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3 years ago

I need help with this topic

Mathematics

1 answer:

madreJ [45]3 years ago

4 0

Let's go through these, sentence by sentence:

a. Henry is driving at a constant speed.

If the y-axis represents the speed of Henry's car, this portion of the graph should be a straight horizontal line, as his speed doesn't change at all. This already eliminates the first and second options.

He then slows down to pass an accident. After passing it, he goes back to his original speed and continues driving at that speed.

We should see a downward-sloped line segment during the period when he slows down, and then an upward-sloped line segment during the time where he speeds up. Graphically, this would look like a V. Finally, the graph would again become a straight horizontal line as he returned to and maintained his original speed. Graph #4 is the only one which represents this description.

b. Teresa is driving to work. She drives at a constant speed for several miles, then stops to pick up breakfast.

On all of the graphs of this situation, the y-axis represents Teresa's distance to work. We have to be careful here, because the further she drives, the further down the graph goes. The y-coordinate starts at some positive fixed position (the total distance to work) and works its way down to 0.

There's only one graph which represents this scenario - a downwards trend - and that's graph #3.

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What is 6 divided by a number that will give you 12

AURORKA [14]

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$Hello$ $There!$

We need to find what number, if 6 is divided by it, gives us 12.

This isn't hard, I promise! I am going to take you through a few steps, and we'll find the answer.

Are you ready? Then let's begin...:)

$Procedure:$

✦ We can think about it as: 6 divided by what number gives us 12?

(let the number be z)

$^6/_z=12$

Multiply both sides by z

$6=12z$

Divide both sides by z

$z=^6/_{12}$

Which is the same as

$z=^1/_2$

hope it's helpful!
wishing you a great day!

$Greetings!$

6 0

2 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

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

5 0

3 years ago

Write the formula for the following sequence. 5.8, 3.6, 1.4,... PLEASE SHOW YOUR WORK.

sladkih [1.3K]

$tn = t(n - 1)- 2.2$

common difference is

3.6 - 5.8 = -2.2

let T1 = 5.8

T2 = 3.6

assuming we are to find T2

T2 = T(2-1) -2.2

= T1 -22

= 5.8 -2.2

= 3.6

3 0

3 years ago

The graph of is shown below. What is the value of y when x = 0?

jarptica [38.1K]

$\huge\text{Hey there!}$

$\huge\textsf{This equation/question should be easy to}\\\\\huge\textsf{to solve for because, we have everything you}\\\\\huge\textsf{you will need on your graph and somewhat in}\\\\\huge\textsf{the question because it already gave you the}\\\\\huge\boxed{\mathsf{x-intercept}}\huge\textsf{.}$
$\mathsf{y = |x| - 2}$