<p>Hey everyone..
I will post some problems that I got a different solution for than the McGraw Hill's SAT Math level 1 book, and that the explanation of the book did not convince me.</p>
<p>Diagnostic test:</p>
<h1>24:</h1>
<p>A bike has wheels with radii of 8 inches. How far
does the bike travel in two complete revolutions of
its wheels?
(A) 8π inches
(B) 16π inches
(C) 32π inches
(D) 64π inches
(E) 128π inches</p>
<h1>27:</h1>
<p>What is the equation of the line containing the point
(1, −2) and perpendicular to the line y = −3x + 7?
(A) y = −3x − 2
(B) y = −3x + 1
(C) y = − (1/3) x
(D) y = (1/3)x + 7
(E) y = (1/3x)- (7/3)
Now I am mostly SURE that this has to be a typing mistake, since the only answer close to the correct answer "(1/3)x- (7/3) is (E) "(1/3x)- (7/3)".. but of course "(1/3)x" is NOT EQUAL TO (1/3x).. no?</p>
<h1>42:</h1>
<p>The area of the rhombus TUVW in Figure 8 is
(A) 64
(B) 32
(C) 5.7
(D) 45.3
(E) 64 2
(the rhombus has a side of length 8, that is opposite to an angle of degree 135</p>
<h1>43:</h1>
<p>Which of the following is equal to (sec θ)(cot θ)?
(A) sin θ
(B) cos θ
(C) sec θ
(D) csc θ
(E) cot θ
(this one I quite frankly don't get at all cuz I didn't take it before, and again the book doesn't provide enough explanation, so )
It would be an amazing plus point if anyone actually HAS THE BOOK.. but anyway.. any help would be more than appreciated :D</p>
<p>24) A wheel travels 1 revolution when it traces out the circumference of the wheel. So 2 revolutions is 2(16pi) = 32pi, choice (C).</p>
<p>27) The given line has slope -3, so a line peroendicular to this line has slope m = 1/3. We can now write an equation of the line in point-slope form:</p>
<p>y+2=(1/3)(x-1) = (1/3)x - (1/3)
So y = (1/3)x - (1/3) - 2 = (1/3)x - 7/3</p>
<p>This is choice (E).</p>
<p>Note: If you don’t know how to do this problem you can eliminate choices by plugging in the point.</p>
<p>42) The area of a parallelogram is absin C, where C is included between a and b. Since all sides of a rhombus are congruent, the area is (8)(8)sin(135) ~ 45.254834. this rounds to 45.3, choice (D).</p>
<p>43) sec θ = 1/cos θ, and cot θ = (cos θ)/(sin θ).</p>
<p>So sec θ cot θ = (1/cos θ)(cos θ)/(sin θ) = 1/sin θ = csc θ, choice (D).</p>
<p>24) But don’t 2 wheels travel twice the distance of one wheel? ok, it may sound stupid, but this is my reasoning… Anyway… That’s what confused me, but I know drew it and sorta understood.</p>
<p>27) as you can see, this was my answer but not the one written in the book… which was “(1/3x)- (7/3)”… </p>
<p>42) I totally forgot that a rhombus had all sides equal… and the figure didn’t have equal sides and didn’t have “Not drawn to scale” written under it so…</p>
<p>43) The book’s explanation was that too… I guess it’s a rule I have to memorize then. I thought there was reasoning behind it (which of course I’m sure there is but I suppose not one that would help me solve or remember… )</p>
<p>Anyway… thanks!! :D</p>
<p>Here’s another Math question that isn’t SAT though:</p>
<p>Given a polynomial and one of it’s factors, find the remaining factors of the polynomial: </p>
<p>16x^5-32x^4-81x+162; the factor: 2x-3</p>
<p>Using synthetic division I got:
(16/3)x^4 - (8/3)x^3 -4x^2- 6x- 36</p>
<p>I put it all equal to zero and tried to factor, but I am stuck!!</p>
<p>Does this need a scientific calculator?! Cuz I lost mine 
Does it have imaginary numbers cuz this is why I am stuck on the normal calculator maybe?!
If no, can you give me a hint?</p>
<p>Thank you XD</p>
<p>16x^5-32x^4-81x+162
= 16x^4(x-2)-81(x-2)
= (x-2)(16x^4-81)
= (x-2)(4x^2-9)(4x^+9)
= (x-2)(2x+3)(2x-3)(4x^2+9)</p>
<p>I haven’t done synthetic division for this one because I could obviously pick out the common terms.</p>
<p>So that gives real roots 2, 3/2, -3/2 and imaginary roots 3i/2, -3i/2</p>
<p>^ ohhh… was that THAT obvious!!! wooow…
um… what about synthetic? </p>
<p>How do you easily do exercises like these?!</p>
<p>In response to your questions:</p>
<p>24) But don’t 2 wheels travel twice the distance of one wheel? ok, it may sound stupid, but this is my reasoning… Anyway… That’s what confused me, but I know drew it and sorta understood.</p>
<p>The question isn’t asking how far the wheels travelled - it’s asking how far the bike travelled. The wheels move at the same time.</p>
<p>27) as you can see, this was my answer but not the one written in the book… which was “(1/3x)- (7/3)”… </p>
<p>Yep - a typo in the book</p>
<p>42) I totally forgot that a rhombus had all sides equal… and the figure didn’t have equal sides and didn’t have “Not drawn to scale” written under it so…</p>
<p>Very bad on the book’s part - an actual SAT question would either have the figure drawn to scale, or would say “Note: Figure not drawn to scale.”</p>
<p>43) The book’s explanation was that too… I guess it’s a rule I have to memorize then. I thought there was reasoning behind it (which of course I’m sure there is but I suppose not one that would help me solve or remember… )</p>
<p>I didn’t use any rules here - just definitions!</p>
<p>By definition, csc θ, sec θ, and cot θ are the reciprocals of sin θ, cos θ, and tan θ, respectively.</p>
<p>^oh okay… thanks a million Dr. Steve… was very helpful…</p>
<p>what about the synthetic one?! I dunno… It seems rather difficult… I mean how am I supposed to know the answer… It takes a lot of trial and error in my opinion… and counts somehow on luck i guess.</p>
<p>And then there is this one…</p>
<p>c^4+2c^3+2c^2-2c-3; factor: c-1
@hopingforbetter, there is no “direct way” for this one…
After synthetic division I got:
c^3+3c^2+5c+3.</p>
<p>I know here I could just see what number when substituted in “c^3+3c^2+5c” would give me -3… as in possible roots… but in this specific problem it maybe easy cuz it is 3 after all so not many possibilities… the answer will be (c+1)(c^2+2c+3)…but again… trail and error! So, each polynomial is its own case then?!</p>
<p>I am not here to say that I am a math genius!:p</p>
<p>As soon as I saw the equation, I could see the first step. I don’t know how that idea came up instantaneously. We do practice a lot. So maybe that’s why…</p>
<p>Since the given factor is x=3/2, synthetic division seem quite problematic. So I chose normal division(I hope you know this). The quotient I got is:
8x^4-4x^3-6x^2-9x-54 (the first two terms have 4x^2 in common and the last two terms have 9 in common)
= 4x^2(2x^2**-x**)-6x^2+9(-x-6)</p>
<p>Do you see those bold terms–the terms which are equal in both the brackets. To make the terms in the brackets equal, the first bracket misses “-6” while the second bracket misses “2x^2”. After that I checked if I can get this equation into this form:
4x^2(2x^2-x-6)+9(2x^2-x-6)
Luckily, I got it right. sigh.
So the equation becomes
(2x^2-x-6)(4x^2+9)
= (x-2)(2x+3)(4x^+9) – You know to factorize 2x^2-x-6, right?</p>
<p>It’s not at all easy to factorize a fourth degree polynomial!! God!!</p>
<p>If you want to use synthetic division, use the rational roots theorem. The theorem says that the possible rational roots of the polynomial are of the form p/q where p divides the constant term and q divides the coefficient of the leading term. This works for polynomial equations with integer coefficients. </p>
<p>By the way, the other method that was used is called “factoring by grouping.”</p>
<p>haahaha… now i know you think I’m stupid… dude… I know all what you just did… My problem is that “<em>Luckily</em>”… I did mention in the very beginning that I had already gotten “8x^4-4x^3-6x^2-9x-54”… but what ****ed me off was that there is no certain way you take to factor AFTERWARDS, u know… you are just left with this equation with no factor to help…
and each equation is different than the other… I guess I just need more practice… </p>
<p>and… um… I think I might know that “normal division” lol… u know since I’m not a complete ■■■■■■ and all?! ;)</p>
<p>I know the factoring by grouping thingy yeah… but my point is HOW to use them efficiently… Like when to use which!? that trial and error thing is driving me crazy…
especially the last one I posted: post #13…</p>
<p>you literally have to try out a huge number of numbers sometimes if you use the method I used in #13… the only method that could be used to solve it…</p>
<p>No. By the rational roots theorem there are only 4 numbers to try. The only possible rational roots of c^3+3c^2+5c+3 are 1, -1, 3 and -3.</p>
<p>Yeah, but my point is if the number is 36 for instance… it’ll take so MUCH TIME…</p>
<p>^He is right. That’s the reason I didn’t use that method.</p>
<p>BTW I never meant to say that you don’t know anything. Sorry dude if sounded like that!
It was 3AM then and my eyes were shutting down beyond my control. So I skipped a few steps assuming that you would know how to do. Those seemingly rhetoric questions were more like a formality!!</p>