Trevor has an investment worth $6,774. 50. He made his only deposit in it 22 years ago. Given that the investment yields 2. 7% simple interest annually, how big was the initial deposit? a. $2,524. 50 b. $4,024. 03 c. $4,250. 00 d. $11,404. 88 Please select the best answer from the choices provided A B C D.

Answers

Answer 1

[tex]~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$6774.50\\ P=\textit{original amount deposited}\\ r=rate\to 2.7\%\to \frac{2.7}{100}\dotfill &0.027\\ t=years\dotfill &22 \end{cases} \\\\\\ 6774.50=P[1+(0.027)(22)]\implies \cfrac{6774.50}{1+(0.027)(22)}=P\implies 4250=P[/tex]


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Answer:

NM = [tex]\sf \sqrt{70}[/tex]

Step-by-step explanation:

Let NM = y

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NM = [tex]\sqrt{70}[/tex]

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NM = [tex]\sqrt{70}[/tex]

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Answers

Answer:

Information

General form of an exponential function:  [tex]y=ab^x[/tex]

where:

[tex]a[/tex] is the y-intercept (or initial value)[tex]b[/tex]  is the base (or growth factor)[tex]x[/tex]  is the independent variable

Also, growth rate [tex]r=b-1[/tex]

Domain:  input values - independent variable ⇒ [tex]x[/tex]

Range:  output values - dependent variable ⇒ [tex]y[/tex]

y-intercept:  when [tex]x=0[/tex]

Asymptote:  the line(s) that the curve approaches as it tends towards infinity

Given function:  [tex]f(x)=200(1.08)^x[/tex]

Solution

Domain:   [tex]-\infty < x < +\infty[/tex]

Range:   [tex]y > 0[/tex]

y-intercept ([tex]a[/tex]):  [tex](0, 200)[/tex]

Asymptote:   [tex]y=0[/tex]

Growth Rate ([tex]b-1[/tex]):   [tex]0.08[/tex]

Growth Factor ([tex]b[/tex]):   [tex]1.08[/tex]

End behaviors:

[tex]\textsf{as } x \rightarrow -\infty, f(x) \rightarrow0[/tex]

[tex]\textsf{as } x \rightarrow +\infty, f(x) \rightarrow+\infty[/tex]

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