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2 years ago

If a and b are two integers then closure property states then

Mathematics

1 answer:

Margarita [4]2 years ago

4 0

Your question has been heard loud and clear.

If a and b are two integers , then closure property states that the product of the two integers a and b , which is ab , is also an integer.

Thank you

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Can somone give me the answer to 2 x 0

marshall27 [118]

Any number times 0 is going to equal 0

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2 years ago

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Which of the following is the expansion of (3c + d2)6?

vovangra [49]

Answer: Option A) $729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$ is the correct expansion.

Explanation:

on applying binomial theorem, $(a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r$

Here a=3c, $b=d^2$ and n=6,

Thus, $(3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r$

⇒ $(3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6$

⇒ $(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}$

⇒ $(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}$

⇒ $(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$

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2 years ago

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What is 1/999 as a decimal

juin [17]

Answer:0.001001001001001

Step-by-step explanation: I have no idea. just put it in a calculator

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PSYCHO15rus [73]

We can start solving this problem by first identifying what the elements of the sets really are.

R is composed of real numbers. This means that all numbers, whether rational or not, are included in this set.

Z is composed of integers. Integers include all negative and positive numbers as well as zero (it is essentially a set of whole numbers as well as their negated values).

W on the other hand has 0,1,2, and onward as its elements. These numbers are known as whole numbers.

W ⊂ Z: TRUE. As mentioned earlier, Z includes all whole numbers thus W is a subset of it.

R ⊂ W: FALSE. Not all real numbers are whole numbers. Whole numbers must be rational and expressed without fractions. Some real numbers do not meet this criteria.

0 ∈ Z: TRUE. Zero is indeed an integer thus it is an element of Z.

∅ ⊂ R: TRUE. A null set is a subset of R, and in fact every set in general. There are no elements in a null set thus making it automatically a subset of any non-empty set by definition (since NONE of its elements are <u>not</u> an element of R).

{0,1,2,...} ⊆ W: TRUE. The set on the left is exactly what is defined on the problem statement for W. (The bar below the subset symbol just means that the subset is not strict, therefore the set on the left can be <u>equal</u> to the set on the right. Without it, the statement would be false since a strict subset requires that the two sets should not be equal).

-2 ∈ W: FALSE. W is just composed of whole numbers and not of its negated counterparts.

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Margie divided 24 by 3

Levart [38]

Answer:

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3 goes into 24, 8 times

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If a and b are two integers then closure property states then​

If a and b are two integers then closure property states then