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3 years ago

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 They spent $18 on tickets to the carnival and $36 on food.  They spent the rest of the money on g They spent $18 on tickets

to the carnival and $36 on food.  They spent the rest of the money on games. ames.

1 answer:

Reptile [31]3 years ago

8 0

Answer:

93 = g + 36 + 18

Step-by-step explanation:

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Beth solved the system of equations by making a substitution for y in the second equation. What equation did she get as a result

Slav-nsk [51]

Hello there!

The correct answer is option C

3x + 20x - 55 = -9

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2 years ago

State the domain and range in set notation

antiseptic1488 [7]

Answer:

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The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). Only the elements "used" by the relation or function constitute the range. Domain: all x-values that are to be used (independent values).

Step-by-step explanation:

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3 years ago

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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

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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

Select the correct answer.

likoan [24]

Answer:

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Step-by-step explanation:

Traingle ABC = Traingle DEC

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3 years ago

1. Use the following graph to determine the coordinated of the y-intercept.

jarptica [38.1K]

Answer:

I believe it's C. (0.4)

Step-by-step explanation:

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2 years ago