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vivado [14]
2 years ago
11

100000000.9996 x 6969781446.913

Mathematics
1 answer:
elena-s [515]2 years ago
3 0

Answer:

6.9697815e+17 (e+17 means 17 0's at the end of the problem. So, the exact answer would be 6.969781500000000000000000)

Step-by-step explanation:

Hope this helps! :D

Also, if it is no trouble, can I have brainliest? This equation took awhile.

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This is homework. please help with the answer for both 11a and 11b :))
RideAnS [48]
11a:
3x+5=5x-57
3x-5x=-57-5
-2x=-62
x=31
11b:
2x+2x+4x+150+4x+150=360
12x+300=360
12x=360-300
12x=60
x=5

hope this helped !!
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3 years ago
Mary has a recipe for cranberry bread. she uses 6 and 2/3 cups of flour to make 2 loaves of cranberry bread.
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Answer: it’s the third one

Step-by-step explanation:

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2 years ago
WILL GIVE BRAINLIEST
steposvetlana [31]

Answer:

Page 3 - the y-intercept would remain at 5 and the slope would be 3 times the upward slope.

Page 5 - the y-intercept would remain at -2 and the slope would be 1/4 instead of 5

Step-by-step explanation:

In point slope form the slope and y-intercept can be read directly from the equation.

y = mx + b where m is the slope and b is the y-intercept.

4 0
3 years ago
Factor<br> a²-a-12<br> Please
Paha777 [63]

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7 0
2 years ago
Imagine that you would like to purchase a $275,000 home. Using 20% as
vfiekz [6]

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

The monthly payment, M = $1,576.15

C. When t = 30 year and r = 5%, we have;

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

8 0
3 years ago
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