WEBVTT

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We're on problem number 7.

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If the average of x and 3x is
12, what is the value of x?

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So the average, so x plus 3x--
and I'm averaging two numbers

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so divide by 2-- that's going
to be equal to 12.

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So we just solve for this.

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Multiply both sides of the
equation by 2 and you get 2

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times, times 2, this
cancels with this.

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You get x plus 3x
is equal to 24.

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And what's x plus 3x.

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That's 4x, right?

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4x is equal to 24.

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x is equal to 6, and
that's choice C.

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Next problem.

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Problem 8.

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At Maple Creek High School, some
members of the chess club

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are also on the swim team, and
no members of the swim team

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are tenth graders.

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Which of the following
must be true.

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This seems like it'll call
for a Venn diagram.

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So let's say that that
represents the chess club.

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And they say some members
of the chess club

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are on the swim team.

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So some members are
on the swim team.

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Maybe I should put the swim
team in like blue.

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So let's say the swim team.

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That's the swim team.

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And these are the members,
right, that are

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in both right here.

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But then it tells us no members
of the swim team are

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tenth graders.

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So if I draw another circle
for the tenth graders, it

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can't intersect with the swim
team, but it could intersect

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with the chess team.

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I don't know.

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I mean it could be like that.

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That could be tenth graders.

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It could be like that.

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Or it could be out
here some place.

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But we don't know.

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There could be chess and tenth
graders, just not the same

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people who are on
the swim team.

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So let's see.

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So which of the following
must be true?

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No members of the chess club
are tenth graders.

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No.

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This is a situation where you
could have some members of the

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chess club who aren't
on the swim team who

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could be tenth graders.

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B, some members of the chess
club are tenth graders.

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Well some members could be, but
we don't know for sure.

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This could be tenth grade.

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We don't know.

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This could be the tenth grade
kind of set or this could be

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the tenth grade.

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There might be no tenth graders
in either the chess

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team or the swim team.

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We don't know for sure.

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And then choice C, some members
of the chess club are

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not tenth graders.

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This we know for sure.

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How do we know it for sure?

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Because these kids who are on
both, they're in the chess

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club, but they're also
on the swim team.

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The fact that they're in swim
team, we know that they can't

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be tenth graders.

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So this is some members of the
chess club-- this little

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intersection here-- that
are not tenth graders.

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So choice C is the
correct choice.

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Next problem.

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If 3x plus n is equal
to x plus 1, what is

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n in terms of x?

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So we essentially just
solve for n.

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Let's subtract 3x
from both sides.

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You get n is equal to--
what's x minus 3x?

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It's minus 2x.

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And n plus 1.

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And we're done.

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And that choice isn't there, but
if you just switch these

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two terms you just get that
equals 1 minus 2x and

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that's choice D.

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Pretty quick problem, especially
for one that's the

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ninth problem.

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They normally get a little
harder by this point.

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Problem 10.

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If k is a positive integer, let
k be defined as a set of

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all multiples of k.

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So k with a square around it
is equal to the set of

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multiples of k.

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All of the numbers in which of
the following sets are also in

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all three of the set-- OK.

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All of the numbers in which of
the following sets are also in

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all three of the sets
of 2, 3 and 5?

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So the what they're saying is 2,
3, 5, this donates all the

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multiples of 2.

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This is all multiples of 3.

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This is all multiples of 5.

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So what they're essentially
saying is let's find a number

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where all of its multiples, all
of this number's multiples

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are also going to be multiples
of each of these.

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So it has to be a multiple--
so every number that--

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whatever this mystery number
is, let's call it x-- every

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multiple of x has to be a
multiple of 2, 3 and 5.

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Well the simple way is if x is a
multiple of 2, 3 and 5, then

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every multiple of x
is going to be a

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multiple of 2, 3 and 5.

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So what's 2 times 3 times 5?

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It's 2 times 3 times 5.

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That's 6 times 5, that's 30.

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So 30 is a multiple of all of
them, so any multiple of 30

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will be a multiple
of all of these.

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When we look at the choices
we don't see 30.

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But do we see any other
number that is a

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multiple of 2, 3 and 5?

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Well sure, 60 is, right?

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We just multiply by 2 again.

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But 60 is still a multiple
of 2, 3 and 5.

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If you were to do 2, 4, 6, 8 all
the way you'd get 60, if

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you go 3, 9, 12, 15 all the
way, you'd get to 60.

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You go 5, 10, 15, 20,
25, you'd get to 60.

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So 60 is a multiple
of all of them.

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So what we're saying is-- so
what's the set of all the

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multiples of 60?

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It's 60, 120, 180, 240,
et cetera, right?

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And all of these numbers are
in each of these sets.

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Because all of these numbers are
multiples of 2, 3 and 5.

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So our answer is 60.

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If you look at the other
choices, some of them are

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divisible by 5, some are
divisible by 2 or 3,

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some are 3 and 5.

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But none of them are divisible
by 2, 3 and 5, only 60 is.

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Next problem.

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That problem was a little hard
to read initially though.

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That's how they confuse you.

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So we're going to go from A to
D-- I should have drawn all

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the lines first. Let me draw the
lines first. It's like a

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hexagon kind of.

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The top, the outside of
the hexagon there.

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A, B, C, D, E, F.

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And then this is the origin.

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And the figure above,
AD is equal to BE.

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Oh, no, no.

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They don't tell us that.

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I'm hallucinating.

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In the figure above AD, BE,
and CF intersect at 0.0.

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The intersect's here
at the origin.

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If the measure of AOB, the
measure of that, is 80

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degrees, and CF bisects
BOD, so it

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bisects this larger angle.

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CF bisect BOD, that angle.

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So that tells us that
this angle has to be

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equal to this angle.

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That's the definition of
bisecting an angle.

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You're splitting this larger
angle in half.

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So these angles have to be
equal to each other.

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So what is the measure of EOF?

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So we want to figure
out this angle.

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Well this angle is opposite to
this angle, so they're going

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to be equal.

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So if we can figure out
this angle we're done.

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So let's call this angle x.

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If that angle's x this
angle is also x.

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This x, this x, and this
80 degrees, they're all

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supplementary because they all
go halfway around the circle.

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So x plus x plus 80 is going
to be equal to 180 degrees.

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2x plus 80 is equal to 180.

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2x is equal to 100,
x is equal to 50.

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And as we said before, x is
equal to 50, the angle EOF,

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which you're trying to figure
out, is opposite to it so it's

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going to be equal.

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So this is also going
to be 50 degrees.

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And that's choice B.

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Next problem.

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I don't know if I have time
for this, but I'll try.

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Problem 12.

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k is a positive integer.

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What is the least value
of k for which the

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square root of-- OK.

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So what is the least value
of k for which 5k

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over 3 is an integer.

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So this has to be a whole
number, right?

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So essentially if we want to
find the least value of k, we

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essentially want to say, well
what's the least integer that

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this could be?

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And they're telling us that
k is a positive integer.

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So first of all, in order for
the square root to be an

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integer, this whole thing has
to be an integer, right?

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So let's see, k has to
be a multiple of 3.

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In order for this expression to
be an integer, k has to be

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a multiple of 3.

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If k is 3, we get square root
of 15 over 3-- well that

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doesn't work.

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If k is 3 we just
get 5 in there.

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Actually, let me continue this
into the next problem because

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I don't want to rush this.

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I'll see you in the
next video.