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Mortgage Payments Common Core Algebra 2 Homework Answers
Mortgage Payments Made Easy with Common Core Algebra 2 Homework Solutions
If you are a homeowner with a mortgage, you know how important it is to make your monthly payments on time and in full. But did you know that you can use your Common Core Algebra 2 homework skills to help you save money on your mortgage payments? In this article, we will show you how to use some of the concepts and techniques from Algebra 2 to calculate your mortgage payments, compare different loan options, and plan ahead for the future.
Mortgage Payments Common Core Algebra 2 Homework Answers
How to Calculate Your Mortgage Payments
One of the most basic skills you need to manage your mortgage is to know how to calculate your monthly payments. This is where your knowledge of quadratic functions and the quadratic formula comes in handy. A quadratic function is a function of the form f(x) = ax^2 + bx + c, where a, b, and c are constants. The quadratic formula is a formula that allows you to find the roots or zeros of a quadratic function, which are the values of x that make the function equal to zero.
Why are quadratic functions and the quadratic formula relevant to your mortgage payments? Because your mortgage payments are determined by a formula that involves a quadratic function. The formula is:
P = L[c(1 + c)^n]/[(1 + c)^n - 1]
where:
P is the monthly payment
L is the loan amount
c is the monthly interest rate (annual interest rate divided by 12)
n is the number of monthly payments (loan term in years multiplied by 12)
This formula can be rewritten as:
P = Lc + Lc[(1 + c)^n - 1]/[(1 + c)^n - 1]
which is a quadratic function of (1 + c)^n.
To find out how much you need to pay each month for your mortgage, you can plug in the values of L, c, and n into this formula and simplify. Alternatively, you can use the quadratic formula to solve for (1 + c)^n and then plug it back into the formula for P.
For example, suppose you have a loan amount of $200,000, an annual interest rate of 4%, and a loan term of 30 years. Then:
L = 200000
c = 0.04/12 = 0.003333...
n = 30 * 12 = 360
P = 200000 * 0.003333... + 200000 * 0.003333...[(1 + 0.003333...)^360 - 1]/[(1 + 0.003333...)^360 - 1]
P = 666.67 + 200000 * 0.003333...[(1 + 0.003333...)^360 - 1]/[(1 + 0.003333...)^360 - 1]
P = 666.67 + 200000 * (3.243 - 1)/(3.243 - 1)
P = 666.67 + 200000 * (2.243/2.243)
P = 666.67 + 200000 * (1)
P = $866.67
So your monthly payment for this loan is $866.67.
How to Compare Different Loan Options
Another skill you need to manage your mortgage is to know how to compare different loan options and choose the best one for your situation. This is where your knowledge of linear systems and solving systems algebraically or graphically comes in handy. A linear system is a set of two or more linear equations with the same variables. Solving a linear system means finding the values of the variables that satisfy all the equations in the system.
Why are linear systems and solving systems relevant to your mortgage payments? Because you can use them to compare different loan options and see how they affect your monthly payments and total interest paid over time. For example, suppose you have two loan options: Option A has a loan amount of $200,000, an annual interest rate of 4%, and a loan term of 30 years; Option B has a loan amount of $200,000, an annual interest rate of 3%, and a loan term of 15 years.
To compare these two options, you can set up a linear system with two equations: one for Option A and one for Option B. The equations are:
Option A: $P_A = L_Ac_A + L_Ac_A[(1 + c_A)^n_A - 1]/[(1 + c_A)^n_A -
1]$where:
$P_A$ is the monthly payment for Option A
$L_A$ is the loan amount for Option A ($200,000)
$c_A$ is the monthly interest rate for Option A ($0.04/12$)
$n_A$ is the number of monthly payments for Option A ($30 * 12$)
Option B: $P_B = L_Bc_B + L_Bc_B[(1 + c_B)^n_B - 1]/[(1 + c_B)^n_B -
1]$where:
$P_B$ is the monthly payment for Option B
$L_B$ is the loan amount for Option B ($200,000)
$c_B$ is the monthly interest rate for Option B ($0.03/12$)
$n_B$ is the number of monthly payments for Option B ($15 * 12$)
To simplify these equations, you can plug in the values of $L_A$, $L_B$, $c_A$, $c_B$, $n_A$, and $n_B$, and round them to two decimal places:
$P_A = \textbf866.67 + \textbf6666.67x$where $x = (1 + c_A)^n_A$
$P_B = \textbf1386.45 + \textbf4159.35y$where $y = (1 + c_B)^n_B$
Now you have a linear system with two equations and two variables: $x$ and $y$. To solve this system, you can use either algebraic or graphical methods.
One algebraic method is called substitution: you solve one equation for one variable and then substitute it into the other equation to eliminate that variable and solve for the other one.
For example, you can solve equation (2) for $y$:
$$y = \fracP_B - \textbf1386.45\textbf4159.35$$
Then substitute it into equation (1):
$$P_A = \textbf866.67 + \textbf6666.67x$$
$$P_A = \textbf866.67 + \textbf6666.67\left(\fracP_B - \textbf1386.45\textbf4159.35\right)$$
Then simplify and solve for $P_B$:
$$\textbf4159.35P_A = \textbf3625000 + \textbf27750000 - \textbf27750000x$$
$$\textbf27750000x = \textbf27750000 - \textbf4159.35P_A$$
$$x = \frac{\textbf27750000 - \textbf{4159
How to Plan Ahead for Your Mortgage Payments
A final skill you need to manage your mortgage is to know how to plan ahead for your future payments and adjust them according to your changing needs and goals. This is where your knowledge of exponential functions and logarithms comes in handy. An exponential function is a function of the form f(x) = ab^x, where a and b are constants and b > 0. A logarithm is the inverse of an exponential function, which means that it undoes the effect of the exponential function. The logarithm of a number y with respect to a base b is the exponent x that makes b^x = y. The logarithm is written as log_b(y).
Why are exponential functions and logarithms relevant to your mortgage payments? Because you can use them to calculate how long it will take you to pay off your mortgage, how much interest you will pay over time, and how much you can save by making extra payments or refinancing your loan.
For example, suppose you have a loan amount of $200,000, an annual interest rate of 4%, and a loan term of 30 years. You already know that your monthly payment is $866.67. How long will it take you to pay off your loan if you make an extra payment of $100 every month?
To answer this question, you can use an exponential function to model the balance of your loan over time. The function is:
B(t) = L(1 + c)^t - P[(1 + c)^t - 1]/c - E[(1 + c)^t - 1]/c
where:
B(t) is the balance of the loan after t months
L is the loan amount ($200,000)
c is the monthly interest rate ($0.04/12$)
P is the monthly payment ($866.67)
E is the extra payment ($100)
To find out how long it will take you to pay off your loan, you need to solve for t when B(t) = 0. This means that you need to find the zero or root of the exponential function. To do this, you can use logarithms to undo the exponential function and isolate t. The steps are:
Solve for t when B(t) = 0:$0 = L(1 + c)^t - P[(1 + c)^t - 1]/c - E[(1 + c)^t - 1]/c$Add $P[(1 + c)^t - 1]/c + E[(1 + c)^t - 1]/c$ to both sides:$P[(1 + c)^t - 1]/c + E[(1 + c)^t - 1]/c = L(1 + c)^t$Multiply both sides by $c$:$P[(1 + c)^t - 1] + E[(1 + c)^t - 1] = Lc(1 + c)^t$Factor out $(1 + c)^t - 1$ from both sides:$(P + E)[(1 + c)^t - 1] = Lc(1 + c)^t$Divide both sides by $(P + E)$:$(1 + c)^t - 1 = \fracLcP + E(1 + c)^t$Subtract $(1 + c)^t$ from both sides:$- 1 = \fracLcP + E(1 + c)^t - (1 + c)^t$Factor out $(1 + c)^t$ from both sides:$- 1 = (1 + c)^t\left(\fracLcP + E - 1\right)$Divide both sides by $\left(\fracLcP + E - 1\right)$:$\frac- 1{\left(\fracLc{P
Take the logarithm of both sides with base 1 + c:$\log_1 + c\left(\frac- 1\left(\fracLcP + E - 1\right)\right) = \log_1 + c\left((1 + c)^t\right)$Use the property of logarithms that $\log_b(b^x) = x$:$\log_1 + c\left(\frac- 1\left(\fracLcP + E - 1\right)\right) = t$Solve for t:$t = \log_1 + c\left(\frac- 1\left(\fracLcP + E - 1\right)\right)$
Now you can plug in the values of $L$, $c$, $P$, and $E$ into this formula and round to the nearest month:
$$t = \log_1 + 0.003333...\left(\frac- 1\left(\frac200000 * 0.003333...866.67 + 100 - 1\right)\right)$$
$$t \approx 312.5$$
So it will take you about 313 months or 26 years and 1 month to pay off your loan if you make an extra payment of $100 every month.
To compare this with the original loan term of 360 months or 30 years, you can see that you will save 47 months or almost 4 years of payments by making extra payments. You can also calculate how much interest you will save by multiplying the monthly payment by the number of months saved and subtracting the extra payments:
$$I = P * (360 - t) - E * (t - n)$$
$$I = 866.67 * (360 - 312.5) - 100 * (312.5 - 360)$$
$$I \approx 41333.75 - (-4750)$$
$$I \approx 46083.75$$
So you will save about $46,083.75 in interest by making extra payments of $100 every month.
You can use similar methods to compare different loan options and see how they affect your monthly payments and total interest paid over time. For example, you can compare how refinancing your loan with a lower interest rate or a shorter loan term will affect your payments and interest.
Conclusion
In this article, we have shown you how to use some of the concepts and techniques from Common Core Algebra 2 to help you manage your mortgage payments. We have shown you how to use quadratic functions and the quadratic formula to calculate your monthly payments, linear systems and solving systems algebraically or graphically to compare different loan options, and exponential functions and logarithms to plan ahead for your future payments and adjust them according to your changing needs and goals. We hope that this article has helped you understand how math can be useful and relevant in real life situations. d282676c82
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