HarvardBound1123

The question which said there was an isosceles triangle with sides 5,10,10 asked you to calculate the largest angle right?

So...

c^2 = a^2 + b^2 - 2abcosine(C)

5^2 = 10^2 + 10^2 - 2(10)(10)cosine(c)
25 - 200 = -2(100)cos(c)
-175/-200 = cos(c)
cos(c) = .875
Do inverse cos(.875) you get = 28.955 deg. which is...the SMALLEST angle. so
180 - 28.955 = 151.044
since its isosceles, divide this angle by 2 to get the largest angle
151.0444/2 = 75.522 degress. Which was an answer i belive. thats what i put. im pretty sure this is right.

elaslawek

Harvard- it wasnt the largest angle, it was the angle between both congruent sides, which was the angle between both 10s which was the smallest ones which was 29.

DMOC

It was 29, and I did no math. All the answers were 60 degrees or higher except 29, so I picked 29.

blastie

need an approximate score for my 7 skipped and probably none wrong

mabsjenbu123

790-800 blastie

hummerd

the answer to the: x^2+k was k<0 not <=0 because when you have a multiplicity of 2, the value does not cross the x-axis. Therefore, all values have to be less than 0. If it were 0, the graph would be touching the x-axis, and not crossing.

mabsjenbu123

@hummerd

"The graphical concept of x- and y-intercepts is pretty simple. The x-intercepts are where the graph crosses the x-axis"

x- and y-Intercepts

So shoudn't (0,0) which is the x-intercept count as crossing?

doctor92

I agree with hummerd, but you have to agree that this question is pretty stupid. How does it show understanding of X and Y intercepts?

hummerd

I don't think it counts as crossing. It is an x-intercept, no doubt, but the question specifically asks if it crosses the axis and when there is a multiplicity of a zero, it doesn't count as crossing. The equation, y=x^2 has the zeroes 0 and 0 (multiplicity of 2). In Calculus right now, I am learning how to find the maximum and minimum values of a graph just by looking at the derivative. When the graph of the derivative crosses from positive to negative or from negative to positive, you have a relative extrema. But when the graph only touches a certain solution-(0,0) in our case, it has no relative extrema because it is not considered crossing the x-axis. Hopefully someone who has taken calculus knows what I am talking about and I hope I justified my answer enough.

o1xDanny

I have to agree with ^^. Initially, I was convinced I was wrong by the consensus that it was k <= 0, but I do not believe touching to be an intersection.

"x-Intercepts
The roots of a polynomial function correspond to the x-intercepts of its graph.
• If a root has odd multiplicity, the graph crosses the x-axis at the corresponding x-intercept.
• If a root has even multiplicity, the graph touches but does not cross the x-axis at the corresponding x-intercept."

loserbich

where would 13 skipped leave me... lol

o1xDanny

I`m tired of people asking about their score, it's not that hard to find a conversion chart. Stop being lazy. Sparknotes gives a rough estimate.

SparkNotes: SAT Subject Test: Math Level 2: Math IIC Scoring

loserbich

o1xDanny - um i'm not being lazy, maybe you're the one being lazy... i had already stated on page 7 that i had looked in the collegeboard book for the official conversion chart, but with the variables of difficulty for each test and the resulting curve, i wanted to know what people's estimates would be this time.

stop being rude BRO.

o1xDanny

With 13 skipped, you're bound to have a couple wrong and the curve difficulty only fluctuates your scores plus/minus 10 while one wrong answer could drop you just as much. An answer from anyone here will simply be a restatement of the conversion chart. Hence, use the chart.

With a username like "loserbich" and comments such as "lol you need to shut up" and "lol", I wouldn`t be accusing other people of being rude. And this imaginary post on page 7, I'd like to see it.

loserbich

anyways harvard here i come