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scoundrel [369]
4 years ago
12

Mr. Nelson bought a car for $20,000 with a 5% interest rate.

Mathematics
1 answer:
LekaFEV [45]4 years ago
6 0

Answer:

A ) Hence The interest for the car loan is $ 4,310

B ) Hence The interest for the car loan is $ 3,152

Step-by-step explanation:

Given as :

The loan for the car = $ 20,000

The rate of interest of car loan = 5 % compounded annually

A ) The period of loan = 48 months = 4 years

Let the interest of the loan = CI

<u>From compounded method</u>

Amount = Principal × (1 + \frac{\textrm Rate}{100})^{\textrm time}

Or, Amount = $ 20,000 × (1 + \frac{\textrm 5}{100})^{\textrm 4}

Or,  Amount = $ 20,000 × (1.05)^{4}

Or, Amount = $ 20,000 × 1.2155

∴ Amount = $ 24310

So, Interest = Amount - Principal

Or, CI = $ 24,310 - $ 20,000  

∴  CI = $ 4,310

Hence The interest for the car loan is $ 4,310

B ) The loan for the car = $ 20,000

The rate of interest of car loan = 5 % compounded annually

The period of loan = 3 years

Let the interest of the loan = CI

So,

<u>From compounded method</u>

Amount = Principal × (1 + \frac{\textrm Rate}{100})^{\textrm time}

Or, Amount = $ 20,000 × (1 + \frac{\textrm 5}{100})^{\textrm 3}

Or,  Amount = $ 20,000 × (1.05)^{3}

Or, Amount = $ 20,000 × 1.1576

∴ Amount = $ 23,152

So, Interest = Amount - Principal

Or, CI = $ 23,152 - $ 20,000  

∴  CI = $ 3,152

Hence The interest for the car loan is $ 3,152

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Answer:

The system has infinitely many solutions

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Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

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To solve the following system

\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}

Step 1: Transform the augmented matrix to the reduced row echelon form

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

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

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The reduced row echelon form of the augmented matrix is

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

which corresponds to the system

\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}

The system has infinitely many solutions.

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

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3 years ago
The distance between two locations, A and B, is calculated using a third location C at a distance of 15 miles from location B. I
Jobisdone [24]
We will first calculate ∠A:
∠∠A = 180° - ( 105° + 20 ° ) = 180° - 125° = 55°
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AB · 0.81915 = 15 · 0.342
AB · 0.81915 = 5.13
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AB = 6.26 ≈ 6.3 miles
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The final answer is 26 1/6~
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