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3 years ago

Which equation is represented by the number line

Mathematics

1 answer:

Pavel [41]3 years ago

5 0

6 + (-10) = -4 hope that helps

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Which expression is equivalent to this expression? (41-6) 4

leva [86]

Answer:

35x4

Step-by-step explanation:

do the equation in the parenthesis to simplify it

7 0

3 years ago

7 wholes and one half divided by 1 whole and nine tenths

Deffense [45]

Answer:

3.947

Step-by-step explanation:

7 wholes = 7

one half = 0.5

1 whole = 1

nine tenths = 0.9

7.5/1.9 ≈ 3.947

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3 years ago

Solve each equation by factoring 6k2+21=27k

ikadub [295]

$6k^2+21=27k\\\\ 6k^2-27k+21=0\ \\\\3(2k^2-9k+7)=0\\\\3(2k^2-2k-7k+7)=0\\\\3[2k(k-1)-7(k-1)]=0\\\\3(k-1)(2k-7)=0\ \ \ \Leftrightarrow\ \ \ k=1\ \ \ or\ \ \ k=3.5$

4 0

3 years ago

Read 2 more answers

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

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

5 0

3 years ago

All of the teachers at Chapel Glen are working together to make a HUGE snow fort! They need a total of 118 snow bricks, each wei

pogonyaev

Step-by-step explanation:

interested girls only

id : 495 842 5906

pass: TkD16c

5 0

3 years ago