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NeTakaya
3 years ago
5

Which is the value of the 6 in 20.896?

Mathematics
1 answer:
OLEGan [10]3 years ago
3 0

20.89<u>6</u>

The number 3 places after the decimal is the thousandth place.

.8 → tenth place

.8<u>9</u> → hundredths place

.89<u>6</u> → thousandth place

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After Halloween, a variety store had some costumes regularly priced at $20.50 on sale for 20% off the regular price. A few weeks
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20.50 * 0.8 = $16.40

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Add the two expressions.<br><br> m + 6 and 4m + 2<br><br> Enter your answer in the box
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Question 1
Alexxandr [17]

Answer:

No, it is NOT a function.

Step-by-step explanation:

A functions' inputs are assigned to exactly one output.

We are given the points:

{(8,2) (4,-3) (6,4) (8,1)}

The input of 8 is repeating.

The given relation is not a function.  

Hope this helps!

5 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
2 years ago
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