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IrinaVladis [17]
3 years ago
14

Simplest form 7/15 - 1/5

Mathematics
2 answers:
Lunna [17]3 years ago
8 0

Answer:

4/15

Step-by-step explanation:

7/15 - 3/15 = 4/15

tekilochka [14]3 years ago
4 0

Answer:

4/15

Step-by-step explanation: i did the math

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magazine subscription cost $29.88 for 12 issues or $15.28 for six issues which subscription cost more per issue
Shtirlitz [24]

Answer:

$15.28 for 6 issues cost more

Step-by-step explanation:

$29.88 divided by 12 is $2.49

$15.28 divided by6 is $2.546

3 0
3 years ago
Consider the expressions 7y + 5 − 3 and 7y + 2. Which statement is true?
romanna [79]
The two expressions are equivalent because they have the same value regardless of the number of substituted in for y
4 0
2 years ago
All of the following are terminating decimals except _____.<br><br> 1/8<br> 2/5<br> 2/3<br> 17/50
klemol [59]
2/3 is the answer Hope this helps. :)
5 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
The sum of two times a number and four
Leokris [45]
Twice the sum of a number and 4 is 14. In this case the numbers are added first, and the ANSWER is doubled.
8 0
3 years ago
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