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Sauron [17]
3 years ago
5

An album sells for $17.00 through an online music service. If the album is 20% off, and sales tax is 6%, what is the total price

of the album including tax?
Mathematics
1 answer:
V125BC [204]3 years ago
5 0

Answer:

$14.42

Step-by-step explanation:

Price of Album - $17.00 * .20 = 3.4 (this is the discount)

Then subtract 3.40 from 17.00 to get 13.60

Add the sales tax in by doing $13.60 * 0.06 = $0.82

Add that to the 13.60 to get $14.42 for the total after discount and after tax.

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Suppose there is a bag containing 5 red marbles ​(Upper R​), 5 pink marbles ​(Upper P​), 5 green marbles ​(Upper G​), and 5 blac
Radda [10]

Answer:

In the Explanation

Step-by-step explanation:

(a)From the attached probability tree, the possible outcomes are:

RR,RP,RG,RB,PR,PP,PG,PB,GR,GP,GB,GG,BR,BP,BG,BB

(b)Probability Distribution of Drawing Pink Marbles

\left|\begin{array}{c|c}Result&Probability\\----&----\\0&\frac{9}{16}\\----&---\\1& \frac{3}{8}\\----&---\\2&\frac{1}{16}\end{array}\right|

3 0
3 years ago
Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
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