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lisabon 2012 [21]
3 years ago
15

The conjugate is used in _____ of complex numbers.

Mathematics
2 answers:
creativ13 [48]3 years ago
8 0
A. quadratic i’m pretty sure
Nikolay [14]3 years ago
7 0
Quadratic is the correct answer.
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Work out the area of this circle. Take pie to be 3.142 and give your answer to 2 decimal places.
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Answer:

Work out the area of this circle. Take pie to be 3.142 and give your answer to 2 decimal places.

Step-by-step explanation:

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3 years ago
What is 6 divided by 1/3
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6 / 1/3 = 18

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4 years ago
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Drag the tiles to the correct boxes to complete the pairs.
GenaCL600 [577]

The values have being matched to the variables that they represent here:

  • x-variable: the number of hours Sam works at the airport
  • y variable:  the amount in dollars that Sam has after paying back Daniel
  • 12: amount in dollars that Sam earns per hour.
  • 40: the amount in dollars Sam owes Daniel

<h3>How to solve for the equation.</h3>

The equation of this form y = 12x-40

The variable y has been said to be the amount sam has after Daniel has been paid.

X is the amount that has to be multiplied by what he earns in dollars by the hour.

40 dollars is the amount that he owes which is going to be subtracted from what he earns by the hour.

Read more on functions here:

brainly.com/question/17043948

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8 0
2 years ago
Can someone please help me ?? ​
Marat540 [252]

C, the higher a negative is, the lower its value gets

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3 years ago
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PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS
Reil [10]

Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

7 0
3 years ago
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