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3 years ago

How can you make your answer (41-13i)/(25) into a+bi form?

Mathematics

1 answer:

KengaRu [80]3 years ago

5 0

Hello, do you know that for any a, b and c real numbers with c different from 0 the following is correct?

$\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$

Then,

$\dfrac{41-13i}{25}=\dfrac{41}{25}-\dfrac{13}{25}i$

So,

$a=\dfrac{41}{25}\\\\b=\dfrac{-13}{25}$

Thank you.

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In a recent semester at a local university, 490 students enrolled in both General Chemistry and Calculus I. Of these students, 6

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3 years ago

Read 2 more answers

The amount of money Chaz earned for walking dogs is given in the table. Can the relationship be described by a constant rate? Ex

mojhsa [17]

Answer:

Yes, the relationship can be described by a constant rate of $18.75 per dog

Step-by-step explanation:

see the attached figure to better understand the problem

Let

x ----> the number of dogs

y ---> the amount of money earned

we have the points

$(6,112,50), (8,150), (11,206.25)$

step 1

Find the slope with the first and second point

$(6,112,50), (8,150)$

$m=(150-112.50)/(8-6)=18.75$

step 2

Find the slope with the first and third point

$(6,112,50), (11,206.25)$

$m=(206.25-112.50)/(11-6)=18.75$

Compare the slopes

The slopes are the same

That means, that the three points lies on the same line

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2 years ago

Topic: The Quadratic Formula

Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

$x= \frac{-b +- \sqrt{b^2-4ac} }{2a}$

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² = x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

$x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\= -4 or 1$

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

$x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\$

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

$x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2$

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

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3 years ago