Math, asked by aracelymtz005, 3 months ago

Damian invested $880 in an account paying an interest rate of 5.2% compounded annually. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $1,290?

Answers

Answered by royreyes2002
6

Answer:

Step-by-step explanation:

A=P(1+r)^t

A=P(1+r)  

t

 

A=1290\hspace{35px}P=880\hspace{35px}r=0.052

A=1290P=880r=0.052

Given values

1290=

1290=

\,\,880(1+0.052)^{t}

880(1+0.052)  

t

 

Plug in values

1290=

1290=

\,\,880(1.052)^{t}

880(1.052)  

t

 

Add

\frac{1290}{880}=

880

1290

​  

=

\,\,\frac{880(1.052)^{t}}{880}

880

880(1.052)  

t

 

​  

 

Divide by 880

1.4659091=

1.4659091=

\,\,1.052^t

1.052  

t

 

\log\left(1.4659091\right)=

log(1.4659091)=

\,\,\log\left(1.052^t\right)

log(1.052  

t

)

Take the log of both sides

\log\left(1.4659091\right)=

log(1.4659091)=

\,\,t\log\left(1.052\right)

tlog(1.052)

Bring exponent to the front

\frac{\log\left(1.4659091\right)}{\log\left(1.052\right)}=

log(1.052)

log(1.4659091)

​  

=

\,\,\frac{t\log\left(1.052\right)}{\log\left(1.052\right)}

log(1.052)

tlog(1.052)

​  

 

Divide both sides by log(1.052)

7.5449219=

7.5449219=

\,\,t

t

Use calculator

t\approx

t≈

\,\,7.5

7.5

Answered by aaronr407
0

Answer:

Step-by-step explanation:

in deltamath is 7.5

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