From April to December =9 months
the depreciation expense
3,600×(9÷12)=2,700
Calculate the interest in a year
Multiply the annual percentage of interest by the investment to find the interest.
interest in a year = 2% × 1,000
interest in a year = 0.02 × 1,000
interest in a year = 20
The annual interest is $20
Calculate the interest after 10 years
Multiply the annual interest by 10
interest 10 years = 20 × 10
interest 10 years = 200
The sum of interest after 10 years is $200
Find total investment
investment = first investment + interest in 10 years
investment = 1,000 + 200
investment = 1,200
The investment will be worth $1,200 in 10 years
at first we write each one as a sequence
and find its equation
the first table
-19, -11, -3, 5
8x-19
the third table
15,12,9,6
-3x+15
the second table
-1. 5,1.5,3,4.5
+3 +1.5 +1.5
so the third table is the nonlinear function
Answer/Step-by-step explanation:
Let's work out the subtraction of the polynomial to find out where the error is and the correct result we should have.
6x² - 4x - 5
- 3x² - 7x + 2
= [6x² - 3x²] [-4x -(-7x)] [-5 -(+2)]
= 3x² + 3x - 7
The error that was made in the subtraction given in the question was that the negative signs before the terms, 7x and 5 were ignored while subtracting.
The sign is very important, it defines the value of the term.
We can choose to solve the problem in another way shown below:
(6x² - 4x - 5) - (3x² - 7x + 2)
Distributive the negative sign in the second polynomial
6x² - 4x - 5 - 3x² + 7x - 2
Combine like terms
6x² - 3x² - 4x + 7x - 5 - 2
3x² + 3x - 7
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