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Tatiana [17]
3 years ago
8

Which of the following is a perfect square trinomial ?

Mathematics
2 answers:
Ivahew [28]3 years ago
4 0

For this case, we have that by definition, a perfect square trinomial, is a polynomial of three terms (also called trinomial) that results from squaring a binomial.

(a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2\\(a-b) ^ 2 = a ^ 2-2ab + b ^ 2

We have:

4x ^ 2-20xy + 25y ^ 2\\(2x) ^ 2 = 4x ^ 2\\(5y) ^ 2 = 25y ^ 2

So:

2 * 5y * 2x = 20xy

So, 4x ^ 2-20xy + 25y ^ 2we can write it as:

(2x-5y) ^ 2

Answer:

Option A

Cerrena [4.2K]3 years ago
3 0

Answer:

The answer is A

Step-by-step explanation:

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Answer:

The dependent variable:  the total cost

The independent variable: the amount of books


If the equation is  y = 8x + 200

y would be the dependent variable and x would be the independent variable


I hope this helps.

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Line p passes through A(-2, -4) and has a slope of 1/2 What is the standard form of the equation for line p?
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Point slope form is y-(-4) = 1/2(x-(-2)

y+4 =1/2 (x+2)

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Read 2 more answers
You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first
Tomtit [17]

Answer:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=3, p=0.1)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

Y \sim Binom(n=5, p=0.05)

The probability mass function for the Binomial distribution is given as:

P(Y)=(nCy)(p)^y (1-p)^{n-y}

Where (nCx) means combinatory and it's given by this formula:

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0
3 years ago
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