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3 years ago

Which of the following is a perfect square trinomial?

Mathematics

1 answer:

shutvik [7]3 years ago

5 0

Answer:

Step-by-step explanation:

Take a look at what happens when squaring either of these...

$(ax-by)^2=(ax-by)(ax-by)=a^2x^2-2abxy+b^2y^2\\\\(ax+by)^2=(ax+by)(ax+by)=a^2x^2+2abxy+b^2y^2\\$

Notice a couple of patterns.

1. The last term has a positive coefficient. That rules out answer choices C and D.

2. The coefficient of the middle term is either $+2ab \text{ or } -2ab$ . So what are a and b? a is the square root of the x^2 term and b is the square root of the y^2 term.

$25x^2 \Rightarrow a=5\\\\9y^2 \Rightarrow b=3\\\\ab=15\\\\2ab=30$

The middle coefficient needs to be either +30 or -30. The answer is choice A.

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geniusboy [140]

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5 0

2 years ago

Read 2 more answers

A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e

My name is Ann [436]

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

$n = 2, r=1$ .

Therefore,

$\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

$n = 20, r=10$ .

Therefore,

$\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

$n = 2 \times 10^{23}, r = 10^{23}$ .

Therefore, we obtain that the number of ways is

$\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$