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3 years ago

Yusef drove 130 miles in 2.5 hours. Which equation can be used to calculate r, the average speed at which Yusef drove?

Mathematics

2 answers:

Nezavi [6.7K]3 years ago

6 0

130/2.5 is 52
You are very welcome

sergejj [24]3 years ago

4 0

130m = 2.5h (h= hours)

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Vsevolod [243]

Answer:

Isn't that just soild liquid and gas atoms?

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3 years ago

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£10 is divided between Alex, Peter &

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Answer: Peter gets £3

Step-by-step explanation:

Let Angad get x: equation will be 2(3x) + 3x + x = 10

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Peter get 3x = 3*1 = 3

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Yan is climbing down a ladder. Each time he descends four rungs on the ladder, he stops to see how much farther he has to go

anastassius [24]

Question:

Yan is climbing down a ladder. Each time he descends 4 rungs on the ladder, he stopped to see how much farther he has to go. If Yan made eight stops with no extra steps, which expression best shows another way to write the product of the number of ladder rungs that Yan and climbed?

4+4+4+4

8+8+8+8

(-1)(4+4+4+4)

(-8) + (-8) + (-8) + (-8)

Answer:

$Total = (-8) +(-8) +(-8) +(-8)$

Step-by-step explanation:

Given

<em></em> $Rungs = -4$ <em> --- It is negative because he is climbing down</em>

$Steps = 8$

Required

An expression for the number of rungs climbed

To do this, we simply multiply the number of rungs by the steps taken.

$Total = Rungs * Steps$

$Total = -4 * 8$

This can be rewritten as:

$Total = -8 * 4$

The product, when written as sum is:

$Total = (-8) +(-8) +(-8) +(-8)$

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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

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

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3 years ago

In the slope-intercept equation of a line, which quantity is given by the coefficient of the x-term? A. The y-intercept of the l

-Dominant- [34]

In any linear equation the coefficient of the x-term represents ALWAYS the slope.

So it is B. The slope of the line

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3 years ago