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Stella [2.4K]
2 years ago
6

Answerrrrrr If anyone knows just reply

Mathematics
1 answer:
lina2011 [118]2 years ago
6 0

Answer:

C ) 6 units

Step-by-step explanation:

I' just guessing but based off of the other one is 4 then this must be at least 4.

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Solve the equation: 13W - 2(4W + 1) = W - 58 *
Sergio [31]

Answer:

W = - 14

Step-by-step explanation:

13W - 2(4W + 1) = W - 58

13W - 8W - 2 = W - 58

13W - 8W - W = 2 - 58

4W = - 56

W = - 56 : 4

W = - 14

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3 years ago
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Algebra 2. pls help:(
emmasim [6.3K]
The correct answer is C
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-2.2 + 0.3z= 3 -0.5z -0.8z<br><br> what does z equal
sineoko [7]
Add 2.2 to each side so 0.3z = 5.2 - 0.5z -0.8z
add 0.5z to each side so 0.8z = 5.2 - 0.8z
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4 0
3 years ago
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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



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3 years ago
For every 3 apples in a
Irina18 [472]

Answer:

18 apples equals 6 bananas

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