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3 years ago
Hi slimesWelcome to the forum.  See if this helps...
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3 years ago
Edited: 3 years ago, slimes
Hey bio man thanks for the reply, but I think you're incorrect. The answer to 6. is 12 and the answer to 7. is 1/4.  Another thing to add, one of the questions that I got was lim t -> 3 6+4t/t^2 + 1. I factored the top part which is 2(2t + 3), denominator isn't factorable, plugged in 3 into x both in numerator and denominator, and got 1 4/5. As I checked the solution, apparently you aren't supposed to factor the numerator, just plug in 3, and your answer would be -3/5. Do you know the reason for this?
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3 years ago
Hey bio man thanks for the reply, but I think you're incorrect. The answer to 6. is 12 and the answer to 7. is 1/4. I disagree. Where did you come up with these conclusions? Another thing to add, one of the questions that I got was lim t -> 3 6+4t/t^2 + 1. I factored the top part which is 2(2t + 3), denominator isn't factorable, plugged in 3 into x both in numerator and denominator, and got 1 4/5. As I checked the solution, apparently you aren't supposed to factor the numerator, just plug in 3, and your answer would be -3/5. Do you know the reason for this? Which question is this?
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3 years ago
Edited: 3 years ago, slimes
1. For class we're supposed to look up practice worksheets with solutions on limit questions, so I found this. https://tutorial.math.lamar.edu/problems/CalcI/ComputingLimits.aspx the solutions are given for each of the questions, I'm not even sure if ALL of them are even correct. 2. And the question I'm talking about is question 2 of that website.
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3 years ago
You wrote them wrong!!! Make sure you type them in correctly - you may take screen shots next time. That being said, do you still need help with the solutions? 2. And the question I'm talking about is question 2 of that website. 
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3 years ago
Edited: 3 years ago, slimes
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Ah! Sorry, I will definitely be taking screenshot's next time. And yes please.
I have another question from one of the questions of the website. For question 4, if you look at the solution. I kinda forgot how this works again but how did the denominator, 8 - z go from that to -( z - 8)?. Even though 8 - z is similar to (z - 8) of the numerator, they can't cancel each other out because they aren't exactly the same.
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wrote...
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3 years ago
I have another question from one of the questions of the website. For question 4, if you look at the solution. I kinda forgot how this works again but how did the denominator, 8 - z go from that to -( z - 8)?. Even though 8 - z is similar to (z - 8) of the numerator, they can't cancel each other out because they aren't exactly the same. For this, you need to start a new topic. A forum rule of ours is one question per thread. In the meantime, I'm working on the originals.
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3 years ago
Edited: 3 years ago, slimes
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Will do that, thanks for letting me know.
Looking at the solutions, I'm confused with 7, what did you do first to get each answer? That would be very helpful. Not getting the same answer!
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3 years ago
Edited: 3 years ago, bio_man
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I used a technique known as "multiplying by the conjugate". It is an algebraic technique that shuffles terms around, making the new expression more suitable to work with.
Say you have the expression (3x + 2) at the numerator, by multiplying the top and bottom by its conjugate, 3x - 2, it alters the expression while keeping its meaning mathematically the same.
A demonstration of this technique is shown in the video:
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wrote...
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3 years ago
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oh wow thanks! And so, you would always do that when you have square root type equations, is that correct?
And would you always have to take the new numerator (for ex, z - 4) and use both denominators, z -4 and √2 +2, like what you did after getting z - 4?
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3 years ago
oh wow thanks! And so, you would always do that when you have square root type equations, is that correct? I won't say "always", but it's a good technique whenever you're stuck -- especially with square roots. And would you always have to take the new numerator (for ex, z - 4) and use both denominators, z -4 and √2 +2, like what you did after getting z - 4? No, as you see in example 9, I multiplied the top and bottom by the conjugate of the denominator.
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3 years ago
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With questions like 7 or 8, that technique would definitely work.
For 7. I used FOIL to get z - 4. How did you do that with (√2x + 22 - 4)(√2x + 22 + 4)? Not very good with factoring and working with square root equations unfortunately.
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