@AGoodFloridian thanks…lol 
Yeah, I feel like I know what you’re getting at – the original rectangle has dimensions 6/sqrt(5) and 12/sqrt(5) right? My brain also starts going to sleep around now, and I tend to make silly mistakes either before bed or during exams.
goodluck lol
@MITer94
Just tried your question out.
Not sure if I did it right but is it D?
Work is here
http://imgur dot com/DDSWsjm replace dot with .
There is probably a more elegant solution but this answer took me long enough.
@NotBadNotGood answer is correct. But no trig is required. The segment that you drew with length 1 - (1/2) sqrt(x^2 + 1) also has length x/2, so you can set the two expressions equal.
A slightly easier solution (IMO) is to let the radius of the circle be r, then you have a right triangle with legs 1, 2(1-r), and 2r (where x = 2(1-r)). The circle is tangent to CD at some point M, so the angle formed by D, M, and the center of the circle is a right angle.
where is the question, i couldnt find
If 5 <2x+3<11, what is the possible range of values of -4x-6
hmm? not sure why the answer would be any value greater than -22 and less than -10 not any value greater than -10 or less than -22
When you multiply by -2 the inequalities reverse. So you get (-2)(5)>(-2)(2x+3)>(-2)(11), or equivalently -10>-4x-6>-22. Rewriting the other way, we have -22<-4x-6<-10.
Please help. On the blue book, page 547 #13 and
page 544 #3
Seriously, why not look in https://sat.collegeboard.org/practice/sat-study-guide-owners-area?excmpid=CBP6-ST-2-guide, Test 3, Section 8?
Question: Let $C[x1, x2, …, x_n]$ be the ring of complex polynomials over $n$ variables. Is there always a bijective correspondence between the maximal ideals of the ring and points in n-dimensional complex space?
(Source: A slightly weaker version of HN. Got the easy direction, just wondering how to show things are in the kernel.)
@KinglyBill unfortunately, I don’t think algebra is covered on the SAT.
(abstract algebra, that is)
Can anyone help out with a SAT math question from the new format. It’s in practice test 4 calculator section number 25:
F(x)= 2x^3+6x^2+4x
G(x)= x^2+3x+2
The polynomials f(x) and g(x) are defined above. Which of the following polynomials is divisible by 2x+3
A) h(x)= f(x) + g(x)
B) p(x)= f(x) + 3g(x)
C) r(x)= 2f(x) + 3g(x)
D) s(x)= 3f(x) + 2g(x)
The explanation in the book made sense but I was wondering if anyone knew a simpler way to solve it. Thanks
I would use the Remainder Theorem for this one.
Fact 1: A polynomial is divisible by 2x+3 if and only if it is divisible by x+3/2 (because 2x+3 = 2(x+3/2) ).
Fact 2 (Remainder Theorem) p(x) is divisible by x - r if and only if p® = 0.
So we plug -3/2 (or -1.5) in for x in the two functions (using our calculator).
f(-1.5) = .75 and g(-1.5) = -.25
Now just glance at the answer choices and note that .75 + 3(-.25) = 0. So the answer is B.
Notes: (1) Of course if you don’t see that B gives you 0 right away, just check each answer choice, starting with choice B or C.
(2) You can skip Fact 1 completely, and simply observe that 2x+3 = 0 when x= -3/2 = -1.5 to get the value that you need to plug in.
(3) My guess (with evidence) is that the Remainder Theorem will be very useful for the new SAT. I recommend that any student going for an 800 memorize it and practice using it.
Thank you very much! @DrSteve
I used another approach to solving the problem. I subbed in k+5 for x and 72 for f(x) and got the following equation 72=k^2 + 11k +30 and found that 0 = (k +14)(k-3). Therefore the positive value of k is 3. Is that an appropriate alternative approach?
@youcanleadahorse which question is this?
It’s an old one. If x is positive and f(x) = x(x+1) and f(k+5) = 72. What is k? That’s paraphrased but its basically that.
@youcanleadahorse yes that solution is fine. Slightly easier IMO is to solve f(x) = 72, and since 8*9 = 72, you know that f(8) = 72, and k = 3 is a solution. (the other solution is x = -9 but it is negative).
k stands for the kangaroo constant.
on the actual SAT would anyone suggest doing the “free response” grid questions on both math sections first? my brain is always fried by the time i get to the end…