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4 years ago

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Reptile [31]4 years ago

7 0

Step-by-step explanation:

18% compounded monthly

charges 18% compounded semi-annually.

#1 is worse than #2 because...

Since the nominal rates are the same, the card with the more frequent compounding so it would have the higher effective rat.

Let's compare them!:

(1 + .18/12)^12 = 1.19562 ---> effective rate is 19.562 per annum

(1+.18/2)² = 1.1881

1.1881 has the effective rate of 18.88% per annum

(1 + /17/4)^4 = 1.18114.. --> effective rate is 18.114% per annum

18% monthly is .18 over 12 each month = 0.015/month

so every month multiply by 1.015

total cost = $607.99 * 1.01^number of months

18% semi annual = .18/2 every six months = .09 /half year

so every half year (6 months) multiply by 1.09

total cost = $607.99 * 1.09^ (number of months/6)

17% quarterly = .17/4 every 3 months = .0425 /quarter year

so every quarter year (3 months) multiply by 1.0425

total cost = $607.99 * 1.025^(number of months/3)

HOPE THIS HELPS YOU HAVE A GREAT GREAT GREAT FANTASTIC FRIDAY!

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The sum of eight times a number and fifteen is seven.Find the number. How do I work this out like these?

CaHeK987 [17]

(8x n) + 15=7 : 7-15= (-8), (-8)divided by 8= (-1) : n=(-1) : When you have problems like this it helps to work them backwards. Hope you understand this.

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4 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

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