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rewona [7]
3 years ago
10

Questions E Given EA = 8 and AT = 10 Find ET

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer:

ET = 80/A^2

Step-by-step explanation:

First, isolate the variables that you need, which is E and T, using the two equations you were given.

EA = 8

E = 8/A (divide A on both sides to isolate E)

AT = 10

T = 10/A (divide A on both sides to isolate T)

Now, substitute.

ET --> (8/A) x (10/A) = 80/A^2

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Complete the​ table, for the following​ investments, which shows the performance​ (interest and​ balance) over a​ 5-year period.
sergiy2304 [10]

Answer:

After 5 years of investment, Suzanne will have $ 4,400 in her account, while Derek will have $ 4,416.32 in his account.

Step-by-step explanation:

Since Suzanne deposits $ 4000 in an account that earns simple interest at an annual rate of 2%, and Derek deposits $ 4000 in an account that earns compound interest at an annual rate of 2% and is compounded annually., Both for a period of 5 years, to determine how much money each one will have in their account after 5 years, the following calculation must be made:

Suzanne:

((4,000 x 0.02) x 5) + 4,000 = X

(80 x 5) + 4,000 = X

400 + 4,000 = X

4,400 = X

Derek:

4,000 x (1 + 0.02 / 1) ^ 5x1 = X

4,000 x 1.02 ^ 5 = X

4,416.32 = X

Thus, after 5 years of investment, Suzanne will have $ 4,400 in her account, while Derek will have $ 4,416.32 in her account.

5 0
3 years ago
Given: 9x > -36
GarryVolchara [31]
Hello,

<span>B.) {x | x > 4}

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8 0
3 years ago
Read 2 more answers
How do you compare 1/5 and 0.210
strojnjashka [21]
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3 years ago
7 times a number is 8 less than the square of that number. Find the negative solution.
charle [14.2K]

Answer:

  -1

Step-by-step explanation:

The required relation is ...

  7n = n^2 -8

  0 = n^2 -7n -8 . . . . put in standard form

  0 = (n -8)(n +1) . . . . factor

Solutions are n=8 and n=-1.

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7 0
3 years ago
Linear Algebra: Permutation Matrices Let M be the matrix {{0,0,0,1,0}, {0,0,1,0,0}, {0,1,0,0,0}, {0,0,0,0,1}{1,0,0,0,0}}. What i
FromTheMoon [43]

Multiplying M by any matrix A would return new matrix, B, in which

• the 1st row of B is equal to the 4th row of A,

• the 2nd row of B is equal to the 3rd row of A,

• the 3rd row of B is equal to the 2nd row of A,

• the 4th row of B is equal to the 5th row of A, and

• the 5th row of B is equal to the 1st row of A.

The pattern here is

1 => 4 => 5 => 1

2 => 3 => 2

Let {4, 3, 2, 5, 1} denote the matrix M, where each number refers to the row of the identity matrix, I.

Using this notation, the pattern above gives

M² = {5, 2, 3, 1, 4}

M³ = {1, 3, 2, 4, 5}

M⁴ = {4, 2, 3, 5, 1}

M⁵ = {5, 3, 2, 1, 4}

M⁶ = {1, 2, 3, 4, 5}

so that <em>n</em> = 6.

(Notice that the first cycle has length 3 and the second one has length 2; the minimum <em>n</em> needed here is then LCM(2, 3) = 6.)

6 0
3 years ago
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