<p>Hey, what did u get for the problem that asked gave u six numbers and asked you if an integer is added to the list, then what would be the new median??</p>
<p>the choices for answers were<br>
I.6
II. 6.5
III. 7</p>
<p>a. I only
b. II only
c. I and III only
d. II and III
e. I and II and III</p>
<p>I put A but I probably got it wrong…</p>
<p>it was I and III</p>
<p>i put 1 and 3 only not sure if correct though…but i know its not b/d/e becuase when adding n you get an even amount of numbers</p>
<p>Hey, guys for the pyramid problem, the answer is most definitely (root 3 times m) over 2. </p>
<p>here is the clear explanation.</p>
<p>The problem stated that the base of the square was m and all four lines of the triangles that went up to vertex v were also m. </p>
<p>which means, that all four triangles were equilateral triangles (therefore 60-60-60 degrees per triangle). Then that means from the height, you draw a line to one of the sides of the triangle and make it a 30-60-90 triangle. (When you bisect a 60 degree angle, you get 30 degrees). </p>
<p>So you know for sure that you have to use the 30-60-90 triangle formula.
For 30-60-90, the sides are m, root 3 times m, and 2m.
which means then the length of the base of the square has to be m/2 since the hypotenuse is m. (m/2 times 2 is m). then which means the h has to be
(m/2) times (root 3) which is (root 3 times m) over 2.</p>
<p>most people have agreed it was m/sq2. I am going to leave the explanation up to the other people.</p>
<p>I put m/sqrt2, but I can’t seem to find a flaw with what jae6forever said…can anybody else?</p>
<p>Look back a few pages to see why it’s m/sqrt(2)</p>
<p>I’m too lazy to explain it…</p>
<p>i already read them but i just cant seem to understand why that’s the answer.</p>
<p>That doesn’t work physically. You cannot draw a line from the height to one of the internal angles of the traingles.</p>
<p>for that m/root 2 i just plugged in an answer</p>
<p>
</p>
<p>I did this exact same thing and was confused why everyone else had a different answer. And then I realized what you and I did was we drew a line from the height (center of the base square of the pyramid) to one of the sides of the square, which would be a length of m/2. You actually have to draw the line from where the height touches the base of the pyramid to a corner of that square and then the base of the triangle is no longer equal to m/2. </p>
<p>Sorry, I’m really bad at explaining these online, but basically our base of the triangle was wrong. It wasn’t m/2 because it goes to the corner of the square not just one of its sides.</p>
<p>the base of the triangle was half the diagnol of the square – m x sqrt2 over 2</p>
<p>Ok, so the base of the pyramid was m and so were the edges. The edge of the pyramid is the hypotenuse of the triangle containing h.
Let x be the other side of the triangle.
So h^2+X^2=m^2</p>
<p>so now we need to find x
x is one half the diagonal of the square base.
Since the side of each square is m you get the diagonal to be
2m^2=d^2
so d/2= (sq 2*m)/2=x</p>
<p>Ok now h^2=m^2-x^2
so plugging in (sq 2*m)/2 for x
h^2=m^2/2
take the square root of both sides and you get
h=m/(sq 2)</p>
<p>You are not supposed to infer anything about the angles of the triangle. You don’t know if they are 30-60-90 or not. Gotta go finish my homework now!</p>
<p>quick question — does anyone remember the answer choices for this problem: </p>
<p>f(x) = ka^x and it asks you to solve for a</p>
<p>I know a was supposed to be 4, but i’m scared that I was stupid and put 1/4</p>
<p>darth raider i think you found the hight of the side of the pyramid, not the height that goes through the pyramid</p>
<p>What exactly was the grasshopper question?</p>
<p>there was a grasshopper question?</p>
<p>Can anyone tell me what the heck the school population question was? As in, what was the question like?</p>
<p>was school population an experimental section question?</p>