in-46-60-write-each-quotient-in-ab-i-form-fracfrac15-frac15-ifrac53-4-i

Question

In $46-60,$ write each quotient in $a+b i$ form.
$$
\frac{\frac{1}{5}-\frac{1}{5} i}{\frac{5}{3}-4 i}
$$

Heather Zimmers · Accepted Answer

all right, so we want to change this quotient into standard form. And it makes sense to me looking at this numerator that we should start by factoring 1/5 out of the numerator and that will just make our lives so much easier as we move forward. So we have 1/5 times one minus I over three, minus four I. The next step is to multiply by a special form of one. And that would be the congregate of the denominator. So that would be three plus four I over three plus four I. That&#x27;s because when we multiply congregates, we&#x27;re just going to get a real number. So what do we have now? We have our 1/5 and we&#x27;re going to multiply. The numerator is using the foil method and the same with the denominators. So multiplying enumerators. The first we get three for the outside we get for I for the inside we get negative three I and for the last we get negative. Four. I squared and on the bottom, multiplying. The first gives us nine. The insides and outsides cancel. That&#x27;s what happens with congregates and the last Give us negative 16. I squared, but we want to do now is replaced the I squared with negative one and combine our like terms. So we have 1/5 times three plus I minus four times negative. One over nine minus 16 times. Negative one. Now, let&#x27;s simplify that. So we have 1/5 times three plus I plus four over nine plus 16. Okay, so now we have 1/5 times seven plus I over 25. Now we can multiply the 1/5 in the 25 we have 11 25th. And we can break this up into the real part and the imaginary part. So the real part will be 7/1 25 plus the imaginary part 11 25th I

Heather Zimmers · Answer

first, I&#x27;m going to change the quotient in default into the following form five minus 15 I over three minus I. And then I think it might be helpful to factor of the five Sweet five times the quantity one minus three I over three minus I. Now, if we want to change this to standard form, we&#x27;re going to need to multiply by the congregate of the denominator over itself. This is our special form of one. And so let&#x27;s multiply the numerator. So we have five times the quantity we&#x27;re going to use the foil technique. Multiply the first and the outsides and the insides and the last. So we get three plus I minus nine I minus three. I squared and on the bottom we&#x27;re gonna multiply using the foil technique. But remember, because we&#x27;re multiplying con jackets, the insides and outsides will cancel. So we&#x27;ll have the 1st 9 and the last minus. I squared. So now what do we have so we can simplify our numerator? We can combine the like terms I minus nine I that would be minus eight I and we can change. I square to negative one. So we have five times three minus eight I minus three times negative. One over on the bottom. Let&#x27;s also change the I scored a negative one, so we have over nine minus negative one. Okay. Coming back over to the left side. We can continue to simplify because negative three times negative one is plus three. And we can add that to the three. So we have five times the quantity six minus eight by on the top and on the bottom nine minus negative one is going to be 10. So it was kind of handy that we left the five on the outside. That&#x27;s going to reduce with the 10 and notice that we can also factor a two out of the top. So we have five times two times three minus four I over 10 so we can cancel the 10 from the top in the bottom, and we&#x27;re left with three minus four. I

Jeremy Strong · Answer

in this problem we were asked to do about the complex number five. Modest, too. About the conflicts Number five plus Chua. So you begin. What we&#x27;re going to need to do is we&#x27;re going to need to multiply the top and the bottom or the numerator denominator of this fraction by the complex con jacket of the denominator, which is going to be five Montecito. Remember, we always flip the imaginary component of the complex. Number two take the contract, so we&#x27;re going to need to foil both the numerator and denominator here. So starting in the noon, where we have five times, Sof, which is 25 our outer is five and modest us. So it&#x27;s going to be modest. 10 are our inner is modest two and five to give us another modest, huh? And then our last, It&#x27;s modest to our times a minus two out, which is gonna give us a plus for ah squared. Now in the denominator, we&#x27;re going to need to do the same thing. So we&#x27;re gonna have five times five again, which is 25. Our outer is modest. You and positive five to give us I&#x27;m honest enough our inner is plus two out Tom&#x27;s plus five, which is gonna give us plus 10. Uh, and then our last storm&#x27;s gonna give us Monas for squared. Now, these square terms are equal to modest one because squared is the square root of negative one squared, which will leave behind the negative one. So in the numerator, we have plus four r squared, so that is actually plus four times minus one, and that is going to turn into a modest four block was in the denominator. We have met negative four r squared. So that&#x27;s a negative four times negative one which is going to leave us with a plus four now moving right along and the Newman in the numerator. We have 25 minus four, which is 21 and then we have modest in a modest tonight, which is minus 20. Uh, so that is our simplified numerator. Now, in the denominator, we have modest in not plus a Jenna, and that cancels out to give us zero, which doesn&#x27;t matter. So now we have the 25 plus four, which is 29 now to get our answer in the final form, which is the a plus b ah form that is required for answer. To be correct, we&#x27;re going to need to split this fraction into two separate parts. So we have 21 of Mona&#x27;s 20. Ah, divided by 29 and this may be split apart into 21 over 29. Modest 20 over 29. Uh, in that is the final and correct.

Jeremy Strong · Answer

and this problem. We&#x27;re asked to divide the complex number 10 plus five out by the complex number one plus two on. So what we&#x27;re going to need to do to simplify the fraction is we&#x27;re going to need to multiply the numerator and the denominator by the conjugal of the denominator, which in this case, is going to be one mon ist too, huh? One modest, too. All right. Now we&#x27;re going to need to foil both the numerator and the denominator. Said to begin, we&#x27;re gonna have 10 Tom&#x27;s one, which is 10 and then our outer are modest. Try and positive tend to give us monos 20 on our inner components air plus father and plus one to give us plus five. Uh, and then our last components plus fathered Tom&#x27;s modest to ah is modest chin squared. All right. And that is our numerator Now for the denominator. Want to need to foil yet again. So one times one is one and the outer components are going with modest to Ah, our inner components were gonna give us plus two. Ah, and then our last components are gonna give us Monas for ah squared. OK, so Let&#x27;s take a minute and let&#x27;s take a look at these ice squares. So I squared is negative one cause the square root of negative one squared is negative one so we can rewrite each of these are squared as minus one. So we had negative 10 Tom&#x27;s negative one. So this is going to go away and we can write Plus 10 and lock was down here. We had negative four squared and we can rewrite that as plus for now. Okay, so in the numerator, we need to combine our lock terms. So our real component is 10 plus chin, which is gonna give us 20. And then our magic opponent is modest. 20 odd plus fava, which is honest 15 now and the denominator, we have one plus four. So it&#x27;s gonna give us five and then negative to AA plus two is gonna give us zero. So right now we simply have 20 minus 59 over five. But it is not in the simplified a plus b ah form. And it has to be for our answer to be correct. So we have 20 monness. 15 are divided by five, which is equal to four monos three. Ah, and that is the final and

Heather Zimmers · Answer

all right, so we want to rewrite this quotient in standard form. And the first thing I would see is that we could factor of five out of the top five times a quantity to minus I, and we can factor two out of the bottom two times a quantity one plus three I and that will make the next steps a little bit easier. So the next thing we want to do is multiplied by a special form of one. The conjugated of one plus three. I that would be one minus three I because that&#x27;s going to simplify the denominator quite a bit. So here&#x27;s what we have. Now we have the five over to We&#x27;ll just leave that out and then we&#x27;re going to multiply using the foil technique first outside, inside last, and we end up with two minus six. I minus I plus three. I squared over same thing on the bottom. First outside, inside last. But because we&#x27;re multiplying congregates the outsides and insides will cancel. So first will be one and then last will be minus nine. I squared. Okay. There&#x27;s a lot we can do to simplify. First of all, we have negative. Six i and negative one. I will add those together and get negative. Seven I and every time we have a nice squared will replace it with negative one. So here&#x27;s what we have now. Five halves times Tu minus seven I plus three times negative. One over one, minus nine times negative one. Let&#x27;s simplify that a little more. We have five halves times Tu minus seven. I minus three over one plus nine and then let&#x27;s combine like terms two and negative three. So we have five halves times negative. One minus seven I over 10. Okay, let&#x27;s go ahead and reduce the five and 10. That gives us a two on the bottom in a one on the top. So what we have at this point is negative one minus seven I over four. And then if we want to break that down into two separate parts, the rial part would be negative. 1/4 and the imaginary part would be plus 7/4. I

Jeremy Strong · Answer

R and this problem we were asked about the complex number 1/7 plus by the complex number five divided by. So the simplest way to simplify this is to acknowledge that we have a fraction and the denominator of the entire fraction. So to undo this fraction will need to multiple both the numerator and the denominator by ah, and that will cancel the in the denominator. So moving right along, we have I toms 17 plus I, which is gonna give us ah, divided by seven plus squared and the numerator and then five plus on Tom&#x27;s I the eyes they&#x27;re gonna cancel. It&#x27;s just gonna leave us with a five. All right now, moving right along in the numerator we have won over seven I and then squared is the same thing as a negative one because it&#x27;s the square root of negative one squared, which just leaves the negative one. So instead of squared, we have one son with a modest one and then in the denominator wish tohave five. Now we need our final answer to be in the form a plus B I for answer to be correct. So is the final step we have one. Oh, we need to be careful here because we actually have written the imaginary component first. Um, but technically speaking, the rial component, the A component is what needs to come first. So we have a modest one divided by five star riel. Component is modest. 1/5 and then our imaginary component is 1/7. Ah, divided by five. But this could be further simplified to give us a final answer of negative 1/5. All right. And then we&#x27;re going to add one over 35 I and we do this by multiplying the top and the bottom of this fraction by seven. And it cancels this seven, and it makes this denominator 35. And our final answer, therefore, is negative. 1/5 plus one over 35.

Problem

In $46-60,$ write each quotient in $a+b i$ form. …

Problem 57 Hard Difficulty

In $46-60,$ write each quotient in $a+b i$ form.
$$
\frac{\frac{1}{5}-\frac{1}{5} i}{\frac{5}{3}-4 i}
$$

Answer

$=\frac{1}{5} \cdot \frac{7+1}{25}=\frac{7}{125}+\frac{1}{125} i$

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$=\frac{1}{5} \cdot \frac{7+1}{25}=\frac{7}{125}+\frac{1}{125} i$

Algebra 2 and Trigonometry

Grace H.

Heather Z.

Kayleah T.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

Ann Xavier GantertAlgebra 2 and Trigonometry

Grace H.

Heather Z.

Kayleah T.

Maria G.

Ann Xavier Gantert

Algebra 2 and Trigonometry