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3 years ago

Raul has 40 toy cars in his collection.

Mathematics

1 answer:

rosijanka [135]3 years ago

6 0

Answer:

Step-by-step explanation:

40-1-2=37 silver cars

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The mean score on a physics test was 75 points. Amy’s score was 67 points, which was 2 standard deviations below the mean.

Marrrta [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

Ten minutes after a father left for work on the bus, he noticed that he had left his briefcase at home. His son left home, drivi

valkas [14]

Answer:

12 minutes-ish

Step-by-step explanation:

60 m/h = 1 mile per minute

<u>- 48 m/h </u>= 4/5 miles per minute

12 m/h = 1/5 miles per minute

5 0

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

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

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3 years ago

A sampling of 20 inpatients in Hospital X's intensive Care Unit revealed an average lenght of stay of 3.5 days with a standard d

solmaris [256]

Answer:

cghgvjbhkjknlm:L<";jhku.gyj,tfhdmrgeszfhxv bjnkmlk;l<"l/mkjhgjfhtdgrxcvh bjhkml;,';./

Step-by-step explanation:

ahxchigfxjfhjjm jk kj ub'u 'kj ibl

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4 years ago

Arc Length and Radians question- please help! Will mark brainliest! Is 20pts!

Sav [38]

Answer:

Answer is option 2) pi/3

Step-by-step explanation:

$ra dian = \frac{theta \times \pi}{180} \\ here \: theta = 60 \\ on \: substituting \: the \: values \: \\ radian = \frac{60 \times \pi}{180} \\ from \: dividing \: the \: expression \: we \: get \\ = \frac{\pi}{3}$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION.

4 0

3 years ago