<p>Blue book page 532/7</p>
<p>Edna and Nacy leave the house of a common friend at the same time and walk for 4 hours. Edna walkes due east at the average rate of 4 kilometers per hour and Nancy walks due eat at the average rate of 3 kilmoters per house. What is the straightline distance between them at the end of the 4 hours?</p>
<p>A) 4
B) 5
C) 12
D) 16
E) 20</p>
<p>Obviously A should be the correct answer, but CB says it's C...</p>
<p>Eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee</p>
<p>Read the question carefully.</p>
<p>Sorry, I figured it out now. It's just the panic of approaching SAT...</p>
<p>do u know how to type? furthermore, do u know how to read??
the way u interpret, is both walking east >>>>
the way u type, is both walking east >>>>
are u looking at the book??
read again.
there's loads of typing errors, 3 KILOMETERS PER HOUSE??!</p>
<p>you wrote something wrong!!!!!!!</p>
<p>^c'mon. Easy on newcomers.</p>
<p>I think it is 20 km ?Answer E</p>
<p>Here's how the problem should correctly read:</p>
<p>"Edna and Nancy leave the house of a common friend at the same time and walk for4 hours. Edna walks due east at the average rate of 4 kilometers per hour and Nancy walks due north at the average rate of 3 kilometers per hour. What is the straight-line distance between them, in kilometers, at the end of 4 hours?"</p>
<p>Solution: </p>
<p>If Edna goes east for 4 hours at 4km/hr, she'll travel 16 km.
If Nancy goes north for 4 hours at 3km/hr, she'll travel 12km.</p>
<p>starting from a common point(the house), draw a horizontal line to the right that represents 16 km, and draw a vertical line from that common point upward that represents 12 km.
Now you can make a right triangle by connect the top of the vertical line to the right of the horizontal line. Then either use Pythagorean Theorem to solve for the hypotenuse or recognize it's a multiplier of 4 on a 3-4-5 triangle, and the hypotenuse would be 20.</p>
<p>So, answer is E. 20</p>
<p>My solution was easier ;] I drew a line - 16 cm long and another one - 12 cm long .Another line,that connects that fitst ones in 20 cm long ;]</p>
<p>answer is 4</p>
<p>***> that''s so easy! just find how far each walks then use pythag theorem! but yea sry good luk anyways</p>
<p>Ivan, you're right that is much easier =] i dont know how i didnt think of it.
to make the problem even more realistic, i think we should draw the lines in km ;)</p>
<p>What about the problem 532/6 ,where you need to express m^-1 in terms of k ?Did you solve it ;} It looks a bit harder ;]</p>
<p>Wow. I have no clue at all... how to solve this.</p>
<p>Say there is a parallelogram ABCD, with a diagonal drawn BD. (A picture was included, but it should be easy to visualize. </p>
<p>If the five line segments in the figure above are all congruent, what is the ratio of the length of AC(not shown) to the length of BD?</p>
<p>A) root 2 to 1
B) root 3 to 1
3) root 2 to 2
D) root 3 to 2
E) root 3 to root 2</p>
<p>the problem Ivan was talking about:
Divide both sides by m^2.
then divide both sides by 100</p>
<p>DONE</p>
<p>xclutch's question:</p>
<p>Is it choice B) ??
its pretty difficult to explain via typing online</p>
<p>It is in fact B</p>
<pre><code> A _______________ B
\ \
\ \
\ \
\ \
D ______________ C
</code></pre>
<p>There's a diagonal BD, which is congruent to AB BC CD AD, all of which are congruent as well.</p>
<p>What is the ratio of AC to BD..</p>
<p>How did u get to answer B?</p>
<p>It is in fact B</p>
<p>.....................A _______________ B
........................ \ ......................\
..................................................\
....................................................\
......................................................\
..............................D______________ C</p>
<p>There's a diagonal BD, which is congruent to AB BC CD AD, all of which are congruent as well.</p>
<p>What is the ratio of AC to BD..</p>
<p>How did u get to answer B?</p>
<p>It was pretty easy ,i will try to explain ;]
So if in a parellelogram all the sides have the same size (or are congruent) ,the parallelogram is аlso a rhombus.If diagonals AC and BD cross at point O ,AO = CO and BO = OD. Let AB = AD = BC = BC = BD = 2a.If BD = 2a => BO = DO = 1/2BD =a.
In triangle ABO ,you have AB = 2a,BO =a and <A0B = 90 degrees.In Rhombus the diagonals are perpendicular to each other.So by pythagorean theorem ,if AB = 2a and BO = a,you will find that AO = aroot 3.So OC = AO = aroot 3 and AC = AO + OC = 2aroot 3.
As you know BD = 2a ,so AC/BD = 2aroot 3 /2a = root3 /1.</p>
<p>Was this problem hard or medium ?</p>
<p>This was the last problem: hard. Unlike other math sections, I found this one to be very hard... I got 3 wrong in the section, and I took all 25 minutes... I usually only take about 20/25 minutes in the 25 minute math section.</p>
<p>I am good with algebra, but pretty weak and slow in geometry, and especially weak with graphs.</p>