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Mariana [72]
3 years ago
11

I need help with 10,11,&12

Mathematics
2 answers:
iVinArrow [24]3 years ago
7 0

10) 2/3 * 7/4 = 14/12 = 7/ 6

11) 7/11 + 8/11 = 15/11 = 1 4/11

12) 38,220 ÷ 15 = 2,548

fgiga [73]3 years ago
4 0

10) Cross simplify:

2/2 = 1

4/2 = 2

1/3 x 7/2 = 7/6, or 1 1/6

-----------------------------------------------------------------------------------------------------------------

11) They have the same denominators, so just add across

7 + 8 = 15

15/11 = 11/11 + 4/11

11/11 = 1

1 4/11 is your answer

-----------------------------------------------------------------------------------------------------------------

12) 38220/15 = 2548

-----------------------------------------------------------------------------------------------------------------

hope this helps

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Which of the following equations has a solution of x = -5?
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Answer:

C)

Step-by-step explanation:

16x - 7 = 11x - 32

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-4x - 10 = 2x + 20

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How does the graph of g(x)=1/x-5 +2 compare to the graph of the parent function f(x)=1/x?
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6 0
4 years ago
Read 2 more answers
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
4 years ago
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