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Mrrafil [7]
3 years ago
10

If the sum of 4040 consecutive integer numbers is 2020, find out the absolute difference between the first number and the last n

umber.
Mathematics
1 answer:
bearhunter [10]3 years ago
4 0

Let x be the smallest of the consecutive integers summing to 2020. Then the largest integer in the sum is x+4039, so our answer is \boxed{4039}.

For completeness, though, let's solve for the value of x that satisfies the conditions of the problem. Recall that the sum of an arithmetic series with first term a_1, last term a_n, and n terms, is \frac{n(a_1+a_n)}{2}. Plugging in our known values gives us the equation \frac{4040(2x+4039)}{2}=2020, which simplifies to 2x+4039=1. Thus, x=-2019, so the 4040 consecutive integers are -2019,-2018,\dots,-1,0,1,\dots,2018,2019,2020. Notice how everything except the 2020 cancels out!

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State the domain of the rational function. f(x) = seventeen divided by quantity five minus x
alexandr1967 [171]

If we are going to write the equation in its mathematical form, we will see that it becomes,

<span>                                        f(x) = 17/(x – 5)</span>

The domain of the function is the number of x that would allow us to solve the equation and get a real value for f(x). In this equation, x can take any real numbers so long as it is not 5. This is because 5 – 5 in the denominator will lead to 0 which in turn makes the equation indefinite.

<span> </span>

4 0
3 years ago
Help out?? Mathematics image.
Brilliant_brown [7]

Answer: (12) ∠1 = 20° (13) ∠2 = 50° (14) ∠3 = 15° (15) UV = 80° (16) AB = 40°  (17) ABC <em>or</em>  180° - CD (18) BC - 140°  (19) ABC = 150°

<u>Step-by-step explanation:</u>

12) \frac{1}{2}(UV - ST) = ∠1

\frac{1}{2}(80 - 40) = ∠1

\frac{1}{2}(40) = ∠1

20 = ∠1

13) \frac{1}{2}(UV + ST) = ∠2

\frac{1}{2}(70 + 30) = ∠2

\frac{1}{2}(100) = ∠2

50 = ∠2

14) \frac{1}{2}(VB - BS) = ∠3

\frac{1}{2}(60 - 30) = ∠3

\frac{1}{2}(30) = ∠3

15 = ∠3

15) \frac{1}{2}(UV - ST) = ∠1

\frac{1}{2}(UV - 20) = 30

UV - 20 = 60

UV = 80

16) ∠1 = arc AB

     ∠1 = 40

              arc AB = 40

17) arc AB + arc BC = arc AC

                               = 180 = arc CD  

18) ∠1 + ∠2 + ∠3 = 180

   20 + ∠2  + 20 = 180

            ∠2 + 40 = 180

                      ∠2  = 140

19) ∠1 + ∠ 2 = arc ABC

     ∠1 + ∠2 + ∠3 = 180

    arc ABC + 30 = 180

            arc ABC = 150


4 0
3 years ago
C) Parbati buys a mobile for Rs 6,300 and sells it to Laxmi at 15% profit. How much
Leni [432]

Answer:

Rs. 7245

Step-by-step explanation:

Given parameters:

Cost price  = Rs. 6300

Percentage profit  = 15%

Unknown:

Selling price = ?

Solution:

If profit is made on a trade, the selling price is higher than the cost price.

 Profit  = Selling price  - Cost price

To find the selling price simply;

  Selling price  =( 1 + \frac{15}{100})  x cost price

  Selling price  = 1.15 x 6300  = Rs. 7245

6 0
3 years ago
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