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3 years ago

Hoy many pairs of parallel lines does a Square have

Mathematics

2 answers:

gavmur [86]3 years ago

8 0

2 because a square has 4 sides and so it has 2 sets of parallel lines.

otez555 [7]3 years ago

7 0

A square has 2 pairs of parallel line because all four sides are equal

Read more about functions at:

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4 0

3 years ago

Read 2 more answers

You can afford a $950 per month mortgage payment. You've found a 30 year loan at 7% interest. a) How big of a loan can you affor

Lubov Fominskaja [6]

Based on the amount to be paid per month, the size of the loan you can afford is $142,792.19.

The total amount you pay to the loan company is $342,000.

The interest on the money is $199,207.81.

The size of the loan is the present value associated with that loan payment that you can afford:

950 = (7/12% x Present value) / ( 1 - (1 + 7/12%)⁻⁽³⁰ ˣ ¹²⁾

950 = (7/12% x Present value) / 0.8767

Present value = $142,792.19

The total amount of money paid is:

= Number of periods x Payment per period

= 30 years x 12 months per year x 950

= $342,000

The interest is:

= 342,000 - 142,792.19

= $199,207.81.

Find out more on interest payments on loans at brainly.com/question/13005100

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4 0

1 year ago

How would 350 percent be written as a simplified fraction

Sauron [17]

Divide by 100. 350/100=3.5 which is 3 and 1/2

4 0

3 years ago

Read 2 more answers

PLS HELP ME ASAP WITH THIS! (LOTS OF POINTS!) (show work OR explain)

riadik2000 [5.3K]

So they tell you to let x represent the cost of Deandra's sofa. So since Selena's sofa costs 30 dollars LESS than HALF the cost of Deandra's, we know we have to divide by 2 and subtract 30.
The equation would be x/2-30. (B)

7 0

3 years ago

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$