One ordered pair (a,b) satisfies the two equations ab^4 = 384 and a^2 b^5 = 4608. What is the value of a in this ordered pair?

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer: a = 3∛2

Step-by-step explanation:

ab⁴ = 384 --> a = 384/b⁴

Substitute a = 384/b⁴ into the second equation to solve for "b".

a²b⁵ = 4608

$\bigg(\dfrac{384}{b^4}\bigg)^2\cdot b^5=4608\\\\\\\dfrac{147,456b^5}{b^8}=4608\\\\\\\dfrac{147,456}{b^3}=4608\\\\\\\dfrac{147,456}{4608}=b^3\\\\\\32=b^3\\\\\\\sqrt[3]{32} =b\\\\\\2\sqrt[3]{4} =b$

Substitute b = 2∛4 into the first equation to solve for "a".

ab⁴ = 384

a(2∛4)⁴ = 384

a = 384/(2∛4)⁴

a = 24/4∛4

= 6/∛4

= 6(∛2)/2

= 3∛2

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Use the binomial expression (p+q)n

zvonat [6]

Answer:

{0.16807, 0.36015, 0.3087, 0.1323, 0.02835, 0.00243}

Step-by-step explanation:

The expansion of (p+q)^n for n = 5 is ...

(p+q)^5 = p^5 +5·p^4·q +10·p^3·q^2 +10·p^2·q^3 +5·p·q^4 +q^5

When the probability p=0.3 and q = 1-p = 0.7 the terms of this series correspond to the probabilities of 5, 4, 3, 2, 1, and 0 favorable outcomes out of 5 trials.

For example, p^5 = 0.3^5 = 0.00243 is the probability of 5 favorable outcomes in 5 trials where the probability of each favorable outcome is 0.3.

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The attachment shows the calculation of these numbers using a graphing calculator. It lists them in reverse order of the expansion of (p+q)^5 shown above, so that they are the probabilities of 0–5 favorable outcomes in the order 0–5.