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Y_Kistochka [10]
3 years ago
11

Round 992,449 to the nearest hundred thousand

Mathematics
1 answer:
guapka [62]3 years ago
3 0
1,000,000 is what it would be rounded
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If you have $20 and spend $12 on food, what fraction of your money do you still have?
Bas_tet [7]

Answer:

2/5

Step-by-step explanation:

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3 years ago
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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
Slove for y show all your work 12y + 4 = 8y-12​
Vedmedyk [2.9K]

Answer:

y = -4

Step-by-step explanation:

12y + 4 = 8y-12​

Subtract 8y from each side

12y - 8y +4 = -12

4y +4 = -12

Subtract 4 from each side

4y +4-4 = -12 -4

4y = -16

Divide by 4

4y/4 = -16/4

y = -4

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3 years ago
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I need friends this is my instragram german_escamilla57​
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The ratio of boys to girls in a school is 9:11. If there are 400 pupils in the school,how many boys are there​
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I believe its 180.

I have seen this question before and the amount of girls was 220, so 400-220 is 180

Hope this helps :)
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