9514 1404 393
Answer:
- $304
- $91.83
Step-by-step explanation:
1. The finance charge is found from the simple interest formula;
I = Prt
where P is the principal amount, r is the annual rate, and t is the number of years.
24 months is 2 years, so the interest charged is ...
I = $1900×0.08×2 = $304
The finance charge is $304.
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2. The monthly payment will be the total amount due, divided by the number of months.
payment = ($1900 +304)/24 = $2204/24 ≈ $91.83
The monthly payment is $91.83.
Answer:
36 ft.
Step-by-step explanation:
The given perimeter of triangle EDF is 12, in which the hypotenuse is 5. This means that it is a 3-4-5 triangle, meaning that ED = 3, DF = 4, EF = 5.
It is given that BC (the other triangle's hypotenuse) is 15. This means that the left triangle is to the scale factor of 3 of triangle EDF. Therefore, multiply 3 to all sides measurements, and then add all the sides to get the perimeter.
Hypotenuse: 5 x 3 = 15
Leg 1: 4 x 3 = 12
Leg 2: 3 x 3 = 9
Perimeter: 15 + 12 + 9 = 36
36 ft. is your answer.
You have to give every thing that the problem says for me to be able to figure it out.
Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=![\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%266%5C%5C2%261%263%5C%5C-2%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%3D0)
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=![\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%26-2%5C%5C-7%261%265%5C%5C4%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
Hence no solution or infinite
Option C
![\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%26-2%5C%5C-2%260%266%5C%5C1%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2836-2%29%3D68)
Hence there will be a unique solution
Option D
![\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2610%265%5C%5C4%261%264%5C%5C-1%26-2%26-1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer
