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Fiesta28 [93]
3 years ago
7

If there is 1/6 of a cake and Mitchell only wants 3/12 of it. how much of the cake does he want?

Mathematics
2 answers:
kherson [118]3 years ago
8 0

he wants 1/24 of the cake

Marta_Voda [28]3 years ago
8 0

Answer:

Mitchell wants \frac{1}{24} of the cake.

Step-by-step explanation:

There is \frac{1}{6} of cake and Mitchell wants \frac{3}{12} of the cake.

<em>Observation: The word "of" means "multiply"</em>

This means, \frac{3}{12} .\frac{1}{6}

Then \frac{3}{12} .\frac{1}{6}=\frac{3}{72}

We can simplify: \frac{3}{72}=\frac{3.1}{3.24}=\frac{3}{3}.\frac{1}{24}=\frac{1}{24}

Then Mitchell wants \frac{1}{24} of the cake.

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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}

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(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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