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andreyandreev [35.5K]
3 years ago
10

Which of the following best describes the set of all numbers of the form a+bi where a and b are any real numbers and i equals sq

uare root of -1?
A. Rational Exponents

B. complex numbers

C. complex fractions

D. Rational Exponents
Mathematics
1 answer:
Andrei [34K]3 years ago
6 0
<h3>Answer: B) Complex numbers</h3>

Complex numbers are always in the form a+bi with 'a' and 'b' as real numbers.

If b = 0 and 'a' is nonzero, then a+bi = a+0i = a which is strictly a real number

If a = 0 and b is nonzero, then a+bi = 0+bi = bi indicating that the number is now purely imaginary

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Find the volume of the following figure using unit cubes.
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3. Rewrite -2x + 3y = 6 in slope intercept form.​
zaharov [31]

The equation in slope-intercept form is:

y=\frac{2}{3}x+2

Further explanation:

The standard slope-intercept form is:

y=mx+b

Given equation is:

-2x+3y=6

In order to bring the equation in slope-intercept form we have to isolate y on left hand side of equation

So,

<u>Adding 2x on both sides</u>

-2x+3y+2x=6+2x\\3y=2x+6

<u>Dividing both sides by 3</u>

\frac{3y}{3}=\frac{2x+6}{3}\\y=\frac{2x+6}{3}

<u>Simplifying</u>

y=\frac{2x+6}{3}\\y = \frac{2x}{3}+\frac{6}{3}\\y=\frac{2}{3}x+2

The equation in slope-intercept form is:

y=\frac{2}{3}x+2

Keywords: Slope-intercept form, linear equation

Learn more about slope-intercept form at:

  • brainly.com/question/12954015
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4 0
3 years ago
Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal
pickupchik [31]

Answer:

The first and second iteration of Newton's Method are 3 and \frac{11}{6}.

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form f(x) = 0 based on the following formula:

x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}

Where:

x_{i} - i-th Approximation, dimensionless.

x_{i+1} - (i+1)-th Approximation, dimensionless.

f(x_{i}) - Function evaluated at i-th Approximation, dimensionless.

f'(x_{i}) - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be f(x) = x^{2}-8 and f'(x) = 2\cdot x, the resultant expression is:

x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}

First iteration: (x_{1} = 2)

x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}

x_{2} = 2 + \frac{4}{4}

x_{2} = 3

Second iteration: (x_{2} = 3)

x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}

x_{3} = 2 - \frac{1}{6}

x_{3} = \frac{11}{6}

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4 years ago
Draw to explain how these to problems are different 35-15 and 43-26
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Well, 
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