(4-5i)/(2+4i) plz help!!

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

-3/5-13i/10

Step-by-step explanation:

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Suppose the names of the months of the year are placed in a hat and a name is drawn at random. (a) list the sample space for thi

ollegr [7]

(a)

The sample space is a set whose elements are all the possible outcomes for the experiment. Since we will extract one of the months of the years, the sample space is the set composed by all the 12 months:

$\Omega = \{ \text{January},\ \text{February},\ldots,\text{December}\}$

(b)

An event is a subset of the sample space. Events are often defined by their properties. In this example, the event E is the subset of the sample space defined as

$E = \{ x \in \Omega: x \text{ starts with the letter J}\}$

So, we have

$E = \{\text{January},\ \text{June},\ \text{July}\}$

(c)

If all outcomes have equal probability, then the probability of an event is the ratio bewteen its cardinality, and the cardinality of the whole sample space:

$P(E) = \dfrac{n(E)}{n(\Omega)} = \dfrac{3}{12} = \dfrac{1}{4}$

In words, since there are three months beginning with J out of 12 months, we have a probability of 3 over 12 to pick a month starting with J, which simplifies to 1 over 4.

3 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :