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3 years ago

Write an expression to represent Vivian's change in elevation from -9.6 feet at the bottom of the Ferris wheel to 12.3 feet at

Mathematics

1 answer:

Effectus [21]3 years ago

5 0

Answer:

<h2><em>ΔE =12.3 - (-9.6) which is equivalent to 21.9 feet </em></h2>

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Step-by-step explanation:

Give the distance covered by Vivian at the bottom of the elevation to be -9.6 feet and the distance covered by Vivian at the top of the elevation to be 12.3 feet, Vivian's change in elevation will be expressed as the difference in the elevation at the bottom and the elevation at the top.

Let the initial elevation Ei = -9.6 feet

Final elevation Ef = 12.3 feet

Change in elevation ΔE = Ef - Ei

Change in elevation ΔE =12.3 - (-9.6)

Change in elevation ΔE = 12.3 + 9.6

Change in elevation ΔE = 21.9 feet

<em>Hence the expression to represent Vivian's change in elevation from -9.6 feet at the bottom of the Ferris wheel to 12.3 feet at the top is </em><em>ΔE =12.3 - (-9.6) which is equivalent to 21.9 feet </em>

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Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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4 years ago

Use the Venn diagram to list the set of elements of A ∩ B in roster form.

artcher [175]

Answer:

A. {e, h}

Step-by-step explanation:

In a Venn diagram, the set of elements in any intersection can simply be visualised. The elements contained in the region where the circles representing different sets overlap, are the set of elements of intersection.

In the Venn diagram given, the set of elements contained in the region where the circles representing A and B overlap are {e, h}.

{e, h} is common to both set A and set B.

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3 years ago

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Alecsey [184]

Answer:

Step-by-step explanation:

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3 years ago

shtirl [24]

A grocery store sells a 6-pack of bottled water for $3.79, a 9-pack for$4.50, and a 12-pack for $6.89. which package costs the least per bottle?

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4 years ago

if a motercycle is moving at a constant speed down the highway of 40 km/hr, how long would it the motorcycle to travel 10 km

ahrayia [7]

Answer:

15 minutes

Step-by-step explanation:

First, the motorcycle goes at a speed of 40 km/hr.

The question asks for how long it would take to travel 10 km.

Well, there are 60 minutes in an hour, since we will be translating into minutes.

Also, 10 km is 1/4 of 40 km, so it would make sense that the time length would be 1/4 of an hour as well.

1/4 of 60 minutes is 15 minutes. So it takes 15 minutes for the motorcycle to travel 10 km.

Now, if all this wordy stuff is too much to comprehend, you can also solve using proportional relationships.

$\frac{40km}{60min}=\frac{10km}{xmin}$

Now cross multiply:

$40km*xmin=10km*60min\\40x=600$

Divide both sides by 40:

$\frac{40x}{40}=\frac{600}{40}\\x=15$

Again, this shows that it wouls take 15 minutes for the motorcycle to travel 10 km.

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3 years ago