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3 years ago

Complete the following exercises by applying polynomial identities to complex numbers. Show your work:

Mathematics

1 answer:

Murrr4er [49]3 years ago

7 0

1.)

=(x-8i)(x+8i)

x^2+8ix-8ix-64i^2

x^2-64i^2

x^2-64(-1)

x^2+64

2.)

=(4x-7i)(4x+7i)

16x^2+28ix-28ix-49i^2

16x^2-49i^2

16x^2-49(-1)

16x^2+49

3.)

=(x+9i)(x+9i)

x^2+9ix+9ix+81i^2

x^2+18ix+81(-1)

x^2+18ix-81

4.)

=(x-2i)(x-2i)

x^2-2ix-2ix+4i^2

x^2-4ix+4(-1)

x^2-4ix-4

5.)

=[x+(3+5i)]^2

(x+5i+3)^2

(x+5i+3)(x+5i+3)

x^2+5ix+3x+5ix+25i^2+15i+3x+15i+9

x^2+6x+10ix+30i+25i^2+9

x^2+6x+10ix+30i+25(-1)+9

x^2+6x+10ix+30i-25+9

x^2+6x+10ix+30i-16

Hope this helps :)

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true

kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 276, p = 0.06$

$\mu = E(X) = np = 276*0.06 = 16.56$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454$

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{27.6 - 16.56}{3.9454}$

$Z = 2.8$

$Z = 2.8$ has a pvalue of 0.9974

99.74% probability that the sample proportion will be less than 0.1

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The concession stand at a ballpark makes $0.43 profit from each hot dog sold. It also makes a profit of $0.65 from each pretzel

kaheart [24]

0.43h + 0.65p + 1.31d = c

explanation:
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each price next to the variable describes the profit per item.

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Answer:

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