Can someone please help me?

Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C

Lesechka [4]

Answer:

Solution : $-\frac{3}{4}-\frac{3}{4}i$

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]$

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

$\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}$

= $-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)$ ÷ $2\sqrt{2}\left(0-1\right)i$

= $3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right)$ ÷ $-2\sqrt{2}i$

= $\frac{3\left(1-i\right)}{\sqrt{2}}$ ÷ $2\sqrt{2}i$ = $-3-3i$ ÷ $4$ = $-\frac{3}{4}-\frac{3}{4}i$

As you can see your solution is the last option.

3 0

3 years ago

Whats 1+1!!!!!!!!!!!!!!!!!!!!!!!!

ryzh [129]

Answer:

♡ madeline here ♡

1+1=2 i'm so sorry i am late! ☆

have a great day! hope this helps.

- madeline/madi

✧･ﾟ: *✧･ﾟ:･ﾟ✧*:･ﾟ✧･ﾟ

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

This luggage scale can measure weight to the nearest

Alinara [238K]

Answer:

hah idk grams? wait no pounds

Step-by-step explanation:

the answer is pounds

6 0

3 years ago

Read 2 more answers

When the temperature in London is 5 degrees celsius, the temperature in New York is - 7 degrees C. If the temperature in London

SOVA2 [1]

Answer:

the temperature in London would be -1 degrees as 5-6=-1

6 0

2 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

7 0

3 years ago