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Komok [63]
3 years ago
13

Walter took out a $6,000 loan for six years. He is being charged 6 percent interest, compounded annually. Calculate the total am

ount he will pay.
Mathematics
1 answer:
cluponka [151]3 years ago
8 0

Answer:

  $8511.11

Step-by-step explanation:

Each year, the amount Walter owes is multiplied by 1.06, so at the end of 6 years, Walter owes 1.06^6 times the amount he borrowed.

  he will pay $6,000×1.06^6 ≈ $8511.11

_____

At the end of the first year, Walter owes the original loan amount plus 6% interest. That total is ...

  $6000 + 0.06×6000 = $6000×1.06

At the end of the following year, he owes 1.06 times that amount, or ...

  6000×1.06²

The amount owed is multiplied by 1.06 each year until Walter pays off the loan.

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What is the value of t=1 ∑³ (4 x 1/2^t-1)
Stels [109]

Answer:

7

Step-by-step explanation:

\sum\limits_{t=1}^{3}(4\cdot(\frac{1}{2})^{t-1})

This is the sum of the first three terms of a geometric sequence, where the first term is 4 and the common ratio is ½.

We can use a formula to find the sum, or, since there's only three terms, we can find the value of each term then add up the results.

4 · (½)¹⁻¹ = 4

4 · (½)²⁻¹ = 2

4 · (½)³⁻¹ = 1

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4 0
3 years ago
the area of a small square is 25 square units what is the exact length in units of a single of a large square
Furkat [3]

Answer:

5

Step-by-step explanation:

The area of any quadrilateral can be determined by multiplying the length of its base by its height. Since we know the shape here is square, we know that all sides are of equal length. From this we can work backwards by taking the square root of the area to find the length of one side.

3 0
3 years ago
Use Cramer's Rule to solve the following system: –2x – 6y = –26 5x + 2y = 13
andriy [413]
\bf \begin{cases}
-2x-6y&=-26\\
\quad 5x+2y&=13
\end{cases}\stackrel{\textit{determinant of the coefficients}}{D=
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\end{bmatrix}}\implies (-4)-(-30)
\\\\\\
D=-4+30\implies \boxed{D=26}\\\\
-------------------------------\\\\

\bf x=\cfrac{D_x}{D}\implies x=\cfrac{
\begin{bmatrix}
\boxed{-26}&-6\\\\ \boxed{13}&2
\end{bmatrix}}{D}\implies x=\cfrac{(-52)-(-78)}{26}
\\\\\\
x=\cfrac{-52+78}{26}\implies x=\cfrac{26}{26}\implies \boxed{x=1}\\\\
-------------------------------\\\\
y=\cfrac{D_y}{D}\implies y=\cfrac{
\begin{bmatrix}
-2&\boxed{-26}\\\\ 5&\boxed{13}
\end{bmatrix}}{D}\implies y=\cfrac{(-26)-(-130)}{26}
\\\\\\
y=\cfrac{-26+130}{26}\implies y=\cfrac{104}{26}\implies \boxed{y=4}
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