Ask question

Ask question

All categories

3 years ago

9

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

1 answer:

Step2247 [10]3 years ago

8 0

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

You might be interested in

Ima just give all my points away.<br><br> P.S<br> I don't like you brainly.

OLga [1]

Answer:

interesting

Step-by-step explanation:

5 0

3 years ago

WRITE A TWO COLUMN PROOF for number 19!

Fittoniya [83]

Answer:

Step-by-step explanation:

The answer is 19

5 0

3 years ago

The cost for b, breakfasts, l, lunches, and d, dinners for a group of people is 6.5b + 9.757 + 17.25d. What is the total cost of

ira [324]

Answer:

417.084

Step-by-step explanation:

9*6.5+12*9.757+14*17.25

58.5+117.084+241.5

⇒417.084

5 0

4 years ago

Read 2 more answers

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.

$I(5)=x^5e^x-5I(4)$

$I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)$

$I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)$

$I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)$

$I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))$

Finally,

$I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C$

so we end up with

$I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C$

$I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C$

and the antiderivative is

$\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C$

8 0

3 years ago

A ladder is placed 30 inches from a wall. It touches the wall at a height of 50 inches from the ground. The angle made by the la

Alex73 [517]

The degree measure is about 59° and the length of the ladder is 10√34 in or about 59.3in

3 0

3 years ago