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xxTIMURxx [149]
2 years ago
10

Ernest is purchasing a $175,000 home and is making a 20% down payment toward the cost. He is also buying 2 points in order to lo

wer his interest rate. The appraisal fee is $400, the title is $300, and the processing fee is $575. Including the down payment, what are his total closing costs?
A. $37,246
B. $33,143
C. $32,315
D. $39,075
Mathematics
2 answers:
nevsk [136]2 years ago
6 0

Answer: $39,075

Step-by-step explanation:

belka [17]2 years ago
5 0
Your answer would be. $39,075.
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I believe the answer would be the second option. The equations given above when graph would intersect only once with each other. They intersect at only point (0,6/5). These are the values of x and y that will agree to the equation. Two equations in a set will always intersect at only one point.
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Tulio Fernandez purchased a treadmill with an installment loan that has an APR of 12%. The treadmill sells for $1,672. The store
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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
musickatia [10]
(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}
\implies r^4-5r^3+6r^2+4r-8=0
\implies (r-2)^3(r+1)=0\implies r=2,r=-1

So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

So the particular solution to the recurrence is

h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n

(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n
G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)
G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)
(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3
G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}
G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of x^n, ideally matching the solution found in part (a).
3 0
2 years ago
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2 years ago
A caterer charges a $65 equipment fee to set up for a party and $9.50 per p person for food served at the party. Which of the fu
AleksAgata [21]

Answer:

The total cost function of having a party catered  is f(p)  =  65 + 9.50 p

Step-by-step explanation:

The equipment fee of the caterer = $65

The charge per plate = $9.50

Let, Total number of people invited in the party = p

So, the cost of plates of p people  = Number of people x Per plate cost

                                                         =  p  x $9.50

Now, Total Catering Cost = Equipment Fee +  Cost of all plates

or,                       Total cost =  $65  + p  x $9.50

Hence, the total cost function of having a party catered

 is f(p)  =  65 + 9.50 p.

3 0
2 years ago
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