Friday

April 18, 2014

April 18, 2014

Number of results: 994

**Physics**

Fr = ((Vs+Vr)/(Vs-Vo))*Fo = 205 Hz ((336+0)/(336-Vo))*200 = 205 ((336)/(336-Vo))*200 = 205 67,200)/(336-Vo) = 205 205(336-Vo) = 67200 68,880-205Vo = 67200 -205Vo = 67200-68880 = -1680 Vo = 8.2 m/s. = Speed of the train.
*Tuesday, September 3, 2013 at 5:59pm by Henry *

**algebra**

(55+80+70+x)/4<=65 can you help me 205-205+x<=260-205 x<=55 is this right
*Sunday, September 18, 2011 at 10:29pm by babygirl*

**Chemistry**

I agree it probably can be done with a proportion; however, I wouldn't set up the proportion that way. 0.160 is for x 0.205 is not for 1.5 ppm but for 1.5 ppm+xppm. (0.160/x) = (0.205/1.5+x) and solve for x. Another way. 0.205 = 1.5 + x 0.160 = ......x ----------- 0....
*Monday, March 5, 2012 at 4:05am by DrBob222*

**algebra**

Let x = the shortest side x + 10 + x + 2x-5 = 205 4x + 5 = 205 4x = 200 x = 50
*Monday, February 28, 2011 at 3:53pm by Ms. Sue*

**Stats**

1. Z = (s-205)/11 = 205/11 = 18.63??? This one confuses me 2. I have no idea how to do this.... Please help!!
*Saturday, October 15, 2011 at 12:22pm by Jennifer*

**MATH**

I will use A for alpha cot A = tan(A + 40) 1/tanA = (tanA + tan40)/(1 - tan^2A) let tanA = x 1/x = (x + .8391)/(1-x(.8391)) x^2 + .8391x = 1 - .8391x x^2 + 1.6782x - 1 = 0 by the formula x = (-1.6782 ฑ √6.816355)/2 = .4663 or -2.1445 so tanA = .4663, then A = 25ฐ or 205...
*Wednesday, August 31, 2011 at 8:48pm by Reiny*

**math**

93logd = 215. Divide both sides by 93: logd = 2.3118, Exponential form: 10^(2.3118) = d, 205 = d, d = 205.
*Sunday, November 14, 2010 at 3:49pm by Henry*

**physics**

F = mg = 1250kg * 9.8N/kg = 12,250 N. P = F*V = 12250 * 16.8m/s = 205,800 J/s = 205,800 Watts.
*Monday, March 12, 2012 at 3:54pm by Henry*

**maths**

f(208)=208/207+208/207ื206/205+208/207ื206/205ื204/203+
=208 This can be obtained by induction: Let f(n) be the function such that: f(n)=n/(n-1) + n/(n-1)*(n-2)/(n-3) +
n/(n-1)*(n-2)/(n-3)
(2/1) where n is even. Then f(2)=2/1=2 f(4)=4/3+(4/3)*(2/1)=4/3+8/3=4 f(6)=6/5+(6/5...
*Wednesday, June 5, 2013 at 4:38am by MathMate*

**crt 205**

The class is Critical Thinking 205 at University of Phoenix Axia. Jessica, I can only presume to know what the final project is since I have taken the course, but you need to specify what it is by including it in your post.
*Monday, December 22, 2008 at 3:05pm by Suz*

**205**

You must have posted this on the wrong website. We do not know what subject "205" is. We have no access to any college (including Axia) materials. Please pass the word.
*Tuesday, July 6, 2010 at 2:59pm by Ms. Sue*

**CRT 205**

What does CRT 205 stand for? Is this English, law, etc.? Sra
*Friday, April 10, 2009 at 11:31pm by SraJMcGin*

**CRT/205**

I am currently in CRT/205. This class is not that hard if you read all the material and ask questions to your instructor or go to the critical thinking lab located under the library tab.
*Wednesday, March 31, 2010 at 9:07pm by Anonymous*

**Physics 141**

Distance axle above step edge = .33 - .125 = .205 Angle T is angle between straight down from Axle and the top edge of the step so cos T = .205/.33 T = 10.58 deg sin T = .184 Take moments about the top edge of the step. F * .205 clockwise (my step is on the right) m g * .33 ...
*Saturday, April 5, 2014 at 4:30pm by Damon*

**chemistry**

Zenapaxฎ is used as part of immunosuppressive therapy for transplant patients. The dosage is calculated on the basis of 1.0 mg/kg. What is the proper dose for a 175 lb patient? (Given: 1kg = 2.205 lbs) 175*2.205=385.875
*Wednesday, February 1, 2012 at 12:16am by kr*

**math**

1) initial investment = $600 annual % rate = ? time to double = ? amount after 10 years = $19,205.00 19,205 = 600e^(10r) I do not know how to solve for r of find the time to double.
*Monday, February 8, 2010 at 10:03pm by Hannah*

**chemistry**

.........N2 + 2O2 ==> 2NO2 .......191.5..205.0...240 dSrxn = (n*products)-(n*reactants) dSrxn = (2*240) -[(191.5+2(205)]
*Saturday, April 21, 2012 at 11:52am by DrBob222*

**(Please help) Math**

1) initial investment = $600 annual % rate = ? time to double = ? amount after 10 years = $19,205.00 19,205 = 600e^(10r) I do not know how to solve for r of find the time to double.
*Tuesday, February 9, 2010 at 3:50pm by Hannah*

**math(Please please help!!!)**

1) initial investment = $600 annual % rate = ? time to double = ? amount after 10 years = $19,205.00 19,205 = 600e^(10r) I do not know how to solve for r of find the time to double.
*Monday, February 8, 2010 at 4:36pm by Hannah*

**math**

The sum of two consecutive numbersis 205. whatare two numbers? (hint:let n represent the first number. then n + 1 represents the second number. ALternatively: 205/2 = 102.5 making the two numbers 102 and 103.
*Thursday, November 1, 2007 at 11:05am by tchrwill*

**(Please help) Math**

That looks like a misprint, but if the investment really increased that fast, 19,205 = 600 (1 + r)^10 10 ln (1+r) = ln (19,205/600) = ln 32.0 = 3.4657 ln (1+r) = 0.34657 1+r = 1.414 r = 41.4% The time to double is 2.0 years
*Tuesday, February 9, 2010 at 3:50pm by drwls*

**math**

Since there are 123 girls, that should be a ratio of 3:5. 123/3 = 41 41* 2 = 82 Check 123 + 82 = 205 82/205 = .4 = 2/5
*Saturday, March 23, 2013 at 12:10pm by PsyDAG*

**statistics**

1. Suppose a certain instructor is interested in determining the average final grade for all students taking STA 205 in any given semester. Suppose she obtains the academic records for 100 former STA 205 students. For this scenario, determine both the population and the sample...
*Tuesday, January 18, 2011 at 4:03pm by hussain*

**Physics**

Wb = m*g = 21kg * 9.8N/kg = 205.8 N. = Wt. of the block. Fp = 205.8*sin33 = 112.1 N. = Force parallel to the incline. Fe-Fp = m*a = m*0 = 0 Fe - 112.1 = 0 Fe = 112.1 N. = Force exerted.
*Wednesday, February 19, 2014 at 10:56pm by Henry *

**Math**

Determine 208/207+(208/207ื206/205)+(208/207ื206/205ื204/203)+..+(208/207ื206/205ื204/203..4/3*2/1)
*Wednesday, June 5, 2013 at 2:05pm by Clavin*

**MATH**

2.5x-15 = 220 add 15 to each side of the equation to get 2.5x -15 + 15 = 220 + 15 2.5x = 205 Now divide both sides by 2.5 2.5x / 2.5 = 205 /2.5 x = 82
*Wednesday, October 14, 2009 at 9:00pm by MathMate*

**Stats**

1. Z = (183-205)/11 = -22/11 = -2 Z = (227-205)/11 = ? Apply the theorem. 2. 3 = (score-297)/18 -3 = (score-297)/18 Solve for each score.
*Saturday, October 15, 2011 at 12:22pm by PsyDAG*

**maths**

Determine 208/207+208/207ื206/205+208/207ื206/205ื204/203+
*Wednesday, June 5, 2013 at 4:38am by rohit*

**Physics!**

1 gallon of water has a mass of 3.785 kg. Therefore you are heating M = 54.4*3.785 = 205.9 kg. The total enegy required to heat that amount of water 40.2 C is 205.9 kg*40.2 C*4184 Cal/kg*C = 3.463*10^7 J Divide that by the watts of the heater (4910) to get the required time in...
*Sunday, November 11, 2012 at 10:28am by drwls*

**Math**

50+60+95=205 not 210 Further, the 60 Mil. people not working also aren't looking for work, so they would not be counted (at least in the American system), making the answer 0% which is imposable. If they were counted, the only ones you are looking at are them. SO... 205/60 = 3...
*Saturday, February 23, 2013 at 3:56pm by Ken*

**Chemistry**

I'm getting: d = 0.205 D = 0.0000003091 0.205 / 0.0000003091 = 663215.78777094791329666774506632 663215.78777094791329666774506632 / 60 / 60 = 184.22660771415219813796326251842 I'm putting 184 in but it's wrong...any ideas where my math is going wrong?
*Saturday, December 15, 2012 at 4:19pm by Anonymous*

**physics**

F = m*g = 21kg * 9.8N/kg = 205.8 N. Fk = u*mg = 0.5 * 205.8 = 102.9 N.=Force of kinetic friction. Fe-Fk = m*a Fe-102.9 = m*0 = 0 Fe = 102.9 N. = Force exerted. Pe = Fe * d/t = 102.9 * 15/20 = 77.18 N. = Power exerted.
*Thursday, October 24, 2013 at 3:05pm by Henry *

**physics**

A 2.15 kg mass is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 18.7 N is required to hold the mass at rest when it is pulled 0.205 m from its equilibrium position (the origin of the x axis). The mass is now released from rest with an ...
*Wednesday, November 17, 2010 at 8:44pm by Anonymous*

**crt 205 week 5**

Crt 205 week 5
*Wednesday, October 8, 2008 at 2:45am by Anonymous*

**Physics**

Fr = 22,000N @ 0o+80,900N @ 135o = Resultant force. A. X = 22000 + 80900*cos135 = -35,205 N. Y = 80900*sin135 = 57,205 N. Fr^2=X^2 + Y^2 = (-35205)^2 + (57205)^2 Fr = 67,170 N = Resultant force. a = Fr/m = 67170/15700 = 4.3 m/s^2. B. tanAr = Y/X = 57205/-35205=-1.6249 Ar = -58...
*Tuesday, January 29, 2013 at 8:17am by Henry*

**Buffer solution**

Frankly, I would use the Henderson-Hasselbalch equation. I assume your instructor has a reason for working the problem another way. Here is what I would do to follow the ICE method. CH3NH2 is a base. Think NH3. We write that equilibrium as NH3 + HOH ==> NH4^+ + OH^- CH3NH2 ...
*Saturday, March 27, 2010 at 7:34pm by DrBob222*

**Math 1200**

a complex has 205 apartment units. When the rent is $635 per month all 205 units are occupied.Assume that for each $75 increase in rent 15 units becomes vacent. Assume that the number of occupied units is a linear function of the rent. Each occupied unit also requires an ...
*Thursday, April 5, 2012 at 4:27pm by Andrew*

**algebraic reasoning**

the relationship b/w the distance d, in feet, required to stop a vehicle and s, the speed in miles per hour that the vehicle was traveling, is given by the equation d=0.0155s(squared)/f f represents the coefficient of friction b/w the tires and the road *it took a car 205 feet...
*Saturday, February 23, 2008 at 6:34pm by megan*

**AED/205**

I am working on my final for AED/205. The question is "Do students come in to class with a desire to learn or is it the teacher's job to foster a desire to learn?" I feel that when kids are younger they want to learn as they get older maybe not so much, they just want to grow ...
*Wednesday, May 4, 2011 at 9:26am by Jeanette*

**205**

C
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**205**

true
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**205**

true
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**205**

true
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**AED 205**

Thank you.
*Sunday, March 29, 2009 at 11:42am by Nicole*

**geometry**

205
*Monday, July 12, 2010 at 10:29am by nelson*

**physics**

205
*Wednesday, November 24, 2010 at 12:59pm by Anonymous*

**college**

205
*Saturday, February 6, 2010 at 7:30pm by sha*

**Math**

d 205
*Tuesday, March 4, 2014 at 2:08pm by DoItForTheVine*

**Solid Mechanics**

E=205 kN/mm²= 20510⁹ N/m², μ = 0.3 σ=Eε(longitudinal) F/A = Eε(longitudinal) ε=ε(longitudinal) = F/AE= =40000/(2010⁻³)²20510⁹=4.910⁻⁴. If a=0.2 m, b=c=0.002 m, ε=4.910⁻⁴, ...
*Thursday, July 25, 2013 at 11:29am by Elena*

**solid mechanics**

E=205 kN/mm²= 20510⁹ N/m². σ=Eε(longitudinal) F/A = Eε(longitudinal) ε=ε(longitudinal) = F/AE= =40000/(2010⁻³)²20510⁹=4.910⁻⁴. If a=0.2 m, b=c=0.002 m, ε=4.910⁻⁴, μ = 0.3 ...
*Thursday, July 25, 2013 at 12:48pm by Elena*

**Science**

You might find help in some of the following links; http://search.yahoo.com/search?fr=mcafee&p=the+mammal+order+with+the+dental+formula+205%2F105 Sra
*Tuesday, May 4, 2010 at 10:59pm by SraJMcGin*

**Chemistry, acids and bases**

(a) Find pH and pOH of 1.0 M solution of sodium benzoate, NaC6H5COO. (Ka of benzoic acid is 6.2x10^-5) (b) Calculate pH after 0.205 moles per liter of HCl was added (assume volume does not change (c) Calculate pH after 1.0 moles per liter of HCl was added (assume volume does ...
*Tuesday, May 6, 2008 at 8:45pm by Marissa*

**INB 205**

Thank you so much!
*Thursday, September 11, 2008 at 7:06pm by Chrisitne*

**205**

2.c 3.c 4.c 6.c 7.c 8.c 10.c 11.b
*Thursday, May 8, 2008 at 5:37pm by Tia*

**205**

Probably true
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**CRT 205**

And what is your question?
*Wednesday, December 10, 2008 at 9:36pm by bobpursley*

**crt 205**

What is your question? ??
*Sunday, March 8, 2009 at 2:36pm by Writeacher*

**205**

I can say 100% that it IS A.
*Wednesday, July 9, 2008 at 3:26pm by Anonymous*

**crt/205**

what is the answer
*Monday, June 22, 2009 at 10:07pm by pinky*

**CRT 205**

Yes.
*Saturday, September 5, 2009 at 7:23pm by Ms. Sue*

**CRT 205**

Yes.
*Sunday, November 1, 2009 at 8:32pm by Ms. Sue*

**205**

A is the correct answer.
*Wednesday, July 9, 2008 at 3:26pm by anonymous*

**CT. 205**

dkjj
*Wednesday, October 28, 2009 at 6:16pm by Brandie*

**eco 205**

Thank you GuruBlue
*Monday, January 18, 2010 at 11:55am by Mary*

**hum/205**

What exhibition??
*Wednesday, March 31, 2010 at 3:18am by Writeacher*

**205**

1 b 2 c 3 c 4 c 5 b 6 c 7 c 8 c 9 a 10 c 11 b 12 c
*Thursday, May 8, 2008 at 5:37pm by k*

**Hum/205**

What is your question?
*Sunday, October 3, 2010 at 3:36pm by Writeacher*

**AED/205**

Thanks so much!!!
*Wednesday, May 4, 2011 at 9:26am by Jeanette*

**CRT 205**

argument
*Sunday, November 22, 2009 at 5:55pm by dave*

**math**

0.205 * 7700 = ?
*Monday, December 9, 2013 at 2:58pm by Sam*

**chemistry**

205 g of HNO3
*Sunday, January 24, 2010 at 1:51am by Anonymous*

**Calc**

At a time of t hrs, let the position of ship A be P and let the position of ship B be Q Join PQ, and complete the large righ-angled triangle having a base of 100 and a height of 15t + 30t or 45t (the horizontal distance between them is always 100 km PQ^2 = 100^2 + (45t)^2 2 PQ...
*Saturday, November 23, 2013 at 10:52pm by Reiny*

**Chemistry**

Calculate ∆G (in kJ) at 541 K for the following reaction under the given conditions. Use data obtained from the table below and assume that ∆Hfo and So do not vary with temperature. Report your answer to three significant figures in scientific notation (i.e. 1.23E4...
*Sunday, February 17, 2008 at 5:24pm by Matt*

**chemistry**

Assuming 12 mL ethanol + 48 mL H2O = 60 mL (not a true assumption), (12 mL/60)*100 = ? b. 12 mL ethanol has a mass of m = volume x density = 12*0.789 = about 9.46 g convert to mols = g/molar mass = about 0.205 mols. Again, assuming the volume is 60 cc (not true), then M = mol/...
*Tuesday, March 19, 2013 at 4:52pm by DrBob222*

**physics**

This looked a little sloppy imo... Here's my example 1. Find most probably velocity of Krypton v = √[(2*8.314J/K*273K)/(0.083798 kg)] = 237.75 m/s Find the ratio of each velocity 205/237.75 = .8808 The average is .8851 207/237.75 = .8894 Find the ratio of change in ...
*Wednesday, April 18, 2012 at 2:38pm by Erica*

**205**

1 b 2 c 3 c 4 c 5 b 6 c 7 c 8 c 9 a 10 c 11 b 12 c
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**crt 205**

What is your question?
*Saturday, July 11, 2009 at 7:02pm by Ms. Sue*

**crt 205**

Wishful Thinking
*Sunday, October 28, 2007 at 10:13pm by Dave*

**INB 205**

You're very welcome.
*Thursday, September 11, 2008 at 7:06pm by Ms. Sue*

**crt 205**

And your question is??
*Friday, January 9, 2009 at 4:17pm by Ms. Sue*

**CTR 205**

yes. Do you have a question?
*Thursday, January 22, 2009 at 8:21pm by bobpursley*

**205 ctr**

What is your specific question?
*Thursday, February 12, 2009 at 4:47pm by Writeacher*

**205 ctr**

And your question is?
*Thursday, February 12, 2009 at 4:49pm by Ms. Sue*

**205 ctr**

in what categorizing fallacies it is
*Thursday, February 12, 2009 at 4:49pm by dalia*

**205 ctr**

in what categorizing fallacies it is
*Thursday, February 12, 2009 at 4:47pm by dalia*

**205 ctr**

Hasty generalizing
*Thursday, February 12, 2009 at 4:49pm by dalia*

**crt 205**

Final Project
*Sunday, April 12, 2009 at 5:28pm by Jim*

**crt/205**

define utilitarianism
*Thursday, April 23, 2009 at 11:04pm by Anonymous*

**205**

3.c 4.b 6.b 7.b 8.c 10.c 11.b 12.a
*Thursday, May 8, 2008 at 5:37pm by Anonymous*

**CRT 205**

Identify fallacies
*Sunday, May 31, 2009 at 1:46pm by Toni*

**CRT 205**

Taking a position
*Thursday, August 27, 2009 at 2:39am by Lily*

**CRT/205 critical thinking**

Thank you!!!!
*Friday, October 16, 2009 at 5:34pm by Linda*

**crt 205**

Punishment by death.
*Sunday, October 25, 2009 at 11:15pm by MattsRiceBowl*

**crt 205**

what is capital punishment
*Sunday, October 25, 2009 at 11:15pm by precious*

**eco 205**

What exactly do you need help on?
*Tuesday, November 17, 2009 at 8:09am by economyst*

**CRT 205**

proof?
*Sunday, November 22, 2009 at 5:55pm by Ms. Sue*

**CRT 205**

Thank you that was very helpful.
*Tuesday, February 9, 2010 at 9:36pm by Austin*

**CRT 205**

You're very welcome.
*Tuesday, February 9, 2010 at 9:36pm by Ms. Sue*

**crt-205**

thanks, what type of rhetorical is this?
*Friday, March 12, 2010 at 12:23pm by sam*

**Crt 205**

1 and 3 are correct.
*Saturday, March 20, 2010 at 1:51am by Ms. Sue*

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