Can somebody please help fast?

Give an example of a quadratic equation with 2 + 3i as one of its solutions

Anastasy [175]

Answer:

Step-by-step explanation:

roots are (2+3i) and (2-3i)

reqd. eq. is (x-2-3i)(x-2+3i)=0

or (x-2)²-(3i)²=0

or x²-4x+9-9i²=0

or x²-4x+9+9=0

or x²-4x+18=0

5 0

3 years ago

Simone invests 1200 into two savings accounts. One account earns 4% annual interest and the other earns 5.9% annual interest. At

Allisa [31]

Answer: He invested $350 at 4% and $850 at 5.9%

Step-by-step explanation:

Let x represent the amount which he invested in the account earning 4% interest.

Let y represent the amount which he invested in the account earning 5.9% interest.

Simone invests 1200 into two savings accounts. One account earns 4% annual interest and the other earns 5.9% annual interest. This means that

x + y = 1200

The formula for determining simple interest is expressed as

I = PRT/100

Considering the account earning 4% interest,

P = $x

T = 1 year

R = 4℅

I = (x × 4 × 1)/100 = 0.04x

Considering the account earning 5.9% interest,

P = $y

T = 1 year

R = 5.9℅

I = (y × 5.9 × 1)/100 = 0.059y

At the end of 1 year, Simon earned 64.15 in interest. it means that

0.04x + 0.059y = 64.15 - - - - - - - - -1

Substituting x = 1200 - y into equation 1, it becomes

0.04(1200 - y) + 0.059y = 64.15

48 - 0.04y + 0.059y = 64.15

- 0.04y + 0.059y = 64.15 - 48

0.019y = 16.15

y = 16.15/0.019

y = 850

x = 1200 - y = 1200 - 850

x = 350

7 0

3 years ago

A conservative estimate of the number of stars in the universe is 6 x 10^22. The average human can see about 3,000 stars at nigh

Ilia_Sergeevich [38]

$2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

<h3><u>Solution:</u></h3>

Given that, conservative estimate of the number of stars in the universe is $6 \times 10^{22}$

The average human can see about 3,000 stars at night with only their eyes

To find: Number of times more stars are there in the universe, compared to the stars a human can see

Let "x" be the number of times more stars are there in the universe, compared to the stars a human can see

Then from given statement,

$\text{Stars in universe} = x \times \text{ number of stars human can see}$

<em><u>Substituting given values we get,</u></em>

$6 \times 10^{22} = x \times 3000\\\\x = \frac{6 \times 10^{22}}{3000}\\\\x = \frac{6 \times 10^{22}}{3 \times 10^3}\\\\x = 2 \times 10^{22-3}\\\\x = 2 \times 10^{19}$

Thus $2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

5 0

4 years ago

I don’t understand this can someone help? Please

Kobotan [32]

Answer:mkay, also yes he did

Step-by-step explanation:

1.Remove the parentheses

2.put -5z and 4z together which makes -9z(negative plus positive equals negative)

3.-9z+ -6z=-3z and 7+8=15

4.15 decided by -3=5

so yes, z=-5

7 0

3 years ago

Read 2 more answers

Line segment YV of rectangle YVWX measures 24 units.

Anna [14]

Answer: 8√3

<u>Step-by-step explanation:</u>

ΔXVW is a 30°-60°-90° triangle. This special triangle has corresponding sides of lengths: b - b√3 - 2b. So,

VW = b

XW = b√3

XV = 2b

Since, YV = 24, then XW = 24 and we stated above that XW = b√3. So,

24 = b√3

$\frac{24}{\sqrt{3}} = \frac{b\sqrt{3}}{\sqrt{3}}$

$\frac{24}{\sqrt{3}}$ = b

$\frac{24}{\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}})$ = b

$\frac{24\sqrt{3}}{3}$ = b

8√3 = b

We are looking for side YX which is congruent (equal) to VW and we stated above that VW = b, So, VW = 8√3

7 0

3 years ago