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aniked [119]
3 years ago
9

For the investment with interest compounded annually, find the final balance using the formula A = P(1 + r)t.

Mathematics
1 answer:
tatyana61 [14]3 years ago
6 0
To solve for this question using the formula, plug in your known values where r is the rate (2.3%), P is the principal amount ($35,000), and t is the time in years. However, keep in mind when you plug the percent rate into the formula, you must first convert it to a decimal which would be 0.023. Now that you have this, plug your values into the equation:
A=35,000(1+0.023)^20
When simplified, you should get a result of about $55,154.47 or the second answer choice.
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Make t the subject of the formula.<br><br> Please help.
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Answer:

t = \frac{2-3q}{q+4}

Step-by-step explanation:

q = \frac{2-4t}{t+3}\\\\

Cross multiply,

q(t+3) = 2 - 4t        

Use distributive property: a*(b + c) =(a*b) + (a*c)

qt + 3q = 2- 4t

qt + 4t + 3q = 2

  qt + 4t     = 2 - 3q

    t(q + 4)  = 2 -3q

               t = \frac{2-3q}{q+4}

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Step-by-step explanation:


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Represent each of these relations on {1, 2, 3, 4} with a matrix (with the elements of this set listed in increasing order). a) {
Irina-Kira [14]

Answer:

for a  a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]

Step-by-step explanation:

in matrix, arrays are placed in rows , which represents the horizontal sides from left to right, while arrays in the column are placed vertically from top to bottom. Here, we placed the arrays in a 4x4 matrix

for a  a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]

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3 years ago
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