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fiasKO [112]
2 years ago
10

444444+44444444333333333333333

Mathematics
2 answers:
WARRIOR [948]2 years ago
5 0

Answer:

4.4444444e+22

Step-by-step explanation:

Dimas [21]2 years ago
3 0

Answer:

4.444444433 x 10^23

Step-by-step explanation:

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1. How many degrees is x<br> .x<br> 41<br> 82<br> 20.5<br> o<br> 41<br> О<br> 10
Alja [10]

Answer:

Option (1)

Step-by-step explanation:

By the inscribed angle theorem inside a circle,

"Measure of an inscribed angle is half the measure of the intercepted arc"

\frac{1}{2}[m(arc AB)] = m(∠ABC)

m(arc AB) = 2[m(∠ABC)]

x = 2(41°)

x = 82°

Option (1) is the correct option.

5 0
2 years ago
Oblicz ułamki 2/8=.../4
Marina86 [1]

Answer:


Step-by-step explanation:


7 0
3 years ago
Janet is designing a frame for a client. She wants to prove to her client that m ZAGE &gt; mCHE in her sketch. What is the missi
nalin [4]
The question isn’t clear, can you write it better so I can understand?
5 0
3 years ago
Using a multiple of 10 for at least one factor, write an equation with a product that has four zeros
Maslowich
10 times 4000= 40000
3 0
3 years ago
Read 2 more answers
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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