Ask question

Ask question

All categories

4 years ago

10

How to solve: (imaginary number)

ac{8}{3i} " alt=" \frac{8}{3i} " align="absmiddle" class="latex-formula">

1 answer:

Anuta_ua [19.1K]4 years ago

4 0

Multiply both sides of this fraction by i to cancel out the i in the denominator.

8 * i = 8i

3i * i = 3i^2 = -3

-(8i/3) is the simplified expression.

You might be interested in

Savatey [412]

If we want to find ∠XYZ, we have to find the measurement of ∠VUW.
If we want to find ∠VUW, we have to solve for x.

We know that ∠VUW and ∠VUT add up to 90°, so we can do the following...

(6x - 1) + (3x - 8) = 90

Combine like terms...

9x - 9 = 90

Solve for x.

9x = 99

x = 11

Now we have to plug in 11 for x in the equation 3x - 8 to solve for ∠VUW.

3(11) - 8 = ∠VUW = ∠XYZ = 25°

5 0

3 years ago

Read 2 more answers

Solve this linear equation for p: 2.6(5.5p – 12.4) = 127.92 1. Distributive property: 2. Addition property of equality: 3. Divis

Ronch [10]

P= 11.2

Not sure why you took p out of the equation.

2.6(5.5p-12.4)=127.92

14.3p - 32.24 =127.92

14.3p = 160.16

P = 11.2

4 0

3 years ago

Read 2 more answers

Identify a pattern in each list of numbers. Then use this pattern to find the next number.

dsp73

Answer:

-140

Step-by-step explanation:

a₀ = 3

a₁ = 3 - 13*1 = -10

a₂ = -10 + 13*2 = 16

a₃ = 16 - 13*4 = -36

a₄ = -36 + 13*8 = 68

<u>a₅</u> = 68 - 13*16 = <u>-140</u>

* a₀=3 n is integers and n ≥ 1

aₙ = aₙ₋₁ + (-1)ⁿ * 13 * (2)ⁿ⁻¹

3 0

3 years ago

HELP ME OUT PLZZ I WILL GIVE BRAINLISt

BaLLatris [955]

It might be A

Example: I tried to do the math and i came out with 1/20 so try it. i’m very sorry if it’s wrong

4 0

3 years ago

Read 2 more answers

The price of products may increase due to inflation and decrease due to depreciation. Marco is studying the change in the price

inn [45]

Answer:

Product A has a greater percentage change in price.

Step-by-step explanation:

Part A:

The price f product A, f (<em>x</em>) after <em>x</em> years is given by:

$f(x) = 0.69\cdot(1.03)^{x}$

After <em>x</em> = 0 years, the price of product A is:

$f(0) = 0.69\cdot(1.03)^{0}=0.69$

After <em>x</em> = 1 years, the price of product A is:

$f(1) = 0.69\cdot(1.03)^{1}=0.69\cdot (1+0.03)=0.69\cdot (1+3\%)$

After 1 year, the price of product A is 3% times more than the original price.

This means that after one year, the new price is 103% of the original price, which means the price product A is increasing by 3%.

Again after <em>x</em> = 2 years, the price of product A is:

$f(2) = 0.69\cdot(1.03)^{2}=[0.69\cdot (1+3\%)]\times (1.03)$

This implies that after 2 years, the price of product A is 103% of the price after year 1.

This implies that the price of product A is 3% more than the previous year.

Thus, the price of product A is increasing each year by 3%.

Part B:

The data for Product B is as follows:

Time (t) Price [f (t)]

1 10,100

2 10,201

3 10,303.01

4 10,406.04

Product B is clearly increasing in price.

Consider the changes in price of Product B in the following intervals of years:

Year 1 - Year 2:

Price in year 1 = $10,100

Price in Year 2 = $10,201

Compute the increase percentage as follows:

$\text{Increase}\%=\frac{10201-10100}{10100}=0.01=1\%$

Year 2 - Year 3:

Price in Year 2 = $10,201

Price in year 3 = $10,303.01

Compute the increase percentage as follows:

$\text{Increase}\%=\frac{10303.01-10201}{10201}=0.01=1\%$

Year 3 - Year 4

Price in year 3 = $10,303.01

Price in Year 4 = $10,406.04

Compute the increase percentage as follows:

$\text{Increase}\%=\frac{10,406.04-10303.01}{10303.01}=0.09999\approx 0.01=1\%$

It is quite clear that the price of product B increases by 1% each year.

Thus, Product A has a greater percentage change in price.

3 0

3 years ago