1. 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"

(MATH B REGENTS IN LESS THAN 12 HOURS!!!)
for 31 you need bernoulli trials. that's just memorizing and using a formula. you want to take a shot at this one yourself. if you have trouble then i'll do it

2. Originally Posted by Jhevon
$\displaystyle (a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k}$

Calm down! besides, shouldn't you be sleeping now? i told you to get a lot of rest tonight didn't i?
well i went to sleep like at one and woke up now so like 7 hours of sleep isnt baddish?

i guess ill try another regents out, i did pretty okayish on januarary 07, the questions i didnt get i asked and then i understood it. if i have more questions and i hae time ill ask

3. Originally Posted by Jhevon
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. ...
Hello, Jhevon,

the angle A is opposite the smaller side of the two given sides of the triangle. Therefore you can't use your congruent laws(?). Depending on the length of a you can get
no triangle (a is too short)
exactly one triangle (then it is a right triangle)
2 triangles

The last case happens here. So it is obvious that you can't determine the value of angle B and you can't determine the shape of the triangle. Therefore the right choice is (4).

I've attached a sketch to prove my statement.

4. Alrighty, I'm really sorry but I have more questions from the January 2006 Regents Examination that can be found here http://www.nysedregents.org/testing/mathre/b106.pdf

The Answers are found here http://www.nysedregents.org/testing/mathre/bkey106.pdf

On this test, I have problems with numbers 4 (although I guessed it correctly), 9, 15, 16, and 17. Above all, I am completely stymied by number 30 and 31. I would highly appreciate any help that can be given in this direction towards myself. I grant all of you, Good day.

5. Along with Norman I'm facing some difficulties. I was doing the August 2006 regents i got 4 wrong out of the 20 multiple choice so far. The questions i have is with 11, 13 I got right, but i partially guessed/ 17 i was able to break it down altitle bit but i still got the wrong answer/ for now that is it thanks

Question: http://www.nysedregents.org/testing/mathre/b806.pdf

MATH B REGENTS IN 2 AND A HALF HOURS

6. Originally Posted by Norman Smith
Alrighty, I'm really sorry but I have more questions from the January 2006 Regents Examination that can be found here http://www.nysedregents.org/testing/mathre/b106.pdf

The Answers are found here http://www.nysedregents.org/testing/mathre/bkey106.pdf

On this test, I have problems with numbers 4 (although I guessed it correctly), 9, 15, 16, and 17. I would highly appreciate any help that can be given in this direction towards myself. I grant all of you, Good day.
Hi,

to #17:

$\displaystyle \left(\sqrt[3]{m^4}\right) \left(m^{-\frac{1}{2}}\right) = m^{\frac{4}{3}-\frac{1}{2}} = m^{\frac{5}{6}} = \sqrt[6]{m^5}$

7. Originally Posted by Norman Smith
Alrighty, I'm really sorry but I have more questions from the January 2006 Regents Examination that can be found here http://www.nysedregents.org/testing/mathre/b106.pdf

The Answers are found here http://www.nysedregents.org/testing/mathre/bkey106.pdf

On this test, I have problems with numbers 4 (although I guessed it correctly), 9, 15, 16, and 17. Above all, I am completely stymied by number 30. I would highly appreciate any help that can be given in this direction towards myself. I grant all of you, Good day.
Hi,

to #16:

Draw a rough sketch. If the point P lies on a circle line then the sine value is defined by:

$\displaystyle \sin(\theta) = \frac{y_P}{r}$

From the coordinates of the point P you know that r = 5 (Use Pythagorean theorem)
Plug in the values you know and you'll get: $\displaystyle \sin(\theta) = \frac{4}{5}$

8. Originally Posted by Norman Smith
Alrighty, I'm really sorry but I have more questions from the January 2006 Regents Examination that can be found here http://www.nysedregents.org/testing/mathre/b106.pdf

The Answers are found here http://www.nysedregents.org/testing/mathre/bkey106.pdf

On this test, I have problems with numbers 4 (although I guessed it correctly), 9, 15, 16, and 17. Above all, I am completely stymied by number 30. I would highly appreciate any help that can be given in this direction towards myself. I grant all of you, Good day.
Hi,

to #15:

Both hands include 2 hours out of 12. Use the following proportion:

$\displaystyle \frac{\theta}{2 \pi} =\frac{2 h}{12h} \Longrightarrow \theta = \frac{2}{12} \cdot 2 \pi = \frac{1}{3} \pi$ So it's answer (2)

9. Originally Posted by shanegoeswapow
The questions i have is with 11, 13 I got right, but i partially guessed/ 17 i was able to break it down altitle bit but i still got the wrong answer/ for now that is it thanks

Question: http://www.nysedregents.org/testing/mathre/b806.pdf

MATH B REGENTS IN 2 AND A HALF HOURS
Added to the questions i have numbers 27, 28, and 29.

10. 30 MINUTES!

(pretty much how i feel on the inside)

11. Hello, shanegoeswapow!

11. What is the total number of points of intersection
for the graphs of the equations: $\displaystyle y = x^2$ and $\displaystyle y = -x^2$

. . $\displaystyle (1)\;1\qquad(2)\;2\qquad(3)\;3\qquad(4)\;0$
To find the intersections, equate the functions: .$\displaystyle x^2 \:=\:-x^2$

. . and we get: .$\displaystyle 2x^2 \,=\,0\quad\Rightarrow\quad x^2 = 0\quad\Rightarrow\quad x = 0$

Hence, the only point of intersection is: $\displaystyle (0,0).$

. . Answer: .$\displaystyle (1)\;1$

13. If the perimeter of an equilateral triangle is 18,
the length of the altitude of this triangle is:

. . $\displaystyle (1)\;6\qquad (2)\;6\sqrt{3}\qquad(3)\;3\qquad(4)\;3\sqrt{3}$
Since the perimeter is 18, the length of a side is $\displaystyle 6.$
Code:
            *
/|\
/ | \
/  |  \6
/   |h  \
/    |    \
* - - + - - *
3
Does this diagram insprire you to find $\displaystyle h$ ?

17. The expression $\displaystyle \frac{\sin2\theta}{\sin^2\theta}$ is equivalent to:

. . $\displaystyle (1)\;\frac{2}{\sin\theta}\qquad(2)\;2\cos\theta\qq uad(3)\;2\cot\theta\qquad(4)\;2\tan\theta$

You're expected to know the identity: .$\displaystyle \sin(2x) \:=\:2\sin(x)\cos(x)$

We have: .$\displaystyle \frac{\sin2\theta}{\sin^2\theta} \;=\;\frac{2\sin\theta\cos\theta}{\sin^2\theta} \;=\;\frac{2\cos\theta}{\sin\theta} \;=\;2\cot\theta$

. . Answer: .$\displaystyle (3)\;2\cot\theta$

12. Originally Posted by Soroban
Hello, shanegoeswapow!

To find the intersections, equate the functions: .$\displaystyle x^2 \:=\:-x^2$

. . and we get: .$\displaystyle 2x^2 \,=\,0\quad\Rightarrow\quad x^2 = 0\quad\Rightarrow\quad x = 0$

Hence, the only point of intersection is: $\displaystyle (0,0).$

. . Answer: .$\displaystyle (1)\;1$

Since the perimeter is 18, the length of a side is $\displaystyle 6.$
Code:
            *
/|\
/ | \
/  |  \6
/   |h  \
/    |    \
* - - + - - *
3
Does this diagram insprire you to find $\displaystyle h$ ?

You're expected to know the identity: .$\displaystyle \sin(2x) \:=\:2\sin(x)\cos(x)$

We have: .$\displaystyle \frac{\sin2\theta}{\sin^2\theta} \;=\;\frac{2\sin\theta\cos\theta}{\sin^2\theta} \;=\;\frac{2\cos\theta}{\sin\theta} \;=\;2\cot\theta$

. . Answer: .$\displaystyle (3)\;2\cot\theta$
I know that I can speak for both shanegoeswapow and myself here and tell you that although we don't need your help anymore, we appreciate the help (hoping that he is a kind-hearted soul that we on this message board here know him as) and the amazing times you gave for the both of us. Although this sounds cliche, without you guys, we wouldn't have been able to have these opportunities to possibly do well and instead fail and attend Summer School probably. I may be going on a tirade, but I hope you guys have a super summer!!!

13. Originally Posted by Norman Smith
Alrighty, I'm really sorry but I have more questions from the January 2006 Regents Examination that can be found here http://www.nysedregents.org/testing/mathre/b106.pdf

The Answers are found here http://www.nysedregents.org/testing/mathre/bkey106.pdf

On this test, I have problems with numbers 4 (although I guessed it correctly), 9, 15, 16, and 17. Above all, I am completely stymied by number 30 and 31. I would highly appreciate any help that can be given in this direction towards myself. I grant all of you, Good day.
this is too late i'm sure, but this is for #9, i don't think anyone did that and i'm bored

Given $\displaystyle \sin 2 \theta = \frac { \sqrt {3}}{2}$

$\displaystyle ( \cos \theta + \sin \theta )^2 = \cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta$

$\displaystyle = \cos^2 \theta + \sin^2 \theta + \sin 2 \theta$ ........since $\displaystyle \sin 2 \theta = 2 \sin \theta \cos \theta$

$\displaystyle = 1 + \frac { \sqrt {3}}{2}$

14. Originally Posted by shanegoeswapow
Added to the questions i have (August 2006) numbers 27...
simialar triangles have all three sides and angles congruent. we will show each of the angles in one triangle is congruent to one of the angles in the other triangle, this will prove they are congruent

$\displaystyle \angle AEB \simeq \angle ABC = 90^{ \circ}$ since:

for $\displaystyle \angle AEB$, (theorem) angles in a semicircle subtended by the diameter are $\displaystyle 90^{ \circ}$

and for $\displaystyle \angle ABC$, (theorem) the angle formed by the radius of a circle and a tangent to the circle at the point of tangency is $\displaystyle 90^{ \circ}$

Also, $\displaystyle \angle CAB \simeq \angle ABE$ since they are alternate interior angles

Consequently, $\displaystyle \angle ACB \simeq \angle BAE$ since the angles of a triangle add up to $\displaystyle 180^{ \circ}$. since we showed that the two triangles have two angles congruent, the third angle must be congruent as well, since the third angle will be 180 - the other two angles, and we would get the same value for each triangle

Thus the two triangles are similar

15. Originally Posted by shanegoeswapow
Added to the questions i have (August 2006) numbers 28...
The area of the Triangle is given by:

$\displaystyle A = \frac {1}{2}(AB)(AC) \sin A$

So, originally, the area was:

$\displaystyle A_1 = \frac {1}{2}(20)(18) \sin 36.87^{ \circ} \approx 108$

So we want to know how much does $\displaystyle AB$ have to be decreased by so that when $\displaystyle AC$ is changed to $\displaystyle 22.5$ the area remains the same. so we need to solve:

$\displaystyle \frac {1}{2}(22.5)(AB) \sin 36.87^{ \circ} = 108$

$\displaystyle \Rightarrow AB = \frac {108}{\frac {22.5}{2} \sin 36.87^{ \circ}} \approx 16$

So $\displaystyle AB$ must be decreased by 2 feet

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