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Mariulka [41]
3 years ago
6

How many bacteria are in generation 9? O A. 1,180,980 O B. 393,660 C. 1440 O D. 6561

Mathematics
2 answers:
scoundrel [369]3 years ago
5 0

Answer:

B. 393,660

Step-by-step explanation:

lemme get Brainliest please

asambeis [7]3 years ago
5 0

Answer:

B. 393,660

Step-by-step explanation:

I plugged it into the symbolab calculator and attached the steps below! Hope this helps!

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The floor of our classroom is 50 feet by 75 feet. What is the area of our floor?
N76 [4]
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Area of rectangle= length * hight
8 0
4 years ago
Find the cube roots of 27(cos 330° + i sin 330°)
Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

<u>DeMoivre's Theorem</u>

Let z=r(cos\theta+isin\theta) be a complex number in polar form, where n is an integer and n\geq1. If z^n=r^n(cos\theta+isin\theta)^n, then z^n=r^n(cos(n\theta)+isin(n\theta)).

<u>Nth Root of a Complex Number</u>

If n is any positive integer, the nth roots of z=rcis\theta are given by \sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}} where the nth roots are found with the formulas:

  • \sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr] for degrees (the one applicable to this problem)
  • \sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr] for radians

for  k=0,1,2,...\:,n-1

<u>Calculation</u>

<u />z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)

<u>First cube root where k=2</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]

<u>Second cube root where k=1</u>

\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]

<u>Third cube root where k=0</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]

4 0
3 years ago
Please help by Friday !!
Angelina_Jolie [31]

Answer:

Depends on WHAT figure

Step-by-step explanation:

squares and circles have diffrent formulas, and so do 3D shapes!!!!!!!!!!!!!!

3 0
3 years ago
Read 2 more answers
Could anyone tell me if I’m right. This is a gradient of a line question
pshichka [43]

Answer:

See Below (it is correct)

Step-by-step explanation:

First point given ----  A(-3,0)

2nd point given ----- B(4,0)

The slope is the "change in quantity y" divided by "change in quantity x"

The change would be from 2nd point to 1st point. So,

Change in quantity y is 0 - 0 = 0

Change in quantity x is 4 - (-3) = 4 + 3 = 7

So, slope would be:

Slope = 0/7 = 0

The slope is 0 (which means it is a horizontal line)

3 0
3 years ago
I need help asap guys please I’ll give brainlest to help you out
Maru [420]

Answer:

Table c mate

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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