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vodka [1.7K]
3 years ago
8

Based on the information below, choose the correct answers. Note: The monthly loan payment was calculated at 119 payments of $33

0.38 plus a final payment of $329.73. Loan Balance: $25,000.00 Loan Interest Rate: 10.0% Monthly Loan Payment: $330.38 Number of Payments: 120 Cumulative Payments: $39,644.95 Total Interest Paid: $14,644.95 On average, what dollar amount of each monthly payment is interest? $ What percent of the total payments is total interest? %
Mathematics
1 answer:
Margaret [11]3 years ago
6 0

Answer:

  • $122.04
  • 36.94%

Step-by-step explanation:

a) The amount of concern is ...

  (total interest)/(number of payments) = interest/payment

  $14,644.95/120 ≈ $122.04 . . . . amount of interest per payment

__

b) The ratio of concern is ...

  (total interest)/(total payments) × 100% = 14,644.95/39,644.95 × 100%

  ≈ 36.94% . . . . percent of total payments that is interest

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y=x+4
y=1x+4
slope=1
y intercept=4



slope=1 or 1/1 or 1
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Help please its math on this test.
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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei
Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]

a) To find the characteristic polynomial we calculate \text{det}(A-\lambda I)=0 where I is the identity matrix of appropiate size. in this case the characteristic polynomial is

\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0

Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

(4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0

which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are \lambda=4,3,-1

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression A-\lambda I and solve the system (A-\lambda I)x =0 by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider \lambda = 4. We get the matrix

\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

(x,y,z) = (5z,5z,z) = z(5,5,1)

So, the base for this eigenspace is the vector (5,5,1).

If \lambda = 3 then the base is (0,4,3) and if \lambda = -1 then the base is (0,0,1)

3 0
3 years ago
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sveta [45]

Answer:

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Step-by-step explanation:

Using the equation of line

y = mx + C

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Using y - y_1 = m(x - x_1)

With point ( 3, -2)

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y - (-2) = -3/1(x - 3)

y + 2 = -3/1(x -3)

Open bracket

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y + 2 = (-3x +9)/1

y = (-3x + 9 )/1 - 2

LCM is 1

y = -3x + 9 - 2

y = -3x + 7

The equation of the line is

y = -3x + 7

5 0
3 years ago
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