math iic questions

<p>i took a practice test and these are the ones my puny brain cannot handle.. since they don't have explanations, could you please explain to me how to solve? help would be much appreciated :)</p>

<p>38) if (6.31)^m = (3.02)^n, what is the value of m/n?</p>

<p>40) if the 20th term of an arithmetic sequence is 100 and the 40th term of the sequence is 250, what is the first term of the sequence?</p>

<p>41) if n distinct planes intersect in a line, and another line L intersects one of these planes in a single point what is the least number of these n planes that L could intersect?
a. n
b. n-1
c. n-2
d. n/2
e. (n-1)/2</p>

<p>47) in how many ways can 10 people be divided into 2 groups, one with 7 people and the other with 3 people?</p>

<hr>

<p>Answers:
38= 0.60
40= -42.5
41= b
47= 120</p>

<p>btw i'm taking this on june 4th.. any suggestions for last-minute cramming?</p>

<p>38:</p>

<p>log both sides so m log 6.31 = n log 3.02
Then solve for m/n</p>

<p>40:</p>

<p>40th - 20th = 150
150 per 20 terms, 15/2 per term.
Then subtract 15/2 from 100 to the first term.</p>

<ol>
<li> I can only think of this pictorally.</li>
</ol>

<ol>
<li>i can't explain clearly.</li>
<li>combination (10, 3)</li>
</ol>

<p>for number 40, you have to know the equation:
A(sub n) = A(sub 1) + (n-1)d
*where, n = the what term, a(sub 1) is the first term, and d is the increment for the arithmetic series.</p>

<p>40) if the 20th term of an arithmetic sequence is 100 and the 40th term of the sequence is 250, what is the first term of the sequence?</p>

<p>So...
A(sub 20) = 100 = A(sub 1) + (100-1)d
A(sub 40) = 250 = A(sub 1) + (40-1)d</p>

<p>System of equations, solve for A(sub 1)</p>

<p>*Hope this triggers some Algy2 memories.</p>

<p>41) if n distinct planes intersect in a line, and another line L intersects one of these planes in a single point what is the least number of these n planes that L could intersect?</p>

<p>For 41, just plug in numbers... say...
n = 2. Two planes intersect in a line, and another line L intersects one of these planes in a single point. What is the least number of these '2' planes that L could intersect.</p>

<p>If the two planes are perpendicular, and L is parallel to one plane, intersecting one plane in one point (going through it) then The least number = 1plane. Choice = n-1.</p>

<p>That's a shoddy way of doing it though, i'm not sure how to explain it mathematically, only logically.</p>

<p>Just realized that gives 1 for both B and D. Thought of another way... if a line is going to go through one plane, then it must go through all the planes except for one (the one it's parallel to) The rest, it must go through.</p>

<p>As for 41. It's rather simple. The minimum number of times a line can intersect the plans is to have the line made parallel to one of the planes, for that is the only way to have the line not intersect a plan. Since all planes interesct at one line (hence no plans are parallel), the minimum number would be n-1, which is n plans without the plane parallel to the line.</p>

<p>can anyone explain the combination one (number 47). also, can anyone clearly explain the difference between a permutation and combination...i am so confused!</p>

<p>permutation= arrangement where order matters. Example: ABC and ACB would be counted differently.
combinations = arrangement where order does not matter, so ABC, ACB, BCA, CAB etc. are altogether 1 combination.
So say, if you are selecting members of a comittee, each position in a comittee is equal, so order does not matter, so you use combinations. </p>

<p>If you are determining no. of arrangements of say, the presidency, vice presidency, and other offices, then you use permutations, because the various positions are not equal.</p>

<p>Wow! This is so helpful and it’s from 7 years ago…</p>