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pickupchik [31]

3 years ago

8

Pls help due today >_<. Thank you

2 answers:

denis-greek [22]3 years ago

6 0

Answer:

4/9

Step-by-step explanation:

1/2 is 0.5

4/3 doesn't matter. It's an improper fraction, so it's above 1.

9/10 is 0.9

So, the only one left is 4/9, which is 0.44 repeating.

ArbitrLikvidat [17]3 years ago

3 0

Answer:

4/9

Step-by-step explanation:

The only other recurring decimal would be 4/3 but that's greater than 1.

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What is the factorization of 729^15+1000?

igomit [66]

Answer:

The factorization of $729x^{15} +1000$ is $(9x^{5} +10)(81x^{10} -90x^{5} +100)$

Step-by-step explanation:

This is a case of factorization by <em>sum and difference of cubes</em>, this type of factorization applies only in binomials of the form $(a^{3} +b^{3} )$ or $(a^{3} -b^{3})$ . It is easy to recognize because the coefficients of the terms are <u><em>perfect cube numbers</em></u> (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).

Let's solve the factorization of $729x^{15} +1000$ by using the <em>sum and difference of cubes </em>factorization.

1.) We calculate the cubic root of each term in the equation $729x^{15} +1000$ , and the exponent of the letter x is divided by 3.

$\sqrt[3]{729x^{15}} =9x^{5}$

$1000=10^{3}$ then $\sqrt[3]{10^{3}} =10$

So, we got that

$729x^{15} +1000=(9x^{5})^{3} + (10)^{3}$ which has the form of $(a^{3} +b^{3} )$ which means is a <em>sum of cubes.</em>

<em>Sum of cubes</em>

$(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})$

with $a= 9x^{5}$ y $b=10$

2.) Solving the sum of cubes.

$(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )$

$(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)$

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