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3 years ago

If a,b,c are all non-zero numbers and a+b+c=0, prove that

Mathematics

1 answer:

Butoxors [25]3 years ago

6 0

Given - a+b+c = 0

To prove that- 

a²/bc + b²/ac + c²/ab = 3

Now we know that

when x+y+z = 0,

then x³+y³+z³ = 3xyz

that means

 (x³+y³+z³)/xyz = 3 ---- eq 1)

Lets solve for LHS

LHS = a²/bc + b²/ac + c²/ab

we can write it as LHS = a³/abc + b³/abc + c³/abc

by multiplying missing denominators,

now take common abc from denominator and you'll get,

LHS = (a³+b³+c³)/abc --- eq (2)

Comparing one and two we can say that

(a³+b³+c³)/abc = 3

Hence proved,

a²/bc + b²/ac + c²/ab = 3

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4 years ago

The following lists the joint probabilities associated with smoking and lung disease among 60-to-65 year-old men. Has Lung Disea

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Answer:

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Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Does not smoke

Event B: Lung disease

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This means that $P(A \cap B) = 0.03$

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This means that $P(A) = 0.03 + 0.7 = 0.73$

What is the probability of the following event: He has lung disease given that he does not smoke?

$P(B|A) = \frac{0.03}{0.73} = 0.0411$

4.11% probability that he has lung disease given that he does not smoke

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3 years ago

What are the partial products for 3.2 x 0.6?

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Answer:

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Answer:

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