All categories

3 years ago

Difficult pre calculus stuff.

Mathematics

1 answer:

leva [86]3 years ago

7 0

When we divide complex numbers we divide their magnitudes and subtract their angles, so this isn't all that difficult:

$\dfrac{10(\cos 45^\circ + i \sin 45^\circ)}{5 (\cos 15^\circ + i \sin 15^\circ)} = \dfrac{10}{5} (\cos(45^\circ - 15^\circ) + i \sin(45^\circ - 15^\circ))$

$= 2(\cos 30^\circ + i \sin 30^\circ)$

Answer: $2(\cos 30^\circ + i \sin 30^\circ)$

That's probably what's called simplified polar form until you get to Euler's Formula. Since it's 30 degrees we can get a nice rectangular form:

$2(\cos 30^\circ + i \sin 30^\circ) = 2\left( \dfrac{\sqrt{3}}{2} + i \dfrac 1 2 \right) = \sqrt{3} + i$

You might be interested in

Lacy has a $1000 bond with a 6.2% coupon that she purchased for $1025. What is the yield of this bond? Round your answer to 2 de

Ede4ka [16]

So the question is asking what is 6.2% of 1000? Then rounding our decimals?

4 0

3 years ago

Read 2 more answers

Somebody please help me thansk

ad-work [718]

Answer:

y=-2x-1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the volume of the cylinder Type an exact answer in terms of pi

ZanzabumX [31]

Answer:

648pi

Step-by-step explanation:

V=pi*r^2*h

V=pi*9^2*8

V=pi*81*8

V=648pi

7 0

4 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

brainly.com/question/1216161

8 0

3 years ago

A community garden with raised beds is being redesigned. The area of the new rectangular beds can be expressed as one-half x (3

Angelina_Jolie [31]

The simplification form of the expression is 3x²/2 + 2x or three-halves x squared + 2 x option second is correct.

<h3>What is the area of the rectangle?</h3>

It is defined as the area occupied by the rectangle in two-dimensional planner geometry.

The area of a rectangle can be calculated using the following formula:

Rectangle area = length x width

We have:

A community garden with raised beds is being redesigned. The area of the new rectangular beds can be expressed as one-half x (3 x + 4).

The expression:

$\rm = \dfrac{1}{2}(3x+4)$

By using the distributive property:

$\rm = \dfrac{1}{2}\times3x^2+\dfrac{1}{2}\times4x$

$\rm = \dfrac{3}{2}x^2+2x$

Thus, the simplification form of the expression is 3x²/2 + 2x or three-halves x squared + 2 x option second is correct.

Learn more about the rectangle here:

brainly.com/question/15019502

#SPJ1

4 0

2 years ago