Sabretooth Posted October 24, 2007 Posted October 24, 2007 Okay, here's me in action: 700/6 = 116.6(recurring), rounded off to 117 700 - 117 = 583, this the amount Mike pays in dollars. 583/5 = 116.6 Therefore, 583 - 116.6 = 466.4 Final Answer: $466.4. If that is right, this was easier than pie.
Commander Obi-Wan Posted October 24, 2007 Posted October 24, 2007 Sabretooth said: Okay, here's me in action: 700/6 = 116.6(recurring), rounded off to 117 700 - 117 = 583, this the amount Mike pays in dollars. 583/5 = 116.6 Therefore, 583 - 116.6 = 466.4 Final Answer: $466.4. If that is right, this was easier than pie. Close enough, if you don't round off until the end, your answer should be $466.66. But, you did it correctly, and probably still deserve full marks. Your turn.
tk102 Posted November 6, 2007 Author Posted November 6, 2007 Using only lower case letters of the English alphabet, how many 4-letter "words" can be created using unique letters with at least one vowel(a,e,i,o,u,y) in each word? No other rules apply.
Darth333 Posted November 6, 2007 Posted November 6, 2007 tk102 said: Using only lower case letters of the English alphabet, how many 4-letter "words" can be created using unique letters with at least one vowel(a,e,i,o,u,y) in each word? No other rules apply. When I saw this thread I could only think of one four letter word with at least one vowel and no repetitions: "nerd" (j/k) Sorry for the unorthodox method but I don't remember my math Total words with 4 letters (26x25x24x23) - Total words with no vowels (20x19x18x17) So the answer should be 242520 if I am correct.
tk102 Posted November 6, 2007 Author Posted November 6, 2007 Darth333 takes the short cut to the right answer! Here's the long method: Words with 1 vowel (V) in them can have the vowel appear in one of the 4 positions with the others being taken by consonants ©. VCCC CVCC CCVC CCCV and there are 6 possible vowels. The first consonant has 20 possibilities, the 2nd has 19, and the 3rd has 18 so: 1 vowel words = (6*20*19*18) + (20*6*19*18) + (20*19*6*18) + (20*19*18*6) = 164160 2 vowel words: VVCC VCVC VCCV CVVC CVCV CCVV six variations of (6*5*20*19) = 68400 3 vowel words: VVVC VVCV VCVV CVVV 4 variations of (6*5*4*20) = 9600 and 4 vowel words: 1 variation of 6*5*4*3 = 360 add them all up: 164160+68400+9600+360=242520 I like D3's way better.
Darth333 Posted November 6, 2007 Posted November 6, 2007 "Why make simple when we can make complicated?" If it was on paper that answer would have taken at least one more tree I didn't had the courage to go through that!
tk102 Posted November 27, 2007 Author Posted November 27, 2007 You have 50 coins from the USA totaling $10.00. You have more nickels ($0.05) than pennies ($0.01), you have more dimes ($0.10) than nickels, and you have more quarters ($0.25) than dimes. How many coins of each type do you have? (two answers)
Boba Rhett Posted November 28, 2007 Posted November 28, 2007 34 Quarters, 14 Dimes, 2 Nickels and 0 Pennies?
tk102 Posted November 28, 2007 Author Posted November 28, 2007 That's one! But you didn't show your work! 25q + 10d + 5n + p = 1000 q + d + n + p = 50 q > d > n > p p must be divisible by 5 in order to get the nice round number of $10.00 Test: can p = 5? pennies nickels dimes quarters COUNT VALUE 5 6 7 32 50 $9.05 No! Even if we maximize the number of quarters, we can't reach $10 with 5 pennies in play. Therefore pennies = 0. Now we can easily reach $10... pennies nickels dimes quarters COUNT VALUE 0 1 2 37 50 $12.00 We'll try shifting the coins from quarters to dimes until we fall below $10.00 pennies nickels dimes quarters COUNT VALUE 0 1 16 33 50 $9.90 So we know we need at least 34 quarters. Try that q=34 (VALUE = $8.50) n+d = 16 5n+10d = 150 5(16-d) + 10d = 150 80 + 5d = 150 d = 14 n = 2 pennies nickels dimes quarters COUNT VALUE 0 2 14 34 50 $10.00 What about 35 quarters? q=35 (VALUE = $8.75) n+d = 15 5n+10d = 125 5(15-d) + 10d = 125 75 + 5d = 125 d = 10 n = 5 pennies nickels dimes quarters COUNT VALUE 0 5 10 35 50 $10.00 What about 36 quarters? q=36 (VALUE = $9.00) n+d = 14 5n+10d = 100 5(14-d) + 10d = 100 70 + 5d = 100 d = 6 n = 8 ! no d must be greater than n. That is all.
Boba Rhett Posted November 28, 2007 Posted November 28, 2007 I don't have any work. Is the other one 35 Quarters, 10 Dimes, 5 Nickels, and 0 Pennies?
Boba Rhett Posted November 28, 2007 Posted November 28, 2007 I'll start out with an easy one - mainly because I'm rusty with the longhand of this too. Calculate the contribution (In Amps) of the 480 V source to Io (the current through the 140 Ohm Resistor). Hint: Super-positioning is the shortcut through this.
tk102 Posted November 28, 2007 Author Posted November 28, 2007 To determine the contribution of the 480V, we isolate it by opening the current source and shorting the other voltage source... 100 140 +-/\/\/\---/\/\/\---+------+ | | | -+- | \ 150 480V | / -+- | \ | | | +-------------------+------+ 240 +----------/\/\/\---+ | | -+- | 480V | -+- | | | +-------------------+ 480V through 240ohms = 2A
tk102 Posted November 28, 2007 Author Posted November 28, 2007 Looking at the 2nd voltage source <---- 720V through 240ohm = -3A 100 140 +-/\/\/\---/\/\/\---+------+ | | | | --- \ 150 | 720V / | --- \ | | | +-------------------+------+ The current source is a little trickier. Need to find the equivalent voltage. 1/R = 1/R1 + 1/R2 1/R = 1/100 + 1/140 R = 58.33 12A * 58.33 Ohm = 700V 100 140 +-/\/\/\-+-/\/\/\---+------+ | | | | | ^ | \ 150 | 12A | / | ^ | \ | | | | +--------+----------+------+ 58.333 58.333 +-/\/\/\---+ +-/\/\/\---+ | | | | ^ | ---- | 12A | ----> 700V | ^ | ---- | | | | | +----------+ +----------+ 100 140 700V through 140 ohms = 5A +-/\/\/\-+-/\/\/\---+------+ | | | | | ---- | \ 150 | 700V | / | ---- | \ | | | | +--------+----------+------+ Io: 2A + -3A + 5A = 4A P.S. Thank you Rhett. I completely forgot all of this stuff the minute I left my EE210 class. Good to know part of my brain could pick it back up again. I'll see if I can come up with another problem unless someone else posts one first.
tk102 Posted November 30, 2007 Author Posted November 30, 2007 On a hot summer's day a guy selling snow-cones runs out of cones and has to use paper cups instead. If the cups are the same height and same radius as the cones, and the snow-cones cost $0.75 each, how much should he charge for the snow-cups?
tk102 Posted November 30, 2007 Author Posted November 30, 2007 Via trigonometry: tan(theta) = 54.663/36.7 theta = arctan(54.663/36.7) theta = 56.123° sin(theta) = 54.663/x x = 54.663 / sin(theta) x = 65.84 or by Pythagoras: x = sqrt(36.7^2 + 54.663^2) = 65.84 but what about the snow-cone guy?
Boba Rhett Posted November 30, 2007 Posted November 30, 2007 (Pi)(R^2)(H) = area of cylinder (1/3)(Pi)(R^2)(H) = area of cone So a comparable cylinder is holding three times as much. (.75)(3) = $2.25 So the answer is that he darn well better still be charging 75 cents because there's no way in heck I'm paying that much for a snow cone.
Ctrl Alt Del Posted December 1, 2007 Posted December 1, 2007 tk102 said: On a hot summer's day a guy selling snow-cones runs out of cones and has to use paper cups instead. If the cups are the same height and same radius as the cones, and the snow-cones cost $0.75 each, how much should he charge for the snow-cups? Assuming that the paper cups are cylindric and having on mind that a cylinder have a volume that's 3 times greater than that of a cone, I would say: Cylinder radius and height = x Cones radius and height = x Therefore, they're equal. But the cylinder got a volume 3x greater. If logic applies, then he should sell the cups 3 times more expensive. Answer: $2.25 Edit: Oh, I didn't saw Rhett's reply...
tk102 Posted December 1, 2007 Author Posted December 1, 2007 Ah you're both right of course. I always thought it was strange that a cone was 1/3 the volume of the equivalent cylinder. After all, if you draw a rectangle and then draw and isosceles triangle inside it, that triangle has 1/2 the area of the rectangle. But if you spin that configuration on the axis, the volume encompassed by the triangle (now a cone) is 1/3 the volume encompassed by the rectangle (now a cylinder). Kind of spooky.
Det. Bart Lasiter Posted December 1, 2007 Posted December 1, 2007 tk102 said: On a hot summer's day a guy selling snow-cones runs out of cones and has to use paper cups instead. If the cups are the same height and same radius as the cones, and the snow-cones cost $0.75 each, how much should he charge for the snow-cups?The invisible hand of the free market will, like the hand of God Himself, set the price of a paper cup at its optimal price and ensure the that our way of life will not be threatened by the sanguine tentacles of Communism
Marius Fett Posted December 1, 2007 Posted December 1, 2007 How about this one: Find the solution to the equation X ³ + X = 40 Express the answer to 1 decimal place...
tk102 Posted December 1, 2007 Author Posted December 1, 2007 To 1 decimal place? Might as well use trial and error to get that rather than use the cumbersome cubic equation. 4^3 = 64 3^3 = 27 so we know it's between 3 and 4. 3.4 -> 42.7 3.3 -> 39.23 so 3.3 close enough.
Sabretooth Posted December 1, 2007 Posted December 1, 2007 tk102 said: To 1 decimal place? Might as well use trial and error to get that rather than use the cumbersome cubic equation. 4^3 = 64 3^3 = 27 so we know it's between 3 and 4. 3.4 -> 42.7 3.3 -> 39.23 so 3.3 close enough. Actually, I did find that out (with the help of Microsoft Windows Calculator), but I thought trial and error was evil in real maths.
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