All categories

2 years ago

Simplify the expression to a + bi form: (2 – i) – (-4+8i)

Mathematics

2 answers:

Ne4ueva [31]2 years ago

8 0

Answer:

6-9i

Step-by-step explanation:

(2-i)-(-4+8i)

2-i+4-8i

2+4-i-8i

6-i-8i

6-9i

If you're satisfied with the answer, then please mark me as Brainliest.

ser-zykov [4K]2 years ago

5 0

Answer:

$→(2 - i) - ( - 4 + 8i) \\ = 2 - i + 4 - 8i \\ = 2 + 4 - i - 8i \\ = \boxed{6 - 9i}✓$

<u>6 - 9i</u> is the right answer.

You might be interested in

On a model castle, 1 inch equals 83 feet. The model is 10.5 inches wide. What is the actual width of the castle?

Furkat [3]

Answer:

871 and 1/2 ft.

8 0

3 years ago

8x – 3 = -19<br><br>Who give it fast they have branliest <br>Give the correct answer step by step

lakkis [162]

Answer:

Step-by-step explanation:

$8x - 3 = - 19$

Collect like terms and simplify

$8x = - 19 + 3 \\ 8x = - 16$

Divide both sides of the equation by 8

$\frac{8x}{8} = \frac{ - 16}{8}$

Simplify

$x = - 2$

4 0

3 years ago

Read 2 more answers

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2