What are all the numbers of the square root of pie?

Identify the quotient in the form a + bi. HELP ASAP PLEASE!!

Step2247 [10]

Solving the expression $\frac{4-5i}{2+3i}$ we get $-\frac{7}{13}-\frac{22i}{13}$

So, Option A is correct.

Step-by-step explanation:

We need to solve the expression: $\frac{4-5i}{2+3i}$

Multiplying and dividing by 2-3i

$=\frac{4-5i}{2+3i}*\frac{2-3i}{2-3i}\\=\frac{4-5i*2-3i}{2+3i*2-3i}\\=\frac{4(2-3i)-5i(2-3i)}{(2)^2-(3i)^2}\\=\frac{8-12i-10i+15i^2)}{4-9i^2}\\We\,\,know\,\, i^2=-1\\=\frac{8-12i-10i+15(-1)}{4-9(-1)}\\=\frac{8-15-12i-10i}{4+9}\\=\frac{-7-22i}{13}\\=-\frac{7}{13}-\frac{22i}{13}$

So, solving the expression $\frac{4-5i}{2+3i}$ we get $-\frac{7}{13}-\frac{22i}{13}$

So, Option A is correct.

Keywords: Complex numbers

Learn more about Complex numbers at:

#learnwithBrainly

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3 years ago

Consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number,

Aleksandr-060686 [28]

Answer:

The expected value of the points earned on a single roll in this game is $\dfrac{1}{6} = 0.1667$ .

Step-by-step explanation:

We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.

Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).

Here, the prime numbers are = 1, 2, 3 and 5

Numbers which are not prime = 4 and 6

This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.

Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.

Note that the probability of getting any of the specific number on the dice is $\dfrac{1}{6}$ .

Numbers on the dice (X) P(X)

+1 $\frac{1}{6}$

+2 $\frac{1}{6}$

+3 $\frac{1}{6}$

-4 $\frac{1}{6}$

+5 $\frac{1}{6}$

-6 $\frac{1}{6}$

Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.

Now, the expected value of X, E(X) = $\sum X \times P(X)$

= $(+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}$