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avanturin [10]
3 years ago
7

Write the ratio: 3 1/6 to 17/6 as a fraction in lowest terms.

Mathematics
1 answer:
slega [8]3 years ago
6 0

Answer:  \frac{19}{6} : \frac{17}{6}

Step-by-step explanation:

Let us first represent the ratio.

3\frac{1}{6} : \frac{17}{6}

Now we need to convert 3\frac{1}{6} into a improper fraction.

\frac{19}{6} : \frac{17}{6}

Now it is in its simplest form.

The final answer is  \frac{19}{6} : \frac{17}{6}

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Sally made a decorative chess board for her beach house rental. If the area of the board is 144 square inches, then how many squ
Colt1911 [192]

Answer:

36

Step-by-step explanation:

When you divide 144 by 1/4 you get 36. hope I helped

4 0
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Draw the line of reflection that reflects AABC onto AA'B'C'.
geniusboy [140]

Answer:

x=-2

Step-by-step explanation:

answer attached

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HALP PLEAAASE I NEED ANWSER I JEED TO HURRY SO I CAN GO TO MY AFTER SCHOOL CLUB- WHICH IS THREATENING TO KICK ME IF I DONT MAKE
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Step-by-step explanation:

6 0
3 years ago
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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
2 years ago
5(2y−3)−4y=8−2y−15<br> Justify this please
juin [17]

Answer:

y=1     (sorry i dont know what justify means lol, i forgot) i know how to solve the answer lol.

Step-by-step explanation:

10y-15-4y=8-2y-15

10y-4y+2y=8-15+15

8y=8

y=1

3 0
3 years ago
Read 2 more answers
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