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4 years ago

Which number is divisible by 2, 3, and 6

Mathematics

2 answers:

enyata [817]4 years ago

6 0

Any multiple of 6 is divisible by 2, 3, and 6.

gladu [14]4 years ago

6 0

Six is divisable by 2 3 and 6 ;D

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3 years ago

In the expansion of (3a + 4b)8, which of the following are possible variable terms? Explain your reasoning. a2b3; a5b3; ab8; b8;

Fiesta28 [93]

We have to find the expansion of $(3a+4b)^{8}$

We will use binomial expansion to expand the given expression, which states that the expression $(a+b)^{n}$ is expanded as :

$(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{2}a^{n-2}b^{2}+........^{n}C_{n}b^{n}$

Now expanding $(3a+4b)^{8}$ we get,

$(3a+4b)^{8}=^{8}C_{0}(3a)^{8}+^{8}C_{1}(3a)^{7}(4b)+^{8}C_{2}(3a)^{6}(4b)^{2}+^{8}C_{3}(3a)^{5}(4b)^{3}+^{8}C_{4}(3a)^{4}(4b)^{4}+^{8}C_{5}(3a)^{3}(4b)^{5}+^{8}C_{6}(3a)^{2}(4b)^{6}+^{8}C_{7}(3a)(4b)^{7}+^{8}C_{8}(4b)^{8}$

$(3a+4b)^{8}=^{8}C_{0}(3)^{8}a^{8}+^{8}C_{1}(3)^{7}(4)(a^{7}b)+^{8}C_{2}(3)^{6}(4)^{2}(a^{6}b^{2})+^{8}C_{3}(3)^{5}(4)^{3}(a^{5}b^{3})+^{8}C_{4}(3)^{4}(4)^{4}(a^{4}b^{4})+^{8}C_{5}(3)^{3}(4)^{5}(a^{3}b^{5})+^{8}C_{6}(3)^{2}(4)^{6}(a^{2}b^{6})+^{8}C_{7}(3)(4)^{7}(ab^{7})+^{8}C_{8}(4)^{8}(b^{8})$

So, the variables are $a^{5}b^{3}$ , $b^{8}$ , $a^{4}b^{4}$ , $a^{8} , [tex] ab^{7}$

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3 years ago

Whats the answer to this i need help please

Elan Coil [88]

51 feet you have to multiple 32 by 8 it will give you 256 and then divide 256 and 5 and you get 51

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3 years ago