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musickatia [10]
4 years ago
14

Write and solve an equation that means "6 less than a number (x) is 5."

Mathematics
2 answers:
krok68 [10]4 years ago
8 0
X-6=5 is the answer that I got. 6 less than a number(x) means you are subtracting 6 and is 5 means it equals 5. Hope this helps:))
Taya2010 [7]4 years ago
8 0
X-6=5
You should read it how it is written. It is basically saying x minus 6 equals 5.
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100,000 beacause of the internet x 2
3 0
3 years ago
4x-29=5x-9 What is the answer to this, I keep getting -20 and i feel like thats wrong. (solve for x)
VARVARA [1.3K]
It’s definitely-20
Add 9 to both sides to cancel it out so it is now
4x-20=5x
Than subtract 4x on both sides it is now
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Because 5-4 equals 1 and 1x is just written as x
5 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
270 qt: 105 gal<br> Simplified
Sidana [21]

Answer:

wouldn't it be 18/7

Step-by-step explanation:

270 ÷ 15 over 105 ÷ 15

7 0
3 years ago
PLEASE HELP,, MARKING BRAINLIEST!!!<br><br> Solve for the value of x.
Eddi Din [679]

Answer:

37

Step-by-step explanation:

all triangles have a total of 180 degrees

set up equation and solve

180=(2x+4)+(2x-9)+x

180=5x-5

185=5x

37=x

5 0
3 years ago
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