# mpcTutor

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1. ## Could someone help us with this Alg 2 problem?

Download detail pdf solution: Train word problem Best regards. Mohan Pawar http://www.mpclasses.com/ContactUS.htm US Central Time:5:14 PM 9/27/2010
2. ## Could someone help us with this Alg 2 problem?

Please verify. I am getting Westward speed = 5 mph ( same as 5 miles per 60 min) Eastward speed = 7 mph (same as 7 miles per 60 min) I will try to post detail answer later if anyone interested. I have to go. Best regards. mpcTutor http://www.mpclasses.com/ContactUS.htm US Central Time:2:28 PM 9/27/2010
3. ## How in the world do you simplify repeating

For solution see: How to represent a recurring decimal number as a rational number. Best regards. MPCtutor http://www.mpclasses.com/ContactUS.htm ---------------------------------------------------------- AP Calculus, AP Physics, IIT JEE Test Prep. ---------------------------------------------------------- US Central Time:1:28 PM 9/17/2010
4. ## Need Compound interest math help

The previous poster already gave the solution. Here are details. Simply follow the same logic that you used for other problems solved correctly. If P = Present amount = 55000 i% = % interest rate per period (Note that I deliberately wrote per period and not per year ) = 8%/4 = 2% n = No. of periods for which P is in the saving account = 2 years x 4 period per year = 8 S = Lump sum amount yielded at the end of n period = To be found yield = interest paid on investment Then S = P( 1 + i%)^n (One can easily derive this formula with 7th grade algebra background. Note 2% = 2 per cent = 2/100 = 0.02)) = 55000 ( 1 + 0.02)^8 = 64441.27 Yield = 64441.27 - 55000 = 9441.27 (Answer) Best regards. mpcTutor http://www.mpclasses.com/ContactUS.htm ---------------------------------------------------------- SAT/ACT, AP Calculus, AP Physics, IIT JEE Test Prep. ---------------------------------------------------------- US Central Time:1:49 PM 9/16/2010
5. ## any geometry proofers

Your son is right and I don't see anything wrong with his proof. Although, problem asks to prove CA > CO, the zipped figure to me showed CA significantly shorter than CO without making any actual measurements. Also, triangles should appear to be congruent without actually measuring any sides or angles. So I am assuming the original diagram could be different from diagram in your attachment. Short Proof: ------------- Given: OCE & ANE as congruent triangles Consider triangle OCE, OC < CE + EO (due to triangle inequality) OC < CE + EA (due to Given, EA = EO) OC < CA (due to betweeness of points CA = CE + EA ) QED Best regards. mpcTutor www.mpclasses.com -------------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 -------------------------------------------------------------------------------- Note: If you replied to my post but didn't get an answer from me in reasonable time, I ask you to check the assumptions in your question. Thank you US Central Time: 11:30 AM 7/29/2010

7. ## Algebra 2 problem---Help

Solution: Given:3x + y + z = 2 ----------(1) 4x + 2z = 1----------(2) 5x - y + 3z = 0----------(3) If y is eliminated from (1) and (3) we get 8x + 4z = 2 ----------(4) The resulting system now has Eq (2) and (4) which represent only one straight line. These equations are called as "dependent equations* in x and z". Values of x and z will depend on y and can be infinitely many depending on value of y. Thus, The strategy is not to eliminate y but eliminate x or z. (1) x 3 - (3) gives 2x + 2y = 3 Thus x = (3 -2y)/2 ----------------(Answer1) Substituting x from Answer1 in (2) 4(3 -2y)/2 + 2z = 1 Thus, z = -5/2 + 2y ---------------(Answer2) and y = y ------------------------ (Answer3) Now for any given value of y the given eqns (1), (2) and (3) will be uniquely satisfied. Dependent Equations: ------------------------- See Eq (2) & (4). 4x + 2z = 1 and 8x + 4z = 2 are dependent because (4) depends on (2) and vice versa. Such equations are ignored when solving system of simultaneous equations and that's what is done above. Best regards. mpcTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 10:58 PM 5/18/2010 Edited at US Central Time: 7:12 AM 5/19/2010

I gave this solution in hurry, please wait for someone to point out if any errors. Your book formula is correct. Use V for volts, N for turns, subscript p for primary and subscript s for secondary. Rewrite the formula as below: Vs/Vp = Ns/Np Hence Ns = Np (Vs/Vp) --------> (1) Given: Vp = 120,000 volts Vs = 120 volts Np = 1000 turns Substituting above quantities in (1) Ns = 1000 (120/120000) = 1 Current in secondary Is = Vs/Rs = 120/10 = 12 ampere c. what is the power in the secondary coil? Power in Watts Ps = Vs x Is = 120 volts x 12 ampere = 1440 watts Power in primary coil = Power in secondary coil Ip x Vp = Is x Vs = 1440 Ip = 1440/120000 = 0.012 ampere (which is comparatively low amperage) P ower loss = (I^2) R. Thus, power loss increases in square proportion of current I for a given resistance R. By reducing I to low level such as 0.012, the power companies avoid power loss during power delivery from power station which is far away fro home or place of power consumption. Rigorous solution to similar question is in math forum . ( See the link ) Best regards. mpcTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 2:43 PM 5/18/2010
9. ## Singapore NEM

Reference to Saxon algebra-II automatically assumes two more separate books as part of curriculum - i) Algebra-I and ii) Geometry. The switch from New Elementary Math (NEM) to Saxon is tricky unless one has already taught before the entire high school math including pre-calculus. The difficulty is due to Geometry/Co-Ordinate Geometry, Trigonometry and Algebra-II being covered all along in the NEM up to book 4A. On the other hand, NEM 1, 2, 3 and 4A should be more than adequate for ACT + SAT and sufficient for subjects tests SAT Math Levels 1 & 2. For a brief inside view of Saxon's Advanced Math you may want to visit Amazon. Best regards. MPCTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 9:58 AM 5/17/2010
10. ## Singapore NEM

That's odd -- I was able to find the table of contents and sample pages for the Singapore books here:Book 3A http://www.singaporemath.com/ProductDetails.asp?ProductCode=NEMT3A&Show=TechSpecs Book 3B http://www.singaporemath.com/ProductDetails.asp?ProductCode=NEMT3B&Show=TechSpecs Book 4A http://www.singaporemath.com/ProductDetails.asp?ProductCode=NEMT4A&Show=TechSpecs Book 4B http://www.singaporemath.com/ProductDetails.asp?ProductCode=NEMT4B&Show=TechSpecs If you were referring to the Saxon math books, then I would agree: I couldn't find similar info on their website. 69 Thanks for the links in support. Best regards. MPCTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 9:46 AM 5/17/2010
11. ## Singapore NEM

I have not seen Saxon math beyond Algebra 1/2 so it is difficult for me to compare. Like Singapore math very few textbooks make their basic information such as table of contents and a sample topic pages available on their websites for the buyer to evaluate the book. Most of word problems in the placement test are comparable to those in New Elementary Math (NEM). In fact, majority of challenging problems from "Challenging Word Problems (CWP) - Primary 6" from Singapore Math are harder than SAT/ACT word problems if one has do them in limited time. But CWP is another book. There are many many problems in NEM so I have not used any additional NEM workbooks. NEM could be too much as supplement. ------------------------------------------- NEM volume 3 & 4 are actually 4 separate books - 3A/3B and 4A/4B with respectively 220, 240, 129 and 255 pages (i.e. a total of almost 850 pages). 4B with 255 pages is entirely problems for practice covering almost all important elementary, middle and high school math. Along with Saxon say, Algebra 1/2 which requires one to do all 30 problems per lesson, the NEM as supplement could be lot of work. Currently, I am using Singapore Math for 6th grade for students with Saxon Algebra 1/2 as their regular curriculum and it is lot of work for them. Unless criterion of evaluation of curriculum effectiveness is known, choosing a curriculum is going to be rather subjective. One can choose any of the NATIONAL level standardized tests as criterion. Most colleges use them as criterion for admissions. So why not use test prep books as supplement? For example, A book titled "Real ACT prep guide" will give almost 180 math problems that once were on real ACT. Similarly, one can choose SAT prep books from College Board or from some other publisher. Best regards. MPCTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 10:18 AM 5/15/2010
12. ## Please define "conceptual math."

Concept is nothing but certain concrete idea (as against a vague notion). It doesn't matter if the concept is in mathematics or not, the success in identifying "the" concept from all other seemingly similar concepts is needed in understanding that particular concept. Example 1: Explain "All squares are rectangles but all rectangles are not squares." The beginning of above explanation is sometime when a child learns to differentiate a table from a chair and learns to ignore the fact that they both have four legs which in fact are not important in differentiating a chair from a table. The same child will however identify both chair and table as furniture distinctly different from say a car. Use of pictures, verbal description and constant acquaintance with these objects make these non-mathematical concepts pretty clear for most children. Unfortunately, with respect to most mathematical concepts, making the child aware of say, number 3 is not that easy. We can show 3 cars, 3 apples and 3 dogs but identifying the common three-ness in the pictures of cars, apples and dogs is largely left to the child. There is not an easy way to tell the child that the "three" is common to 3 cars, 3 apples and 3 dogs. If the child sees them as "things" then he/she missed the concept of "three". Also, important thing to remember is that even grown ups don't always see everything that is shown. So every child doesn't have to see the idea right after it is explained. There could be several valid reasons for a child not see your point including barriers in communication, child's willingness to be attentive etc. Example 2: The commutative law of addition a + b = b + a is introduced to the child by dividing 15 marbles in to two groups say 10 marbles and 5 marbles and then adding 1st group in the 2nd group and vice versa and then counting to show that the result is 15 marbles. The concept here is the sum of two numbers is not affected by the order in which they are added. If this "concept of commutation" is understood then the child doesn't need to be shown 20 + 30 = 30 + 20 again with the help of marbles. Example 3: The commutative law in Example 2 above is normally not questioned with "why" but simply accepted and its reason is known in later grades when field axioms for real numbers are taught only to learn that highly fundamental facts don't have reason. A new concept called "axiom". Bottom line: In earlier years certainly one could learn the basic facts through manipulative and by solving select problems. Particularly after 4th - 5th grades it is through the practice of solving well chosen problems that many hidden concepts can be understood. In my experience, that's the way the mathematics and physics are: not only for reading but largely for solving problems (with paper and pencil). In doing all of these activities a wise goal is to keep the child interested in mathematics and the sought after math mastery will come as consequence in due time. Best regards. MPCTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 2:11 PM 5/14/2010
13. ## SAT 2 math review-help needed from math experts!

The same notation that you are referring to can be written as C(n.r) or specifically C(6,2) and C(8, 4) for ease of typing (which is accepted method of representation in many textbooks) C(n.r) means number of distinct combination (or groups) possible when each group contains r things out of n things. Please feel free to write back if you need detail answer. But in short, C(6,2) = 6!/[2!(6-2)!] = 15 C(8,4) = 8!/[4!(8-4)!] = 70 Thus, total ways = 15 x 70 = 1050 The topic to look for further information on notation is "Permutation and combination from pre-calculus" . We are currently registering students for online class for Pre-Calculus at our website. Sample of topic/chapter wise DVDs may also be available. Please send your request if you are interested. Best regards. MPCTutor www.mpclasses.com Contact for Additional Info --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 11:06 AM 5/7/2010
14. ## physics help needed please!!

This is purely math problem - No electricity/No Physics. (except use of P = V x I) I will rewrite the question and solve it. If interpreted problem is same as the original problem, use the solution, otherwise please post original problem as it is in the original source. Problem: 1) How much power does a light bulb connected to a 120V outlet use if it draws 0.5A of current? 2) One kilowatt-hour is a measure of energy which is equal to 3,600,000J. How many hours would you have to leave on the light bulb from the above problem in order to expend one kilowatt-hour? (remember that 1W=1J/s) The information in blue is not needed. 3) If one kilowatt-hour costs \$.10, how long would you have to leave the light bulb on from the above problem in order to spend \$1.00 on electricity? Solution: 1) Power = 120 V x 0.5 A = 60 Watts = 0.06 KW 2) Let t = time in hours Power consumed = 1 KWH = (0.06 KW) ( t hour) -----------(1) Solving (1) for t t = 1 KWH / 0.06 KW = 16.67 hours ----------(Answer) 3) Given Rate = 0.1 \$/KWH Total bill = \$1.00 Hence Total KWH = 1/0.1 = 10 KWH So required time = 10 KWH / 0.06 KW = 166.67 hours ---------(Answer) Best regards. MPCTutor --------------------------------------------------------------------------- AP Calculus, AP Physics, IIT-JEE, Singapore Math --------------------------------------------------------------------------- US Central Time: 1:19 PM 5/6/2010
15. ## oops found one more

Agree with other poster. F = 2.304 x 10^-28 N (Force of Repulsion) MPCTutor --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math (7-12) --------------------------------------------------------------------------- US Central Time: 11:58 AM 5/6/2010
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