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postnew [5]
2 years ago
11

27 3/10 - 12 5/9 write in old style math

Mathematics
2 answers:
Julli [10]2 years ago
6 0

Answer:

<em>- 15 </em>\frac{23}{50}

Step-by-step explanation:

  1. We need to make \frac{3}{10} and \frac{5}{9} have the same denominator, or bottom half, before we can subtract them.
  2. In order to do this, we need to find a number that both 10 and 9 (the bottom numbers) both go into. (Hint! Multiplying them together usually helps!)
  3. This number is 90. (See? 10 • 9 = 90!)
  4. Since we multiplied 10 by 9 to get 90, we need to multiply the top (3) by 9 too, to get us \frac{27}{90}.
  5. We multiply \frac{5}{9} by 10 to get us \frac{50}{90}.
  6. Now we can subtract 27 \frac{27}{90} and 12 \frac{50}{90}.
  7. 27 - 12 = 15 (big numbers)
  8. 27 - 50 = -23 (fractions)
  9. Now we have - 15 \frac{23}{50}

If I am incorrect in my reasoning, please let me know so that I can plan better for my future answers. Have an amazing day.

Svetradugi [14.3K]2 years ago
3 0

Answer:

273/10 - 113/9

Step-by-step explanation:

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3 years ago
Read 2 more answers
Find the cube roots of 27(cos 330° + i sin 330°)
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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]

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This function is symmetrical in x and y, so will be a maximum when x=y. That is, you wish to maximize the function

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This is a quadratic in x² that has zeros at x²=0 and x²=50. It will have a maximum halfway between those zeros, at x²=25. That maximum volume is

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Word problem:
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Answer:

Below.

Step-by-step explanation:

30 kg 500g + 28kg 700g

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Difference

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= 2kg + 500 - 700

= 2kg - 200g

= 1kg 800g.

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3 years ago
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