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jok3333 [9.3K]
1 year ago
15

Find the principal amount invested if the amount in a monthly compounded account with a 5.1% interest rate is $9,996.32 after 54

months.
Mathematics
1 answer:
slavikrds [6]1 year ago
4 0
Again, there are 12 months in a year, so 54 months is 54/12 years

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$9996\\
P=\textit{original amount deposited}\\
r=rate\to 5.1\%\to \frac{5.1}{100}\to &0.051\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12\\
t=years\to \frac{54}{12}\to &\frac{9}{2}
\end{cases}

\bf 9996=P\left(1+\frac{0.051}{12}\right)^{12\cdot \frac{9}{2}}\implies \cfrac{9996}{\left(1+\frac{0.051}{12}\right)^{12\cdot \frac{9}{2}}}=P
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The grade of a road is the slope of a road. In the U.S., grade is often expressed as a percent by finding the product 100(slope). Approximate the grade of a road that has a rise of 950 ft over 3 mi is :
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1 year ago
Match each type of function with a characteristic of its graph. Enter the letter of the correct characteristic next to the type
Lerok [7]

Answer:

Linear Function: Its graph has a constant slope (D).

Quadratic Function: Its graph is a parabola (B).

Inverse Variation Function: Its graph has both a horizontal asymptote and a vertical asymptote (F).

Square-root Function: Its graph has a closed endpoint (A).

Exponential Function:  Its graph has a horizontal asymptote, but not a vertical asymptote (C).

Logarithmic Function: Its graph is a reflection of the graph of an exponential function in the line <em>y = x  </em>(E).

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2 years ago
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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
2 years ago
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D2-9k+3 for d=10 and k=9
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Answer:

22

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22

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Answer:

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