<p>9. The total cost of a taxi ride is the sum of 1) a basic fixed charge for using the taxi and 2) an additional charge for each 1/4 mi that is traveled. If the total cost to ride 3/4 mi is $4.00 and the total cost to ride 1 1/2 mi is $5.50, what is the total cost, in dollars, of a 3 mi ride?</p>
<p>The difference between 1 1/2 mi and 3/4 mi is 3/4 mi. The difference between the respective costs (5.50 and 4.00) is 1.50, meaning that each additional quarter-mile costs $.50 (1.50/3)</p>
<p>The difference between 3 mi and 1.5 mi is 1.5 mi, or 6(1/4). Hence the additional charge for a 3 mi ride is 6(.50) = $3. Adding the $3 to the cost of a 1 1/2 mi ride, we get 3 + 5.50 = $8.50</p>
<p>10. Let the function s be defined so that s(x) is the area of a semicircle with diameter x. If s(6) + s(8) = s(b), what is the value of b?</p>
<p>s(x) = [pi(x/2)^2]/2
= pi x^2/8</p>
<p>So
pi(6)^2/8 + pi(8)^2/8 = pi(b)^2/8
36pi + 64pi = b^2 pi
100 = b^2
b = 10</p>
<ol>
<li><p>That’s pretty quick. Some students would have gone back after finding the per mile cost to find the base cost, but your way is quicker than that. </p></li>
<li><p>You could cancel out the pis and over 8s very early on (or not even write them out at all). So, you’d presumably start with 6^2+8^2=b^2 and solve from there, without having to write out everything else.</p></li>
</ol>
<p>Yeah I thought of that for #10 afterward, how the semicircle area would be a function with x^2 in it and the constant next to it wouldn’t matter.</p>
<p>Pentagon ABCDE in the figure above has sides of equal length and the five marked angles have equal measure. The measure of angle ACD is 72. If the measure of angle DAE is x, what is the value of x?</p>
<p>What I did: angle E = 108 using the fact that an individual angle in a convex regular polygon is 180(n-2)/n. DAE and ADE are equal since AE = DE. Then 2x + 108 = 180, and x = 36.</p>
<p>I’d like to know how to do this problem without that formula (using the fact that ACD = 72), even though I imagine it’s the fastest way to do it.</p>
<p>I don’t believe there is a way to do it without knowing that the interior angle of a regular pentagon is 108 (or equivalently, that the sum of the interior angles of a regular pentagon is 540). But that is simple to show, triangulate the pentagon into three triangles, the sum of the interior angles of the pentagon is the sum of the interior angles of three triangles, or 180*3 = 540. (The formula follows from generalization, n-gon can be shown to always be triangulated into n-2 triangles, thus total sum is 180(n-2) and per interior angle is 180(n-2)/n).</p>
<p>Outscribe a circle over pentagon.
The measure of <DAE is the half of its intercepted arc DE, which in turn is 360/5=72 deg.
72/2=36 deg.
Similarly, <BAC = <CAD = 36 deg. Nice fact!</p>
<p>In rectangle ABDF above, C and E are the midpoints of sides BD and DF, respectively. What fraction of the area of the rectangle is shaded?</p>
<p>I drew a line from C to what I called point X (the midpoint of AF), treating the rectangle in two parts. On the left part, clearly half is shaded. On the right, by drawing a line from E to the midpoint of CX and from E to X, there appeared four rectangles of equal size. Three of the four are shaded. So…</p>
<p>1/2 of 1/2 of the rectangle shaded (left side) + 3/4 of 1/2 of the rectangle shaded (right side)
= 1/4 + 3/8
= 5/8</p>
<p>R is the midpoint of line segment PT, and Q is the midpoint of line segment PR. If S is a point between R and T such that QS = 10 and PS = 19, what is the length of segment ST?</p>
<p>Doing it timed I did it in a fairly complicated way, but reviewing my work I did it an easier way.</p>
<p>PQ = PS - QS
= 19 - 10
= 9</p>
<p>Hence QR is also 9.</p>
<p>QS = 10, and QR + RS = QS = 10 so RS = 1.</p>
<p>PR = 18 = RT = RS + ST
18 = 1 + ST
ST = 17</p>
<p>=======================================</p>
<p>A phone company charges x cents for the first minute of a call and charges for any additional time at the rate of y cents per minute. If a call costs $5.55 and lasts more than 1 minute, what is the expression that represents the length of the call in minutes?</p>
<p>I let x = 5 and y = 10. So the number of additional minutes is (555 - 5)/10 = 55. Adding on the first minute, we get 56.</p>
<p>The expression then is (555 - x)/y + 1 which is the same as (555 - x + y)/y</p>
<p>P–Q–R-S—T (S is not placed to scale)
QS=10, PS=19. By looking at the above diagram, it shouldn’t be hard to realize that PQ=9, therefore PR is 18, therefore PT is 36. PT-PS=ST=36-19=17. </p>
<p>t=total time, in minutes. (t-1)*y= total cost after 1st minute. x=cost of first minute. So, x+y(t-1)=5.55, t-1=(555-x)/y, so t=[(555-x)/y]+1. Same thing you got, except I didn’t use actual values. Your way might be better, though.</p>
<p>Point O is the center of both circles in the figure above. If the circumference of the large circle is 36 and the radius of the small circle is half the radius of the large circle, what is the length of the darkened arc?</p>
<p>While doing this timed I did the long way, finding big radius and then small radius and small circumference, then determining that the arc darkened is 80/360 of the total circumference.</p>
<p>Another method I figured out: the circumference of the circle is directly proportional to the radius of the circle. If the large circumference is 36 and its radius is twice that of the small circle, then the small circumference is 36/2 = 18. Then 80/360 = x/18 and x = 4.</p>
<p>Because circumference=2<em>pi</em>r, the circumference of the larger circle is twice that of the smaller circle (all that changes in the above equation is radius). Thus, the circumference of the smaller circle is 18. Then, you know the arc will be 80/360 of the entire circumference (360, if you notice, is the angle of the entire circumference, 80 is the angle accross from the desired arc). So, the arc length is 18*(80/360)=180/45=4 units. </p>
<p>I did this before looking at your response… So, it looks like you did it the fastest way for your second method.</p>
