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Aleks [24]
3 years ago
9

Len invests $5200 at 3%/a, while his friend Dave invests $3600 at 5%/a. How long will it take for Dave’s

Mathematics
1 answer:
kupik [55]3 years ago
4 0
Suppose that it will take n years for Dave's investment to be equal to Len's;
thus using the compound interest formula we shall have:
A=p(1+r/100)^n
thus the investment for Len after n year will be:
A=5200(1+3/100)^n
A=5200(1.03)^n

The total amount Dave's amount after n years will be:
A=3600(1+5/100)^n
A=3600(1.05)^n
since after n years the investments will be equal, the value of n will be calculated as follows;
5200(1.03)^n=3600(1.05)^n
5200/3600(1.03)^n=(1.05)^n
13/9(1.03)^n=(1.05)^n
introducing the natural logs we get:
ln(13/9)+n ln1.03=n ln 1.05
ln(13/9)=n ln 1.05-n ln 1.03
ln(13/9)=0.0192n
n=[ln(13/9)]/[0.0192]
n=19.12
thus the amount will be equal after 19 years



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Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a
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<h3><u>Correct Question :- </u></h3>

\sf\:a,b \: are \: the \: roots \: of \:  {x}^{2} + 20x - 2020 = 0 \: and \:  \\  \sf \: c,d \: are \: the \: roots \: of \:  {x}^{2}  -  20x  + 2020 = 0 \: then \:

\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =

(a) 0

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(c) 8080

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\large\underline{\sf{Solution-}}

Given that

\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \:  {x}^{2} + 20x - 2020 = 0}

We know

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

\rm \implies\:ab = \dfrac{ - 2020}{1}  =  - 2020

And

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

\rm \implies\:a + b = -  \dfrac{20}{1}  =  - 20

Also, given that

\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \:  {x}^{2}  -  20x  + 2020 = 0}

\rm \implies\:c + d = -  \dfrac{( - 20)}{1}  =  20

and

\rm \implies\:cd = \dfrac{2020}{1}  = 2020

Now, Consider

\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)

\sf \:  =  {ca}^{2} -  {ac}^{2} +  {da}^{2} -  {ad}^{2} +  {cb}^{2} -  {bc}^{2} +  {db}^{2} -  {bd}^{2}

\sf \:  =  {a}^{2}(c + d) +  {b}^{2}(c + d) -  {c}^{2}(a + b) -  {d}^{2}(a + b)

\sf \:  = (c + d)( {a}^{2} +  {b}^{2}) - (a + b)( {c}^{2} +  {d}^{2})

\sf \:  = 20( {a}^{2} +  {b}^{2}) + 20( {c}^{2} +  {d}^{2})

\sf \:  = 20\bigg[ {a}^{2} +  {b}^{2} + {c}^{2} +  {d}^{2}\bigg]

We know,

\boxed{\tt{  { \alpha }^{2}  +  { \beta }^{2}  =  {( \alpha   + \beta) }^{2}  - 2 \alpha  \beta  \: }}

So, using this, we get

\sf \:  = 20\bigg[ {(a + b)}^{2} - 2ab +  {(c + d)}^{2} - 2cd\bigg]

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\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}

<em>So, option (d) is correct.</em>

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