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zheka24 [161]
4 years ago
13

In 2012 William bought a new boat for $28,500 if the value of the boat declined by 8% each year find the value of the boat in 20

18
Mathematics
1 answer:
nikklg [1K]4 years ago
8 0

Answer:

$17,281.12

Step-by-step explanation:

28,500 * (.92)^x

x - amount of years after 2012

.92 - 100-8 = 92 which is the percent of the price the previous year

Calculator needed.

28,500 * .92^6

28,500 * ~.60636

17,281.1175383

Round up to 17,281.12

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(d). Use an appropriate technique to find the derivative of the following functions:
natima [27]

(i) I would first suggest writing this function as a product of the functions,

\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x

then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.

We then have, using the power and chain rules,

\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}

\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}

For the third function, we first rewrite in terms of the logarithmic and the exponential functions,

h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}

Then by the chain rule,

\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x

By the product rule, we have

\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}

\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x

\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x

\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}

You could simplify this further if you like.

In Mathematica, you can confirm this by running

D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]

The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in

FullSimplify[%]

(% refers to the last output)

If it still doesn't match, you can try running

Reduce[<our result> == %, {}]

and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)

(ii) Given

\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}

differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives

\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)

\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}

\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}

\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}

\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}

which could be simplified further if you wish.

In M, off the top of my head I would suggest verifying this solution by

Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]

but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,

ResourceFunction["ImplicitD"][<our equation>, x]

but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.

3 0
2 years ago
What is the area of a semicircle with a 15 inch radius
Zanzabum

Answer:

353.43

Step-by-step explanation:

find the area of the circle then divide by two

3 0
4 years ago
Given the directrix of y = 6 and focus of (0, 4), which is the equation of the parabola?
krok68 [10]

y = \frac{1}{24} x^{2}

<u>Step-by-step explanation:</u>

Since directrix is a horizontal line, this is a regular vertical parabola, where the x part is squared.  The equation of a vertical parabola is:

(x - h)² = 4p(y - k)    

where,

(h,k) are the coordinates of the vertex

p = distance from the vertex to the focus

So, we need to find out h, k and p and plug those values :

We know that the vertex of a parabola is halfway between focus and the directrix . We know the focus (0,4) and directrix y = 6  therefore, vertex (h,k) = (0, 0)

Now, p = distance from the vertex to the focus

= distance from the (0,0) to the (0,4)

= 4

Plug all the values in equation:

(x-0)^{2} = 4(4)(y-0)\\x^{2} = 24y

Now, we rearrange to write this into standard form (y =ax^{2}+bx+c)

or,y = \frac{1}{24} x^{2}

7 0
3 years ago
Which value make the equation 3x-6=36 true
elena-s [515]

Answer:

x=14

Step-by-step explanation:

To solve, we need to isolate the variable (x)

3x-6=36

Add 6 to both sides

3x=42

Divide by 3 on both sides

x=14

14 is the value that make the equation true

Hope this helps! :)

7 0
4 years ago
Read 2 more answers
-3(e+1)=2(e+11)<br><br><br> It’s e= -5 right
katrin2010 [14]
Yes it is e= negative 5
7 0
3 years ago
Read 2 more answers
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