60 = 58.33

Put the percent sign after it:

58.33%

I hope this helps.

masya89 [10]3 years ago

5 0

Answer:

58.3333333333(it goes on and on and on actually. it's called a *bar notation*)

Step-by-step explanation:

to set the equation up

35 goes over 60

p or % goes over 100

first, you cross multiply

so 35 x 100

and p or % = 60

now that you multiplied 35 by 100 and gotten 3500, you are going to divide it by 60

once you have divided, you should come up with 58.33333333333(you get the idea)

I hope this helped :)

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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

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

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

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