# Mid-Terms Review

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• January 15th 2008, 06:39 PM
Rocher
Mid-Terms Review
I just have a few questions I don't quite understand. Hope you guys can help me.

Calculate the distance between each of the following pairs of points.

1) I(8,2) and J(-2,-3)

This means find the slope of the two points right? My final answer (finding the slopes) was $\frac{1}{2}$. Is that right? Is it even slope at all or is it something completely different.

2) If the line joining A (a,4) and B (2,3) is parallel to the line joining C (-5,3) and D (1,1), find the value of a.

I am guessing to do this you must find the slope then the y or x intercepts?

3) The express $6\sqrt{2} + \sqrt{32}$ is equivalent to...

It's been so long since I did radicals I have no idea why the teacher would put this on the test... anyways, I'm at a complete blank. From mixed radical to the radical that isn't mixed (Lol, I'm that lost, forgot the name...), it's square the mixed number then add it to the one in the square root right? So $6\sqrt{2}$ will become $\sqrt{38}$. Right? Wrong? If right, then thanks I know how to solve the problem. If wrong, please tell me the right way :P

Thanks for the help guys.
• January 15th 2008, 08:57 PM
mr fantastic
Quote:

Originally Posted by Rocher
I just have a few questions I don't quite understand. Hope you guys can help me.

Calculate the distance between each of the following pairs of points.

1) I(8,2) and J(-2,-3)

This means find the slope of the two points right? Mr F says: Wrong! It means find the distance.

My final answer (finding the slopes) was $\frac{1}{2}$. Is that right? Is it even slope at all or is it something completely different. Mr F says: Does the question mention slope? No. So even though that actually is the correct slope, it's the wrong answer. Does the question mention distance? Yes. The formula is $d = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$.

2) If the line joining A (a,4) and B (2,3) is parallel to the line joining C (-5,3) and D (1,1), find the value of a.

I am guessing to do this you must find the slope then the y or x intercepts? Mr F says: Half right. Find the slope of the line passing through C and D. I know you can find slope because you found the correct slope between the points given in question 1 (pity the question wasn't asking for slope). Now find the slope of the line passing through A and B. It will have a in it. Now put the two slopes equal to each other and solve for a.

3) The express $6\sqrt{2} + \sqrt{32}$ is equivalent to...

It's been so long since I did radicals I have no idea why the teacher would put this on the test... anyways, I'm at a complete blank. From mixed radical to the radical that isn't mixed (Lol, I'm that lost, forgot the name...), it's square the mixed number then add it to the one in the square root right? So $6\sqrt{2}$ will become \sqrt{38}[/tex]. Right? Wrong? Mr F says: Wrong for two reasons: 1. It becomes $\sqrt{36 \times 2} = \sqrt{72}$. NOT $\sqrt{36 + 2} = \sqrt{38}$!!! 2. They probably want you to simplify the expression, in which case you need to reduce $\sqrt{32}$.

If right, then thanks I know how to solve the problem. If wrong, please tell me the right way :P Mr F says: $\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$. So $6\sqrt{2} + \sqrt{32}$ is equivalent to $6\sqrt{2} + 4\sqrt{2} = 10 \sqrt{2}$.

Thanks for the help guys. Mr F says: Just hit the thanks button, dude.

..
• January 16th 2008, 11:21 AM
Rocher
1) My final answer is $\sqrt{-15}$. Correct?

2) My final answer is -1. Correct?
• January 16th 2008, 11:42 AM
colby2152
Quote:

Originally Posted by Rocher
1) My final answer is $\sqrt{-15}$. Correct?

2) My final answer is -1. Correct?

1) No, think... how could you possibly get a negative result from: $a^2 + b^2$. You can't... you did not square the differences.

2) Yes, you correctly matched the slopes.
• January 16th 2008, 11:57 AM
Rocher
Quote:

Originally Posted by colby2152
1) No, think... how could you possibly get a negative result from: $a^2 + b^2$. You can't... you did not square the differences.

It's I(8,2) and J (-2,-3). There are some negatives in there, and so you can... but anyways, I show what I did...

$
d=\sqrt{(-3-2)+(-2-8)}
=\sqrt{(-5)+(-10)}
=\sqrt{-15}
$
• January 16th 2008, 12:05 PM
mathgeek777
Quote:

Originally Posted by Rocher
It's I(8,2) and J (-2,-3). There are some negatives in there, and so you can... but anyways, I show what I did...

$
d=\sqrt{(-3-2)^2+(-2-8)^2}
=\sqrt{(-5)^2+(-10)^2}
=\sqrt{25 + 100}
=\sqrt{125}
=5\sqrt{5}
$

I fixed it for ya ;) Checked the above quote, you forgot to square each term. That's why you ended up with negative numbers.
• January 16th 2008, 12:30 PM
colby2152
Quote:

Originally Posted by Rocher
It's I(8,2) and J (-2,-3). There are some negatives in there, and so you can... but anyways, I show what I did...

$
d=\sqrt{(-3-2)+(-2-8)}
=\sqrt{(-5)+(-10)}
=\sqrt{-15}
$

You still haven't squared the differences.
• January 17th 2008, 06:40 PM
Rocher
Thanks, but $(-5)^2+(-10)^2$ equals out to a negative number on my calculator o.O Also, I have two more questions I hope you can help me with...

1) $32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}$
How in the world are you supposed to do fraction exponents!?!

2) $(\frac{27x^{-5}}{125y^2})^{-\frac{2}{3}}$
Again, how are you supposed to do fraction exponents?
• January 17th 2008, 06:45 PM
Jhevon
Quote:

Originally Posted by Rocher
Thanks, but $(-5)^2+(-10)^2$ equals out to a negative number on my calculator o.O Also, I have three more questions I hope you can help me with...

hehe, i guess this scores one for the people who say calculators should be banned from calc classes...

you entered the numbers wrong or something. (-5)^2 > 0 and (-10)^2 > 0, there is no way their sum is negative

if your calculator has gone haywire, or you're not sure you're entering the numbers correctly, note that (-5)^2 + (-10)^2 = 5^2 + 10^2 = 25 + 100 = 125
• January 17th 2008, 06:49 PM
Jhevon
Quote:

Originally Posted by Rocher
1) $32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}$
How in the world are you supposed to do fraction exponents!?!

note that $\frac {x^a}{x^b} = x^{a - b}$

you think you can try it now?

Quote:

2) $(\frac{27x^{-5}}{125y^2})^{-\frac{2}{3}}$
Again, how are you supposed to do fraction exponents?
just treat fractional exponents as you would regular numbers

here you need the rule: $\left( x^a \right)^b = x^{ab}$ in addition to the rule i gave you earlier

you think you can take it from here?

you really should post new questions in a new thread

for the meaning of fractional exponents, see post #3 here. (i don't suppose you have trouble with negative exponents?)
• January 17th 2008, 10:00 PM
mr fantastic
Quote:

Originally Posted by Jhevon
hehe, i guess this scores one for the people who say calculators should be banned from calc classes...

you entered the numbers wrong or something. (-5)^2 > 0 and (-10)^2 > 0, there is no way their sum is negative

if your calculator has gone haywire, or you're not sure you're entering the numbers correctly, note that (-5)^2 + (-10)^2 = 5^2 + 10^2 = 25 + 100 = 125

It's a sure bet that the expression was keyed in as -5^2 + -10^2, that is, no brackets. The sad part is the uncritical acceptance of what the calculator spits out ....
• January 18th 2008, 12:00 PM
Rocher
Quote:

Originally Posted by Jhevon
note that $\frac {x^a}{x^b} = x^{a - b}$

you think you can try it now?

just treat fractional exponents as you would regular numbers

here you need the rule: $\left( x^a \right)^b = x^{ab}$ in addition to the rule i gave you earlier

you think you can take it from here?

you really should post new questions in a new thread

for the meaning of fractional exponents, see post #3 here. (i don't suppose you have trouble with negative exponents?)

No lol, I still have no idea what to do... There are three exponents on the side, so how are you supposed to add / subtract them?!
• January 18th 2008, 12:32 PM
colby2152
Quote:

Originally Posted by Rocher
Thanks, but $(-5)^2+(-10)^2$ equals out to a negative number on my calculator o.O Also, I have two more questions I hope you can help me with...

1) $32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}$
How in the world are you supposed to do fraction exponents!?!

2) $(\frac{27x^{-5}}{125y^2})^{-\frac{2}{3}}$
Again, how are you supposed to do fraction exponents?

Add exponents of like bases when there bases are multiplied with each other:

$a^{-\frac{2}{3}} \div a^4 = a^{-\frac{2}{3}}a^{-4}\Rightarrow a^{-\frac{2}{3}-\frac{12}{3}}=a^{\frac{-14}{3}}$
• January 18th 2008, 12:46 PM
Rocher
-14? Isn't it supposed to be +10?
• January 18th 2008, 03:03 PM
Rocher
Agh, I give up... can you guys just show me how to do those two step by step? There are two more questions like them and with the two answers I might be able to do the other ones... I just don't get this stuff.
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