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olganol [36]
2 years ago
4

Savannah took out a 5-year loan for $6500 at a furniture store to be paid back with monthly payments at a 17.3% APR. If the loan

offers no payments for the first 21 months, how many payments will Savannah be required to make?
A.81
B.21
C.60
D.39
Mathematics
2 answers:
Allisa [31]2 years ago
7 0
Since they offer no loans for the first 21 months, cancel out C
A
B
D 
are still somewhat possible 
D is the answer
Ivenika [448]2 years ago
6 0

Answer:

Option D is the right answer.

Step-by-step explanation:

The loan amount - $6500

The APR is = 17.3%

The loan term is = 5 years or 5\times12=60 months

Now given is that the loan offers no payments for the first 21 months. This means Savannah will be required to make = 60-21=39 payments.

Hence, the answer is 39 or option D.

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The total amount of paint that the paint shop stocks is 1800 litres.

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The amount of the remaining paint other than white would be

1800 - 432 = 1368 litres

The shops sells 18% of the white paint. This means that the amount of white paint sold by the shop will be

18/100 × 432 = 0.18 × 432 = 77.6 litres.

The shops sells 7% of the rest of the paint.

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The amount of paint that was sold altogether would be

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An ice cream shop sold 48 vanilla milkshake in a day which is 40% of the total number of milkshake sold that day.What was the to
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Let x represent the total number of

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Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe
sukhopar [10]

Answer:

 x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

First we have to determine the \bar x so that A \bar x = 0

\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]

Now the augmented matrix is

\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]

After this, we decrease this to reduce the formation of the row echelon

R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]

R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]

R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]

R_2 = \frac{R_2}{4},  R_3 = \frac{R_3}{-4}  \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]

R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]

R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]

= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0

x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]

By applying the above matrix, we can easily reach an answer

5 0
2 years ago
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