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Serggg [28]
3 years ago
10

What is 20 dg to milligrams

Mathematics
1 answer:
Alla [95]3 years ago
6 0
1 dg = 10 000 milligrams
20 dg = 200 000 milligrams
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Can a denominator be negative in rational number?
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Answer: Rational Numbers are the Numbers where if p/q, q is not equal to 0 and p is an integer. So according to this questionm denominators can’t have the negative rational numbers. If it has, we need to convert it to standard form, eg:- p/-q = -p/q

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the number of grams a woman should eat in a day to maintain her weight is proportional to her weight. a woman weighing 102 pound
vitfil [10]
Given:
grams of fat : 34 grams TO weight of woman : 102 pounds
grams of fat : ? TO weight of woman : 180 pounds

This is a proportion problem: 34 grams to 102 pounds.

We first have to convert a unit of measure to another to maintain uniformity of measure. let us convert pounds to grams:

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180 pounds * 453.592 grams / pound = 81,646.560 grams

34 grams to 46,266.384 grams = x grams to 81,646.560 grams

proportion: a:b = c:d where ad = bc

34 grams * 81,646.560 grams = x * 46,266.384 grams
2,775,983.04 grams² = x * 46,266.384 grams
2,775,983.04 grams² ÷ 46,266.384 grams = x
60 grams = x 

A woman weighing 180 pounds should eat 60 grams of fat to maintain her weight.
7 0
3 years ago
3.26 round to the nearest tenth
zaharov [31]
It's 3.3 because the 2 is in the tenths place and you look at 6 and say is it over 5? Yes it is so the next number after 2 is 3 so your answer is 3.3
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Find four distinct complex numbers (which are neither purely imaginary nor purely real) such that each has an absolute value of
Luda [366]

Answer:

  • 0.5 + 2.985i
  • 1 + 2.828i
  • 1.5 + 2.598i
  • 2 + 2.236i

Explanation:

Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.

Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.

The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:

  • |C| = \sqrt{a^2+b^2}

Then, the work consists in finding pairs (a,b) for which:

  • \sqrt{a^2+b^2}=3

You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}

I will use b =0.5, b = 1, b = 1.5, b = 2

b=0.5;a=\sqrt{9-0.5^2}=2.958\\ \\b=1;a=\sqrt{9-1^2}=2.828\\ \\b=1.5;a=\sqrt{9-1.5^2}=2.598\\ \\b=2;a=\sqrt{9-2^2}=2.236

Then, four distinct complex numbers that have an absolute value of 3 are:

  • 0.5 + 2.985i
  • 1 + 2.828i
  • 1.5 + 2.598i
  • 2 + 2.236i
4 0
3 years ago
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