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SVEN [57.7K]
3 years ago
13

Marco states that 7.696696669 is a rational number because it is a repeating decimal. is he correct

Mathematics
2 answers:
DochEvi [55]3 years ago
7 0

Hello there! And welcome to brainly . com! Your number one asking and answering site.

Answer: No.

Why?

The said decimal is a string of numbers, but does not have the same number repeated over and over again, nor does it have a triple dot after it. Thus, it is not repeating.

Hope this helped you, and have a great day! ♥

SVETLANKA909090 [29]3 years ago
4 0

Answer:

No, Marco is not correct.

Step-by-step explanation:

The number does not actually repeat because after each 9 there is an additional 6 before the next 9. The number is not a rational number.

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Answer:

true

Step-by-step explanation:

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Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule
Sergeeva-Olga [200]

Answer:

First term: 5

Fourth term: 5 1/2

Tenth term: 6 1/2

Step-by-step explanation:

Let's find the first, fourth and tenth terms of the arithmetic sequence described by the given rule:

A(n) = 5 + (n-1) (1/6)

First term:

A(1) = 5 + (1-1) (1/6)

A(1) = 5 + (0) (1/6)

A(1) = 5

Fourth term:

A(4) = 5 + (4-1) (1/6)

A(4) = 5 + (3) (1/6)

A(4) = 5 + 3/6 = 5 3/6 = 5 1/2 (simplifying)

Tenth term:

A(10) = 5 + (10-1) (1/6)

A(10) = 5 + (9) (1/6)

A (10) = 5 + 9/6 = 6 3/6 = 6 1/2 (simplifying)

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3 years ago
Translate the statement below. "One less than twice a number.."​
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One less than twice a number is seventeen.
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Show how you got the answer
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2 years ago
Which of the following is not one of the 8th roots of unity?
Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1.  Since z=1=1+0\cdot i, then

Rez=1,\\ \\Im z=0

and

|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.

This gives you \varphi=0.

Thus,

z=1\cdot(\cos 0+i\sin 0).

2. The 8th roots can be calculated using following formula:

\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.

Now

at k=0,  z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;

at k=1,  z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=2,  z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;

at k=3,  z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=4,  z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;

at k=5,  z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

at k=6,  z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;

at k=7,  z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

The 8th roots are

\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.

Option C is icncorrect.

5 0
2 years ago
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