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goldfiish [28.3K]
3 years ago
13

A quantity with an initial value of 270 grows exponentially at a rate of 0.1% every 10 decades. What is the value of the quantit

y after 23 years, to the nearest hundredth
Mathematics
1 answer:
NARA [144]3 years ago
6 0

Answer:

270.06

How to solve:

Write a function:

f(t)=270(1+0.001)^{t/10}

23 years→23/10→2.3 decades

Plug in t=2.3

f(2.3)=270(1+0.001)^{2.3/10}

=270.062076106

≈270.06

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Item 10 Find the amount of time. I=$450, P=$2400, r=7.5%
Alex787 [66]
The answer is:  "2.5 years" . 
___________________________________________________
  Note:   I = P * r * t  ;    { " Interest = Principal * rate * time "} ; 

          →     Solve for "t" {"time", in years} ;  

Divide each side of the equation by "{P * r}"  ;  
   to isolate "t" on one side of the equation ;

→  I / (P * r)  = {P * r * t) / (P * r} ; 

to get:  " I / (P * r) = t " ;  

  ↔  t = I / (P * r) ; 

Given:  I = $450 ;  
  
            <span>P = $2400 ; 

            r = 7.5% = 7.5/100 = 0.075 ; 

Plug in these values into the formula to solve for the time, "t" :

        </span>→  t  =  I  /  (P  *  r )   ;  

                 =  $450  /   (<span>$2400 * 0.075) ;

                 =  </span>$450  /   ($2400 * 0.075) ;

                 =  $450 / $180 ; 

                 =  $45 / $18 ; 

                 =  ($45 ÷ 9) / ($18 ÷ 9) 

                 =  $5 / $2 ; 

                 =  2.5  ; 

        →  t  =  2.5 years.
_______________________________________________________
The answer is:  "2.5 years" . 
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