Help please urgently:

Beth is taking out a loan to purchase a new home. She is financing $50,000 for 25 years at an interest rate of 14.25%. What is h

kozerog [31]

The present value (PV) of a loan for n years at r% compounded t times a year where there is equal P periodic payments is given by:

$PV=P\left( \frac{1-\left(1+ \frac{r}{t} \left)^{-nt}}{ \frac{r}{t} } \right)$

Given that <span>Beth is taking out a loan of PV = $50,000 to purchase a new home for n = 25 years at an interest rate of r = 14.25%. Since she is making the payment monthly, t = 12.

Her monthly payment is given by:

$50,000=P\left( \frac{1-\left(1+ \frac{0.1425}{12} \right)^{-25\times12}}{ \frac{0.1425}{12} } \right) \\ \\ =P\left( \frac{1-(1+0.011875)^{-300}}{ 0.011875 } \right)=P\left( \frac{1-(1.011875)^{-300}}{ 0.011875 } \right) \\ \\ =P\left( \frac{1-0.028969}{ 0.011875 } \right)=P\left( \frac{0.971031}{ 0.011875 } \right)=81.770994P \\ \\ \therefore P= \frac{50,000}{81.770994} =\$611.46$

Therefore, her monthly payment is about $611.50
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3 0

3 years ago

Marta exercises 30 minutes each day Greg exercises 40 minutes a each day how many more minutes does Greg exercise than Marta in

ryzh [129]

40-30=10 minutes more per day x 31 days = 310minutes in a month with 31 days.

6 0

3 years ago

Read 2 more answers

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

Please help! Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact valu

Nadya [2.5K]

Answer:

$The \ area \ of \ the \ figure = 4\dfrac{1}{2} \cdot \pi + 18 \ cm^2$