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3 years ago

Which of the following sets is NOT dlosed under addition?

Mathematics

2 answers:

sergij07 [2.7K]3 years ago

4 0

Answer:

Odd numbers.

Step-by-step explanation:

Closed under addition means you will able to take any two numbers from whatever set is mentioned and add them to get a number in that same set.

(If I say counting numbers, I'm still talking about the natural numbers.)

So let's look at the natural numbers {1,2,3,4,5,6,...}.

1+1=2

1+2=3

1+3=4

...

6+19=25

Let a and b be counting numbers.

a+b is still going to be a counting number.

You will always get a counting number when adding two counting numbers. So the counting numbers (also known as the natural numbers) is closed under addition.

Let's skip down to odd numbers because the other sets are similar to the first.

Let's add a pair of odd numbers.

3+5=8

8 is not odd so the odd numbers are not closed under addition because we will not always get an odd number. In fact, you will never get a odd number, but the thing is you just need one example to show it is not closed.

(2k+1)+(2a+1)

2(k+a)+2

2(k+a+1) is even so adding two odds will always give you an even.

weeeeeb [17]3 years ago

3 0

Odd numbers is your answer

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3 years ago

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C=1/21.22.23+1/22.23.24+................+1/200.201.202 . = là dấu nhân

Aneli [31]

It looks like you have to find the value of the sum,

$C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}$

so that the n-th term in the sum is

$\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}$

for 1 ≤ n ≤ 180.

We can then write the sum as

$\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}$

Break up the summand into partial fractions:

$\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}$

Combine the fractions into one with a common denominator and set the numerators equal to one another:

$1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)$

Expand the right side and collect terms with the same power of n :

$1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c$

Then

a + b + c = 0

43a + 42b + 41c = 0

462a + 440b + 420c = 1

==> a = 1/2, b = -1, c = 1/2

Now our sum is

$\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)$

which is a telescoping sum. If we write out the first and last few terms, we have

C = 1/(2×21) - 1/22 + 1/(2×23)

… … + 1/(2×22) - 1/23 + 1/(2×24)

… … + 1/(2×23) - 1/24 + 1/(2×25)

… … + 1/(2×24) - 1/25 + 1/(2×26)

… … + … - … + …

… … + 1/(2×198) - 1/199 + 1/(2×200)

… … + 1/(2×199) - 1/200 + 1/(2×201)

… … + 1/(2×200) - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

C = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

C = 1/42 - 1/44 - 1/402 + 1/404

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3 years ago

Can u help me with this one

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Answer:

SU = 16

Step-by-step explanation:

SU= 8x

ST = 6x

TU = 4

SU = ST + TU = 6x + 4 = 8x

SU - ST = 8x - 6x = 2x = 4

2x = 4

x = 4/2

x = 2

SU = 8x

SU = 8(2)

SU = 16

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