Is there an easy way to write mathematical equations by email ?
There's no universally good way. If you check out any online text-based maths community (Google Groups, USENET, IRC, ...) you'll notice that no one has completely solved it. When most people understand TeX (or LaTeX), people often type those commands, but most people don't understand it. One way of communicating maths equations is to use a fixed-width font (Courier New) and to use other characters to try and space your equations like they would be on a page. For example, copy and paste the following into Word, format it as Courier New and insert a few spaces at the beginning of each line (since Y! Answers removes spaces :( ): / 2 V = | sin 2x dx / This requires more effort if you're writing the email, but is pretty fool-proof.
Help needed to solve this Math-Question?
The word bearing has been misused, abused and overused like no tomorrow, in navigation. There are many definitions and reference points. But for the sake of our problem and simplicity, we can assume that we are talking about true bearing. This means, that the angle is in reference to geometric north pole. So, the direction of motion makes an angle given by the bearing with respect to the geometric north south line. If the ship is pointing north, it will be 0 degrees and increasing clockwise until 359.99.. degrees. Hope that gives a basic idea. For our problem, let's take the north to be the Y axis, and east to be X-axis. And assume that the ship starts at O (0,0) for the sake of simplicity. So, the first line makes 140 degrees wrt Y-axis, or makes 50 degress wrt X-axis and is in the fourth quadrant. Now, you can draw the first line, say OA. Co-ordinates of A: xa = 95 sin(140) = 95 sin(40) = 61.06km ya = 95cos(140) = - 95cos(40) = -72.77km Now, the next leg goes from A to B, having angle of 260 wrt Y axis, or having 80 degrees wrt -Y axis: Co-ordinates of B: xb = xa +102*sin(260) = xa - 102*sin(80) = 61.06 - 100.45= - 39.39km yb = ya + 102*cos(260) = ya - 102*cos(80) =-72.77 -17.71 = -90.48km So, B is in the third quadrant. Return journey is from B to O, BO will have bearing in the first quadrant (If you draw a line from B to O and extend it beyond the origin, you will note the angle lies in the first quadrant) Length of journey L= sqrt(xb ^2 + yb^2) = 98.68 km <== Ans Let T(theta) be the bearing, yb = L cos , xb = L sinT tan T = xb/yb = 39.39/90.48 T = 23.52 degrees. <== Answer NOTE: The other answer is beautiful, except he made a mistake in the final part, since bearing is measured wrt Y axis, the y = L cos beta, and x = L sin beta :)
Please help me with these three standard level IB math questions.?
1)a) This is one is the easiest (probably why they put it first). CD is really CO + OD. You should see that CO is really just the negative of OC (-OC). So CD = -OC + OD I won't give you the answers to b) and c), but the key is to use half of OC and OD. The other questions seem to deal with the Binomial theorem (look it up in Google for info if needed). -------------------------------------- Hey there. Yes, the answer to b) is correct. I find the best way to solve vector problems is to draw them. For c), draw 1/2 OC and 1/2 OD (and you may as well draw their negatives too), and see if you can get a combination of two that will give you AD (vector going straight up). Don't forget to draw arrowheads. 2) It's going to be hard to write the binomial theorem and related formulae here, so I'll use the ones on http://en.wikipedia.org/wiki/Binomial_theorem. The terms obtained from the binomial expansion of the above expression are: x^(n - k) * (2 / x²)^k = x^(n - k) * (2^k / x^[2k]) What we want is for the powers (of the x term) to be the same so that they will cancel each other (because the second term will divide the first). Write the equation that will make the powers the same: n - k = 2k You know n, so solve for k. Once you have k, solve 2^k (that's the rest of the expansion I did that did not get canceled), and then find the binomial coefficient using the formula on the Wikipedia page.
MATH HELP PLEASE.....How do I solve 9x-10=4x?
i am trying my best to explain it... 9x-10=4x u combine the like terms and when u move the like terms together you must change the sign 9x-4x=-10 5x=10 multiplication solved by division x=2 ok 55d-(30d-61) that's a tough question ok so let me explain it insert a 1 between the parenthesis and the sign every time u see this kind of problem..now its 55d-1(30d-61) 55d-30d+61 25d+61 ok 8.7m+0.32n-0.31m+7.4n you just combine the expression that are alike: 8.7m-0.31m+0.32n+7.4n just solve it now 8.39m+16.11n translation means change in math, words into algebraic expression 40 more than B ok whenever it says more than u switch the problem around like u put the last in the first and first to the last b+40 that's the translation of 40 more than B i hope i get to be chosen as the best answer...i tried my best to explain it.. if not its ok ....good luck i hope u under stand the way i explain it.. bye
Who can solve this math riddle?
To figure this out, we first assume that all the ships are already in motion and you are now asking the question about one single ship. A ship from San Francisco will meet a Tokyo ship every 12 hours (½ day). It seems like it should be every 24 hours, but remember that both ships are heading towards each other, so the time is half as long. As the ship first leaves dock, it will immediately encounter the ship that left Tokyo 8 days prior. Then it will meet two ships for every 12 hours (2 per day) of its journey until it docks in Tokyo to see the departing ship there. Altogether over 8 days it will see 17 ships: 1 + 8 * 2 = 17 Here's a break down by time (in days): t = 0 --> meet #1 (left Tokyo 8 days ago) t = ½ --> meet #2 (left Tokyo 7½ days ago) t = 1 --> meet #3 (left Tokyo 7 days ago) t = 1½ --> meet #4 (left Tokyo 6½ days ago) t = 2 --> meet #5 (left Tokyo 6 days ago) t = 2½ --> meet #6 (left Tokyo 5½ days ago) t = 3 --> meet #7 (left Tokyo 5 days ago) t = 3½ --> meet #8 (left Tokyo 4½ days ago) t = 4 --> meet #9 (left Tokyo 4 days ago) t = 4½ --> meet #10 (left Tokyo 3½ days ago) t = 5 --> meet #11 (left Tokyo 3 days ago) t = 5½ --> meet #12 (left Tokyo 2½ days ago) t = 6 --> meet #13 (left Tokyo 2 days ago) t = 6½ --> meet #14 (left Tokyo 1½ days ago) t = 7 --> meet #15 (left Tokyo 1 days ago) t = 7½ --> meet #16 (left Tokyo ½ day ago) t = 8 --> meet #17 (is just leaving Tokyo) P.S. I just found this same problem at "Ask Dr. Math" on the Math Forums and it confirms my answer. Check the link below.
Peanut math problem need help!!!?
In the "Peanuts" cartoon, solve the problem that is sending peppermint patty into an agitated state. How much cream and how much milk, to the nearest hundredth of a gallon, must be mixed together to obtain 50 gallons of cream that contains 12.5% butterfat?? Problem number six..... "How many Gallons of cream contains 25% butter fat and milk contains 3 1/2% butter fat must be mixed to.... ...obtain 50 gallons of cream containing 12 1/2% butter fat? MA'AM, would you settle for twenty push-ups???? HELP ME PLEASE!!!
Can someone help me on math homework? subject: angles pretty simple geometry, 10 pts to best answer!
13) Two angles that form a straight line are supplementary. Their sum is 180degrees. So for this one just set: y + 20 + 110 = 180 and 2x + 40 + 70 = 180 So you have the ordered (x,y) pair of: (35 , 50) Also, if you want another theorem(or corallary, I can't remember exact terms for them), opposite angles are the same. So when you plug in 35 for x, that is the result that is equal to it's opposite of 110. The same with y:70 equals (50) + 20 14) The y in this problem is just basically given to you since you already know the "opposite" rule. So we know right off hand that y equals 168degrees. Now, to solve for x you can do it two ways: 1: x = 5x - 48 4x = 48 x = 12 OR 2: x + 168 = 180 x = 12 Your (x,y) ordered pair is: (12 , 168) 15)From the previous problems, you should be able to find x within seconds: x + 48 = 64 x = 16 For y, the same concept applies: You can do it one way and check it by doing it a different way: 14y - 24 = 8y + 36 6y = 60 y = 10 OR 14y - 24 + 64 = 180 14y = 140 y = 10 Your (x,y) ordered pair is: (16 , 10) 16)This is simple as well. For x, just use supplementary angles: 2x - 5 + 75 = 180 x = 55 For y, we can still double check the work by using supplementary angles and then setting what we found equal to what we found for x: y + 75 = 180 y = 105 (105) = 2(55) - 5 105 = 105 It checks out and the ordered pair for this answer is: (55 , 105) One more piece of advice that I sometimes forgot to follow was that when you find an answer for a variable, make sure that it makes sense. Like, if you are dealing with an obtuse angle, make sure that when you plug in the variable you get an angle greater than 90degrees. Anyway, I hope I helped! Good luck in Geometry!!