Datasets:
NationalTutoringObservatory
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MathEd-PII

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volunteer
did you solve in all four cases?
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student
wait how do i do that?
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student
i was only taught to do 2
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volunteer
that is what i was explaining at the beginning. because there is an abs value on both sides, you have two cases for each of them so there are 4 cases. I'll show on the board
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student
you dont have to solve for all 4
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student
its ok im assuming itll be
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student
+ >= +
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student
+<= +
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student
->= +
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student
-<= +
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student
right?
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volunteer
no
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volunteer
yes for the +/- but not for the other
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volunteer
it's always >=
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student
huh
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volunteer
just showing what signs i will be putting for each side
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volunteer
so it is always >=
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student
so
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student
+>=+
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student
+>=-
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student
and what other 2?
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volunteer
it is just shorthand so i don't have to rewrite the equation so many times
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volunteer
->=-
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volunteer
->=+
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volunteer
i was just writing those little things to make it clearer. they are not a part of the problem. don't get confused by them
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student
no it actually does make sense
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volunteer
ok
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volunteer
good
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student
it makes more sense because thats how i was taught the first 2
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student
but the teacher who made all of our notes no longer teaches so if some pages go missing we just have to do without them
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volunteer
oh no. that's annoying. i'm sorry that you have to deal with that
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student
its ok i feel like im still doing fairly decently
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volunteer
that's good
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volunteer
you seem to know a lot
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volunteer
the important thing here is to check for extraneous solutions
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volunteer
you will see here that x = -1 and x = 2 are actually both extraneous. that is how you get the correct answer
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student
ohhh ok
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student
theres 1 more question im confused on
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student
mostly just because i dont know what its asking from me
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volunteer
i'm sorry but i have to go in 5 mins.
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volunteer
we can work on it until then though
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student
oh ok
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student
no its fine because its a longer one
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volunteer
oh ok
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student
thanks for helping
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volunteer
of course
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volunteer
hi how can i help you today
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student
hello! I'm confused with this problem
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volunteer
ok
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volunteer
is that all that the question asks?
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student
yes
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volunteer
did thye give any other info?
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volunteer
oh ok
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student
the directions say to solve
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volunteer
solve for x?
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student
no like determine at what points the inequality is true
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student
in interval notation
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volunteer
oh i see
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volunteer
well then we can just look at the graph. it is asking where the green line is less than or equal to the blue curve
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volunteer
and you look at the y values
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volunteer
for example, which one is greater at x=0
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volunteer
actually that is hard to tell
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volunteer
try when x=1
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volunteer
they are actually equal at this point
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volunteer
now at x=2
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volunteer
so here the green line is greater than the blue
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volunteer
same at x =2
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volunteer
i mean x=3
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volunteer
can you do the rest?
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volunteer
Hello Jamie! How can I help you today?
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[ { "pii_type": "PERSON", "surrogate": "Jamie", "start": 6, "end": 11 } ]
student
I need help preparing for a unit test and I have practice problems I need help with
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student
the first ? is up
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student
how r u today
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volunteer
Alright!
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volunteer
I'm great! How about you?
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student
doing good
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volunteer
Alright, it looks like you already concluded that the law of cosines is needed to solve this problem, which is correct.
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volunteer
How do you think the measure of angle B can be found?
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student
M<B = 9mm / cos(56deg)
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volunteer
Correct!
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student
wait actually!?
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volunteer
No wait, I'm sorry, I made a mistake. First we need to solve for side a.
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student
oh
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volunteer
After finding the measure of side a we can find the measure of the other angles.
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student
ok
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student
so M<A = a/cos(56deg)
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volunteer
Correct! But it should be 66 degrees, not 56.
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student
oh it looked like 56
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student
2.459
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volunteer
Not quite. We need to use the law of cosines formula to solve for a.
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volunteer
Are you familiar with what the formula is?
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student
isn't it A^2 + B^2 = C^2?
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volunteer
That is the pythagorean theorem. I'll write out the formula for the law of cosines.
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student
got it
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volunteer
b and c represent the two other sides, 9mm and 5mm
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student
a = sqrt( 9mm^2 + 5mm^2 - 2(9)(5) cos (A)
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volunteer
Well done! Now, how can we solve this equation?
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student
yes
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student
a = sqrt(106mm^2-90mm^2cos(A)
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student
is what I got
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