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drek231 [11]
3 years ago
11

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

a4b4; a8; ab7; a6b5
Help please!
Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0

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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Answer:

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Step-by-step explanation:

Given:

Length of a rectangular sand box = 5\frac{1}{3}\ feet

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Substituting values given for length and width.

Perimeter of sand box = 2(5\frac{1}{3}+3\frac{3}{4})\ feet

Simplifying by adding fractions:

⇒  2(5+3+\frac{1}{3}+\frac{3}{4})\ feet (Adding whole numbers and fractions separately)

⇒  2(8+\frac{1}{3}+\frac{3}{4}) feet

Whole number 8 can be written as \frac{8}{1}

⇒  2(\frac{8}{1}+\frac{1}{3}+\frac{3}{4}) feet

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LCD for 3 and 4 = 12 as its the least common multiple of 3,4,1.

Making denominators =12 by multiplying numerator an denominator with the corresponding numbers

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⇒  2(\frac{96}{12}+\frac{4}{12}+\frac{9}{12}) feet

Then we simply add numerators.

⇒  2(\frac{96+4+9}{12}) feet

⇒  2(\frac{109}{12}) feet

⇒  2\times \frac{109}{12} feet

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Simplifying fractions to simplest form.

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