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Gennadij [26K]
3 years ago
12

Mrs. Temps borrows $ 200 from Mrs. Abruzzino and is paying her back $ 20 every month. Describe the graph of a function that desc

ribes the amount Mrs. Temps owes Mrs. Abruzzino over time.
Mathematics
1 answer:
AnnyKZ [126]3 years ago
8 0

Answer:

The equation for this line would be y = -20x + 200. The y-intercept is 200 and represents the amount of money Mrs. Temps owes Mrs. Abruzzino in the beginning. The slope is -20 which represents how much Mrs. Temps pays back Mrs. Abruzzino each month. The x-axis should increase by 1 months.

Step-by-step explanation:

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Answer:

this \: paper \: needs \: to \: be \: folded \:  \\  \underline{ \boxed{twice \: (2 \: times)}}

Step-by-step explanation:

let \: the \: let \: the \: length \: be \to \: x \\when \: folded :  it \: becomes \to \:  \frac{x}{2}  \\hence : \to \\  93.5 \times  \frac{x}{2}  = (93.5 - 1) \\ 93.5x = 2(93.5 - 1) \\ 93.5x = 2 \times 92.5 \\ 93.5x = 185 \\ x =  \frac{185}{93.5}  \\  \boxed{x = 1.9786096257 }\\

6 0
3 years ago
is a 50 percent increase followed by a 33 1/2 percent decrease same less or greatter than the original value
Nitella [24]

Answer:less than

Step-by-step explanation:

3 0
3 years ago
A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How
Mazyrski [523]
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:


In the year 1985 the worker earned <span>$10,500. 

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

                                            $10,500(1+0.06).



In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]=$10,500(1.06)^2.


Similarly we can check that in the year 1987 the worker earned $10,500(1.06)^3, which makes the pattern clear. 


We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}.

Factorizing, we have

=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) (where a is the first term and r is the common ratio) we have:

\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17.



Finally, multiplying 10,500 by 59.17 we have 621.285 ($).


The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating (1.06)^{26}.


Answer: D



4 0
3 years ago
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3 years ago
Please simplify these expressions
Oksana_A [137]

10( - 2 - 9r) + 10(10 + 2r) =  - 70r + 80 \\ 2(7n - 10) - 6(5n + 7) =  - 16n - 62
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3 years ago
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