Mathematics: There are 20 consecutive even numbers. how much bigger is the sum of the 10 larger ones than the sum of the ten smaller ones?

There are 20 consecutive even numbers. how much

Ответ:

molinaemily009

15.03.2022

Let the numbers be:
$2n,2(n+1),...,2(n+19)\\\text{Sum of the 10 biggest ones:}\\2(n+10)+...+2(n+19)=2((n+10)+...+(n+19))\\=2\times10\frac{n+10+n+19}{2}\\=10(2n+29)\text{the sum of an arithmetic sequence}$
The sum of the ten smaller one in the same manner:
$2(n)+...+2(n+9)=2(n+...+(n+9))\\=2\times10\frac{n+n+9}{2}$
Compute the difference of the above two sums:
$\\10(2n+29)-10\(2n+9)=290-90=200$

The bigger sum is 200 more bigger than the sum o the ten smaller sum.

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Ответ:

heythere7373

10.02.2020

Mathematics

y = 2x+3

Step-by-step explanation:

Sorry I'm probably too late since Quizziz doesn't wait unless you have the freeze timer thing.

y = mx+b

b is the y intercept. The y intercept (when x=0) is given in the table.

Now plug a coordinate in to solve for m.

My coordinates: (1,5)

5 = m + 3

Solve.

2 = m

m = 2

Note: never plug in the y intercept when looking for m.

You can also find the slope by (subtracting the y value of one coordinate from another y value of another coordinate)/(subtracting x value from the first y coordinate above from the x value from the second coordinate).

In this case, it'd be

m = (7-5)/(2-1)

m = 2/1

m = 2

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There are 20 consecutive even numbers. how much bigger is the sum of the 10 larger ones than the sum of the ten smaller ones?

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