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Lynna [10]
3 years ago
9

16) The angles of a quadrilateral are in AP, whose common difference is 10°.

Mathematics
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

75°, 85°, 95°, 105°

Step-by-step explanation:

Since the 4 angles form an AP, then the 4 angles are

a, a + d, a + 2d, a + 3d

where a is the first term and d the common difference

The sum of the angles in a quadrilateral = 360° thus

a + a + d + a + 2d + a + 3d = 360, that is

4a + 6d = 360, substitute d = 10

4a + 60 = 360 ( subtract 60 from both sides )

4a = 300 ( divide both sides by 4 )

a = 75

Thus the 4 angles are

75°, 75° + 10° = 85°, 75° + 20 = 95°, 75° + 30° = 105°

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A) 5  3/10

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Step-by-step explanation:

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you find a common denominator and then simplify all fractions to that common denominator. then, you add them up!

2 1/5 + 3 1/10

5 can fit into 10 twice so it becomes...

2 2/10 + 3 1/10 which equals

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3) The algebraic process you use to isolate a variable in order to find what values, numbers, etc. will make an equation true. Y
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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
3 years ago
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