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AlekseyPX
3 years ago
8

What is the relationship between the 6s in 6,600 value of the given digits?

Mathematics
2 answers:
Karolina [17]3 years ago
5 0

The relationship between the 6s in 6,600 value of the given digits

We write the numbers in place value chart

Thousand     Hundreds          Tens      Ones

 6                     6                        0             0

first 6 is in thousands place

next 6 is in hundreds place

6 thousand is 10 times greater than 6 hundred

6000 is 10 times greater than 600


svet-max [94.6K]3 years ago
4 0
<span>In 6,600, the 6s are in the thousands and hundreds spots. The 6 in the thousands spot would be 6000 and the 6 in the hundreds spot would be 600. The first 6 from left to right is the thousands spot and the 2nd 6 from the left is in the hundreds spot. The 0s are in the tens and ones spots from left to right, respectively.</span>
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What is the solution to this equation?
kupik [55]

Answer:

x = 5

Step-by-step explanation:

3x - 7 + 4x = 28

Combine like terms

7x -7 = 28

Add 7 to each side

7x - 7 +7  = 28+7

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Imagine that you would like to purchase a $275,000 home. Using 20% as
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Answer:

The mortgage chosen is option A;

15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470

Step-by-step explanation:

The details of the home purchase are;

The price of the home = $275,000

The mode of purchase of the home = Mortgage

The percentage of the loan amount payed as down payment = 20%

The amount used as down payment for the loan = $55,000

The principal of the mortgage borrowed, P = The price of the house - The down payment

∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000

The principal of the mortgage, P = $220,000

The formula for the total amount paid which is the cost of the loan is given as follows;

Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n -  \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }

The formula for monthly payment on a mortgage, 'M', is given as follows;

M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}

A. When the mortgage term, t = 15-years,

The interest rate, r = 3%

The number of months over which the loan is payed, n = 12·t

∴ n = 12 months/year × 15 years = 180 months

n = 180 months

The monthly payment, 'M', is given as follows;

M =

The total amount paid over the loan term = Cost of the mortgage

Therefore, we have;

220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)

M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1}  \approx 1,519.28

The minimum monthly payment for the loan, M ≈ $1,519.28

The total amount paid over loan term, A = n × M

∴ A ≈ 180 × $1,519.28 = $273,470

The total amount paid over loan term, A ≈ $270,470

B. When t = 20 year and r = 6%, we have;

n = 12 × 20 = 240

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1}  \approx 1,576.15

The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276

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n = 12 × 30 = 360

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1}  \approx 1,181.01

The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163

The monthly payment, M ≈ $1,181.01

The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability

Therefore;

The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;

The total amount paid over the loan term = $270,470

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