August 2004 Help

**Norman Smith** · June 13th 2007, 06:37 PM

Hello everyone. My name is Norman and although I'm asking specifically for help on one question now, I may have others later, so please don't get too angry at me if I start asking millions of questions. For now though, I have absolutely no idea in the name of heaven's sake how to do number 15 and would hope that someone on this nice little message board could direct their attention to me at some point before tomorrow at 11:30 a.m. EST. Thanks and see you all around.

**Jhevon** · June 13th 2007, 06:49 PM

Originally Posted by Norman Smith

Hello everyone. My name is Norman and although I'm asking specifically for help on one question now, I may have others later, so please don't get too pissed at me if I start asking millions of questions. For now though, I have absolutely no idea in the name of heaven's sake how to do number 15 and would hope that someone on this nice little message board could direct their attention to me at some point before tomorrow at 11:30 a.m. EST. Thanks and see you all around.

number 15 of what? you haven't posted the question

**Norman Smith** · June 13th 2007, 06:53 PM

Originally Posted by Jhevon

number 15 of what? you haven't posted the question

My deepest apologies Jhevon. I should've known better than to come to conclusions, and shall now post the question. http://www.nysedregents.org/testing/...mathbaug04.pdf

The answer lies in http://www.nysedregents.org/testing/...tbkeyaug04.pdf

And it is number 15. Thanks for any help you or anyone else can offer.

**Jhevon** · June 13th 2007, 07:02 PM

Originally Posted by Norman Smith

My deepest apologies Jhevon. I should've known better than to come to conclusions, and shall now post the question. http://www.nysedregents.org/testing/...mathbaug04.pdf

The answer lies in http://www.nysedregents.org/testing/...tbkeyaug04.pdf

And it is number 15. Thanks for any help you or anyone else can offer.

$\frac { \left( b^{2n + 1} \right)^3 }{b^n \cdot b^{4n + 3}}$

$= \frac {b^{6n + 3}}{b^n \cdot b^{4n + 3}}$ .......since when we raised a number with a power to a power, we multiply the powers

$= \frac {b^{6n + 3}}{b^{5n + 3}}$ ........since when we multiply numbers of the same base, we add the powers

$= b^{6n + 3 - 5n - 3}$ ...........since when we divide numbers of the same base, we subtract the power of the base in the denominator from the power of the base in the numerator

$= b^{n}$ which is choice 2

So to summarize everything you need to know for this problem:

$\left( x^a \right)^b = x^{ab}$

$x^a \cdot x^b = x^{a + b}$

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

**shanegoeswapow** · June 13th 2007, 07:04 PM

I have a question about the January 07 Regents, number 5
the question: http://www.nysedregents.org/testing/mathre/b-107.pdf
answer: http://www.nysedregents.org/testing/mathre/bkey-107.pdf

i dont understand how i need to find the power of i when its higher than 4

**shanegoeswapow** · June 13th 2007, 07:28 PM

i also have a quesiton with number 14
and for number 20 i got the right answer but i dont know how, i partially guessed. i found the other angle to be about 46 degrees so i knew that the other angle couldnt be a right angle. how would i find out the other angle for it? by using ambigous case? if so how do you do that?

**Norman Smith** · June 13th 2007, 07:34 PM

Originally Posted by Jhevon

$\frac { \left( b^{2n + 1} \right)^3 }{b^n \cdot b^{4n + 3}}$

$= \frac {b^{6n + 3}}{b^n \cdot b^{4n + 3}}$ .......since when we raised a number with a power to a power, we multiply the powers

$= \frac {b^{6n + 3}}{b^{5n + 3}}$ ........since when we multiply numbers of the same base, we add the powers

$= b^{6n + 3 - 5n - 3}$ ...........since when we divide numbers of the same base, we subtract the power of the base in the denominator from the power of the base in the numerator

$= b^{n}$ which is choice 2

I've never seen a question like that before, but I really thank you for it and see it somewhat more clear now. I appreciate the help, and if you have any more time, is it possible if you could also help me with number 24 of August 2004. I found both of the answers, but don't see how it is an inequality. I also don't get numbers 26, 31 and 32 and would greatly appreciate help you could offer for those two questions as well.

**Jhevon** · June 13th 2007, 07:49 PM

Originally Posted by shanegoeswapow

I have a question about the January 07 Regents, number 5
the question: http://www.nysedregents.org/testing/mathre/b-107.pdf
answer: http://www.nysedregents.org/testing/mathre/bkey-107.pdf

i dont understand how i need to find the power of i when its higher than 4

recall that, by definition: $i = \sqrt {-1}$ and $i^2 = -1$

so for 5.

$i^{25} = i^{24} \cdot i = \left( i^2 \right)^{12} \cdot i = (-1)^{12} \cdot i = 1 \cdot i = i$

**Jhevon** · June 13th 2007, 07:57 PM

and for number 20 i got the right answer but i dont know how, i partially guessed. i found the other angle to be about 46 degrees so i knew that the other angle couldnt be a right angle. how would i find out the other angle for it? by using ambigous case? if so how do you do that?[/quote]

for 14

recall, if $a$ and
$b$ are the roots of a quadratic equation, then $(x - a)(x - b) = 0$ (Can you tell me why?)

so we are told the roots are $3 + i$ and $3 - i$

so, $(x - (3 + i))(x - (3 - i)) = 0$ .......expand within the brackets

$\Rightarrow (x - 3 - i)(x - 3 + i) = 0$ ....now expand in general

$\Rightarrow x^2 - 3x + xi - 3x + 9 - 3i - xi + 3i - i^2 = 0$ .....remember, $i^2 = -1$

$\Rightarrow x^2 - 6x + 10 = 0$ ........answer

**Jhevon** · June 13th 2007, 08:07 PM

Originally Posted by shanegoeswapow

i also have a quesiton with number 14
and for number 20 i got the right answer but i dont know how, i partially guessed. i found the other angle to be about 46 degrees so i knew that the other angle couldnt be a right angle. how would i find out the other angle for it? by using ambigous case? if so how do you do that?

20 is a strange question. i disagree with the answer given. maybe i'm making an error that someone will soon point out to me. but by the sine rule, angle B is 45.585... which is an acute angle. so i'd say choice (1) is correct. however, they have choice (4) as the right answer. now i'm confused, lol. we'll get back to this later

**Jhevon** · June 13th 2007, 08:20 PM

Originally Posted by Norman Smith

I've never seen a question like that before, but I really thank you for it and see it somewhat more clear now. I appreciate the help, and if you have any more time, is it possible if you could also help me with number 24 of August 2004. I found both of the answers, but don't see how it is an inequality. I also don't get numbers 26, 31 and 32 and would greatly appreciate help you could offer for those two questions as well.

to 24

we make a profit when $P(x)>0$ (that's where the inequality comes from--do you see why it is an inequality?), so we simply must solve for that

So, $-x^2 + 120x - 2000 > 0$

$\Rightarrow (-x + 100)(x - 20) > 0$

$\Rightarrow (-x + 100)> 0 \mbox { or } (x - 20)> 0$

$\Rightarrow x < 100 \mbox { or } x > 20$

But we have to check this! sometimes by solving inequalities, even if we do it correctly, the signs get messed up. check numbers in the different regions separated by the numbers 100 and 20. so you would check say, 19, 25 and 101. you will realize that our inequality holds when x is between 20 and 100. so our intial assertion was right.

Combining the two inequalities we had, we get $20 < x < 100$ which is the answer

**Jhevon** · June 13th 2007, 08:30 PM

Originally Posted by Norman Smith

I've never seen a question like that before, but I really thank you for it and see it somewhat more clear now. I appreciate the help, and if you have any more time, is it possible if you could also help me with number 24 of August 2004. I found both of the answers, but don't see how it is an inequality. I also don't get numbers 26, 31 and 32 and would greatly appreciate help you could offer for those two questions as well.

26 is just asking for the length of the arc HK

recall the formula for length of arc:

$s = \frac { \theta }{360}2 \pi r$

where $s$ is the length of the arc, $\theta$ is the angle that subtends the arc in degrees, and $r$ is the radius.

So, $HK = \frac {9}{360}2 \pi (3954) \approx 621.1$

**shanegoeswapow** · June 13th 2007, 08:36 PM

again for my regents i have questions on both 26 and 31. binomial expansion im still alittle bit fuzzy on. im not sure when to include 0 when im putting the n term or if that is just for the "middle term"

(MATH B REGENTS IN LESS THAN 12 HOURS!!!)

**Jhevon** · June 13th 2007, 08:50 PM

I just remembered that Soroban and I did some problems from Aug 2004 before. including 31. see here for the solution to 31 and others

Originally Posted by Norman Smith

I've never seen a question like that before, but I really thank you for it and see it somewhat more clear now. I appreciate the help, and if you have any more time, is it possible if you could also help me with number 24 of August 2004. I found both of the answers, but don't see how it is an inequality. I also don't get numbers 26, 31 and 32 and would greatly appreciate help you could offer for those two questions as well.

Here's 32

(BY the way, you guys really should attempt the problems on your own before viewing my solutions--it makes a whole lot of difference when you do that, since it gets your mind used to thinking about the problems and forming ideas)

$\frac { \sin^2 \theta}{ 1 + \cos \theta} = 1$ ....multiply both sides by $1 + \cos \theta$

$\Rightarrow \sin^2 \theta = 1 + \cos \theta$

$\Rightarrow - \sin^2 \theta + \cos \theta + 1 = 0$ .......recall $\sin^2 \theta = 1 - \cos^2 \theta$

$\Rightarrow - \left( 1 - \cos^2 \theta \right) + \cos \theta + 1 = 0$

$\Rightarrow \cos^2 \theta + \cos \theta = 0$

$\Rightarrow \cos \theta ( \cos \theta + 1 ) = 0$

$\Rightarrow \cos \theta = 0 \mbox { or } \cos \theta + 1 = 0$

$\Rightarrow \theta = 90^{ \circ}, 270^{ \circ} \mbox { or } \theta = 180^{ \circ}$ for $0^{ \circ} \leq \theta \leq 360^{ \circ}$

However, $180^{ \circ}$ is extraneous (we see that it doesn't work when we plug it into the original equation), so the answer is just $90^{ \circ} \mbox { and } 270^{ \circ}$

**Jhevon** · June 13th 2007, 08:58 PM

Originally Posted by shanegoeswapow

again for my regents i have questions on both 26 and 31. binomial expansion im still alittle bit fuzzy on. im not sure when to include 0 when im putting the n term or if that is just for the "middle term"

$(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k}$

since k starts from 0, the fourth term is when k is 3

so the fourth term is given by:

${5 \choose 3} (2x)^2 (-y)^3 = -40x^2 y^3$

(MATH B REGENTS IN LESS THAN 12 HOURS!!!)

Calm down! besides, shouldn't you be sleeping now? i told you to get a lot of rest tonight didn't i?

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