Thursday

December 18, 2014

December 18, 2014

Total # Posts: 495

**Algebra**

Since it says "factor completely", I think you are correct with D.
*May 27, 2008*

**math(is this how you work this)**

15-4x = 2(3x+1) 15-4x = 6x + 2 Combine all the 'x' terms on one side and the constant terms on the other side. 15 - 2 = 6x + 4x 13 = 10x So, x=?
*May 22, 2008*

**math (did i do this right)**

1. You substituted OK, but... 2(-1+3) =2(2) =? 2. Again, the substitution is OK, but... 4((-1)^2 - 2(2)(3)/2 + -1 + 3 =4(1) - 12/2 - 1 + 3 =4 - 6 - 1 + 3 =?
*May 22, 2008*

**Calculus**

Instead of radical form, write the problem in exponential form. f(x)=(3x + 15)^(1/2) f'(x)=(1/2)((3x+15)^(1/2 - 1)) 3 f'(x)=(3/2)(3x+15)^(-1/2) f'(x)=3/(2*(3x+15)^(1/2))
*May 21, 2008*

**algebra- FINAL UNIT!!**

Pick some variables to represent what you are trying to find. Here, let: X = smaller positive number Y = larger positive number Algebraically express all the relationships given by the problem. Y = 4X (240 / X) - 15 = (240 / Y) It usually seems to simplify things if you get ...
*May 20, 2008*

**Physics (check)**

The answer is correct. But, be careful in the work. In your work, you are using 1000kg when you mean 1000g or 1kg. You knew what you meant, but you might lose some points in the work.
*May 18, 2008*

**algebra**

X^2 -2X = 8 In "set to zero" form: X^2 - 2X - 8 = 0 factoring: (X + ?) * (X + ?) = 0 The two different numbers that are question marks must equal 8 when multiplying them together. This would suggest trying: 1, 8 or 2, 4 Since 8 is negative, one of the values must be ...
*May 15, 2008*

**algebra2**

It seems a challenge on inspection. If you put the coeffecients into the quadratic equation, you will quickly see that there is no real solution. you can still solve for the imaginary roots. If you did not expect that this might happen, check to be sure that you copied the ...
*May 15, 2008*

**physics- is this correct?**

arccos(0.4375)=+-1.118 radians. cos(-1.118 radians) = cos( (2*pi) - 1.118 ) =cos(5.1652 radians) either of these will give a cosine value of +0.4375
*April 29, 2008*

**physics- is this correct?**

That looks good. Remmeber there will be two phases that have a positive cosine value.
*April 29, 2008*

**physics**

looks OK. Don't forget the units.
*April 29, 2008*

**Physics**

You have the right idea. Yes, the AVERAGE acceleration is just as you stated, (ending velocity - starting velocity)/time. The problem says the starting and ending velocities are the same (always 400m/s). (400m/s - 400m/s) is 0. So the acceleration during this time is 0.
*April 29, 2008*

**Calculus**

rewrite as: f(x)=5 + 6*x^(-1) + 7*x^(-2) The derivative of the first term is 0. For the derivative of the terms with x, multiply the exponent times the coeffecient of the term, the subtract 1 from the exponent.
*April 29, 2008*

**algebra**

Assuming all terms on boths sides are under the radicals, square both sides to get rid of the radical signs. You will get: 7v-4 = 5v+10 Now solve for v.
*April 29, 2008*

**simple math check**

For the formula posted, I got about 1.72 also.
*April 29, 2008*

**algebra**

If grade is defined as rise over run expressed as a percent, then: the rise is: summit - base = 14318 - 9500 the run is the horizontal distance = 15847. Using these definitions, the grade is not 3.3%
*April 28, 2008*

**Algebra**

Train A has a 15 minutes (1/4 hour) "head start". Since it is traveling at 60 miles per hour, it will be how many miles ahead in 1/4 hour? Train B is going 20 miles per hour faster. So, every hour it gains 20 miles on train A. Calculate how far that train A is ahead...
*April 28, 2008*

**physics please help me**

The formula looks suspicious. The left side units are MgH and the rightmost term of the right side is Mg*2. Unless H represents some "unitless" parameter (not impossible, but unlikely), the units do not match.
*April 24, 2008*

**5TH GRADE MATH NEED HELP ASAP**

I forgot to add, the question is not stupid. No one knows what it means the first time the see it. It is always OK to ask what something means.
*April 22, 2008*

**5TH GRADE MATH NEED HELP ASAP**

^ means "raised to the power of" X ^ 2 = X * X X ^ 3 = X * X * X etc.
*April 22, 2008*

**5TH GRADE MATH NEED HELP ASAP**

One possible sequence: 1^1 2^2 3^3 etc. When series get this big so quickly, look for some possible exponent relationship.
*April 22, 2008*

**C++ Programming**

OK, Post the code you just tried. Also, which line is the compiler complaining about?
*April 9, 2008*

**C++ Programming**

Your braces are backwards after your "else" should be: else { } Your 3rd and 4th "if" statements have backward braces also.
*April 9, 2008*

**C++ Programming**

Looking again: Labels cannot begin with a number. 1: is wrong L1: would be OK Try changing your labels ( and goto's).
*April 9, 2008*

**C++ Programming**

Does it compile OK? If so, what is the test input that is failing?
*April 9, 2008*

**C++ Programming**

A few things to check: The first test is true for any number greater than 4. If you enter 5, the first test will go to label 1. Maybe the first test should be less than 4. Another item to check: You are not including the endpoints in the range check. From your code: if ( (nod...
*April 9, 2008*

**C++ Programming**

Just quickly looking... if (nod = 4 < 6) this will set nod true if 4 < 6. Probably this is not what you mean. To test for a range, you will have to separately test for the endpoints. if( (nod >=4) && (nod <= 6) ) { } would test for nod greater or equal to 4 AND ...
*April 8, 2008*

**Java**

My language familiarity is mostly, c, c++, assembler, but here is something to check in your code. In the code, you are inputing to the string variable "models"; however, the switch statement is on the int "model". The variable "model" is ...
*April 4, 2008*

**physics**

I am first pasting the previous response. Then, I will step through the steps. --------------- Break the problem into vertical and horizontal components. Find the initial vertical component (30.0 m/s) * sin(60) because the problem says the missile is launched 30 degress from ...
*April 2, 2008*

**physics**

Did you see the reply on 1 April at 11:44pm about this problem?
*April 2, 2008*

**physics**

I think the horizontal velocity is 10.0m/s not 50.0m/s.
*April 2, 2008*

**physics**

OK, here are the steps to go through to solve the time eqution: Just to be sure... The expression t^2 means t to the second power or t * t. Solving... 50.0m = (0.5)*(9.8 m/s^2)*(t^2) 50.0m = (4.9 m/s^2)*(t^2) dividing both sides by (4.9 m/s^2) (50/4.9)s^2 = t^2 10.2 s^2 = t^2 ...
*April 2, 2008*

**physics(ignore previous)**

OK, but when you get your answers, post them and we'll be sure they look good.
*April 2, 2008*

**physics**

The time that the ball is in the air just depends on how far it has to fall down to the ground. Since the ball was thrown HORIZONTALLY the initial VERTICAL velocity is 0. So, the ball immediately starts to drop. This happens at the same rate as if you just dropped it from the ...
*April 2, 2008*

**Physics - Quidditch**

You have it. Solve that last equation for the mass. I would strongly recommend that you keep the units in your equations--especially ones that are challenging. 90N - m *(10 m/s^2) = m * (1 m/s^2) 90N = m (11 m/s^2) (90N)/(11 m/s^2) = 8.18kg --------------- Your answer to the ...
*April 2, 2008*

**math**

Sketch this out. You will see that is a right triangle problem. Use the Pythagorean theorem: a^2 + b^2 = c^2 where c is the hypotenuse (long side) of the right triangle. a and b are the other sides.
*April 2, 2008*

**Physics**

Your answer looks good! I got the same answer.
*April 2, 2008*

**Physics**

I believe you are using 10.0 m/s^2 for g. Based on that, the acceleration is OK. The accleration is the same for both objects. So put the value you found for acceleration into your calculations for Object 2. Now you have just an equation with m as the only unknown.
*April 1, 2008*

**physics**

11) Correct 12) Separate the velocitie into vertical and horizontal components. Take the vertical velocity component and calculate how long the ball is in the air. Then, using that time and the horizontal component of the velocity, calculate the horizontal distance traveled. I...
*April 1, 2008*

**physics**

5) I did not get the same answer you got. Since there is no vertical component, the time to hit the ground is the same as if th ball had been dropped from a height of 50.0m. Calculate how long it takes for the ball to fall 50.0m. Take that time and multiply by the horizontal ...
*April 1, 2008*

**physics**

I think both answers are correct.
*April 1, 2008*

**Physics(check)**

12: Only A is SI units of speed. B is speed, but it is English units, NOT SI. The correct answer should be A.
*March 30, 2008*

**Physics(check)**

Yes! 2.52 m/s^2 is what I got also.
*March 30, 2008*

**Physics(check)**

Checking your answer for 10: For an initial velocity of 0, final velocity can be expressed as: V = SQRT(2 * acceleration * distance) Using your answer of 0.071 m/s: V = SQRT(2 * (0.071 m/s^2) * 1000m) V = SQRT( 142.0 m^2/s^2) V = 11.92 m/s Uh-oh! Not enough velocity by the end...
*March 30, 2008*

**Physics(check)**

That's the answer I got also.
*March 30, 2008*

**Physics(check)**

7A: The speed of the car is 15.0 m/s. The distance the car cover in 12.0 minutes is: distance = speed * time distance in meters = 15.0 m/s * 12.0 minutes * ( (60 seconds)/minute)) 8: Your answer is OK.
*March 30, 2008*

**Physical Science**

Assuming the top of the bottle is open... If you measured the water pressure at each hole, which hole would have the greatest pressure?
*March 17, 2008*

**Physics Cont'd - Quidditch**

You got it! Using 10m/(s^2) for gravity, 140m is correct.
*March 16, 2008*

**Physics**

You got it! Using 10m/(s^2) for gravity, your answer is correct.
*March 15, 2008*

**Physics**

I got something in that range, but I believe your answer has more error than I would expect. What did you get for time that the ball was in the air? What numbers did you plug into your final equation?
*March 15, 2008*

**Physics**

The ball lands 140m from the base of the building. The balls initial horizontal velocity is 20m/s. 140m = (20m/s) * t Solve for t. Since the time the ball is in flight is now known. Plug that into the "complete" formula for distance: d = (1/2)*(g)*(t^2) + v*t + h Let...
*March 15, 2008*

**math**

11.24 * (32/6) is exactly the same as 11.24 * (16/3). If the answers are different then there was a mistake in calculating either one or possibly both equations.
*March 6, 2008*

**Physics - Quidditch**

OK, here is what was stated for the continuation of the problem. I called this the second part. "Suppose the pilot disregarded the wind and had flown in a direction of 30 degrees N of E. Where relative to his destination would he be when he thought he should be at the ...
*March 2, 2008*

**Physics**

I don't know what you mean by "the angle between the pVw and wVg vector". For the second part: The draw the plane vector which is the same as the desired results. At the end of the plane vector, draw the wind vector. That is where the plane will end up. So, for a...
*March 2, 2008*

**Physics**

Check my reply to this yesterday at 12:27pm
*March 2, 2008*

**Algebra II**

I don't think it should matter. Perhaps your teacher might have some requirement. The answers are -4 and 6. These are both domain values for the function. When expressing domain and range values in a function, the x and y values are shown as (x,y) indicating (domain, range...
*March 1, 2008*

**Algebra II(does it matter what order they go in?)**

I'm not sure what you mean by that.
*March 1, 2008*

**Algebra II**

Factoring x^2 - 2x - 24 =(x + 4)(x - 6) Roots are -4 and 6.
*March 1, 2008*

**Physical science question 2**

Think conservation of momentum here. Relative to the shuttle, the momentum of the astronaut and camera is 0 (zero velocity realtive to the shuttle). After she throws the camera, her momentum plus the momentum of the camera is still 0.
*March 1, 2008*

**Physics**

The heading solved will be a heading relative to east (chosen since the original problem used that). You can adjust that to a heading with 0 degrees at true north if desired.
*March 1, 2008*

**Physics**

One method Three velocity vectors: the wind, the plane, and the desired results. Since the plane must reach the town in 1/2 hour, the results must be 400km/hr at 30 degrees N of E. That defines the speed over the ground. Draw the results vector. Now draw the wind vector as ...
*March 1, 2008*

**Math**

Another approach might be to work work the problem backwards. For a third degree polynomial, Q''' is just a constant. Set that to 12. Inegrate to get Q'' =12x + C Adjust the constant for the desired result. Repeat for Q' and Q.
*February 29, 2008*

**physics**

Find the component of the man's force parallel to the floor. Subtact the force due to kinetic friction to get the net force. Use acceleration=force/mass to find the acceleration. For an initial velocity of 0, use v=sqrt(2*acceleration*distance) to calculate the speed at 2.90m
*February 29, 2008*

**electricity 188**

(6 lamps) * (100W/lamp) * (5 hours/day) * (30 days) You will end up with a number of watt hours. Convert watt hours to kilo watt hours. Multiply that number by $0.04
*February 24, 2008*

**inequality**

A variable can be a symbol or word used to represent some value. For this problem W could be a variable representing the number of additional games that they must win to have won 75%. (number won so far) + (addtional won) >= 75% (total games) substituting... 9 + W >= 0....
*February 24, 2008*

**pre-algebra..7th grade**

Good! That is what I got, also.
*February 24, 2008*

**pre-algebra..7th grade**

If we add up everybody's time, we should get the total time of the CD. Let M represent the total time of the CD. Everything will be in minutes. M= (1/5)*M + 12 + (1/3)*M + 32 Combining terms... (1 - 1/5 - 1/3)*M=44 Solve for M
*February 24, 2008*

**Math: Dot Product**

For equilibrium on the ramp, the sum of the forces parallel to the ramp must be 0. For a ramp inclined 30 degrees, the force component of the box parallel to the ramp is 215N * sin(30). The frictional force is given as 27N up the ramp. Solve for the additional force up the ...
*February 24, 2008*

**Math**

Part 1 I don't think you set up the equation correctly. See me first response
*February 24, 2008*

**Math**

2. I am not sure what 5-/-2/ means. If / is supposed to be a bracket or parenthesis then: 5-[-2] =5+2
*February 24, 2008*

**Math**

OK, I think you are correct 5 - |-2| =5-2 =3
*February 24, 2008*

**Math**

1. The sum of a number and 17 more than twice the same number is 101 should be: n + (2n+17) = 101
*February 24, 2008*

**Math _ Vectors**

Just the last line is wrong. The angle 55 degrees references the resultant from the -x axis. Subtract that value from 180 to find the reference from the positive x axis 180 - 55=125
*February 20, 2008*

**Physics**

You have it! Solve the quadratic. You will get 2 answers. One of them, the lower, is the impact time. The other is a "phantom event" where again the distance 96m when the cars back up because the negative acceleration. That event, of course, does not happen.
*February 20, 2008*

**Math: Calculus - Vectors**

If the boat moves directly across then there is enough of the boats speed vector to counteract the river flow. That is 2m/s. So, the vector parallel to the shore (the river flow) speed is 2m/s. The TOTAL speed of the boat is 5m/s. Now solve this for the speed perpendicular to ...
*February 19, 2008*

**Math**

That is what I got, also.
*February 17, 2008*

**Math**

Yes, that is what I got.
*February 17, 2008*

**Math**

The total weight of each can packed with tuna is the weight of the can plus the weight of the tuna. Multiply that total can weight by 144 to get the weight of 1 gross in OUNCES. Now convert that to POUNDS as requested by the problem.
*February 17, 2008*

**math**

Let D= depth of the pool in inches Amaya's height in inches = (2/5)D - 2 inches From the problem, Amaya's height is 52 inches. 52 inches = (2/5)D - 2 inches
*February 17, 2008*

**math**

20*(3 7/8) + 19*(5/8) =20*(31/8) + 19*(5/8) =(620/8) + (95/8) =715/8 =89 3/8
*February 17, 2008*

**math**

There are 20 rows of brick and 19 rows of mortar. The total height would be: 20*(3 7/8)inches + 19*(5/8)inches =??
*February 17, 2008*

**algebra**

Substitute the value you found for C into either equation to get the one way time. The problem seems to indicate that you need the total time. That is the time it took Ann to swim up and back. Be sure to indicate that time is for one direction and the total time is twice that.
*February 13, 2008*

**algebra**

First find the speed at which Ann swims. 0.4km/20min = 0.02km/m Let C be the speed of the current Ann's upstream time is: 0.25km/(0.02km/m - C) Ann's downstream time is: 0.75km((0.02km/m + C) The problems states that they are equal. 0.25km/(0.02km/m - C) = 0.75km/(0....
*February 13, 2008*

**Physics**

Ohm's law can be stated as: (resistance in ohms)= (voltage in volts)/(current in amps) . For part (a): resistance=(1.5V)/(25A)=0.06 ohms Use the same equation for (b) and (c).
*February 12, 2008*

**Physics**

You are on the right track. Your conversion to m/s looks good. The question is how far does the car move in 0.71s. distance = rate*time = 16.111m/s * 0.71s =11.44m
*February 11, 2008*

**Speed and distance**

You have it figured out. 47(r-5)+9r=5r(r-5) 47r-235+9r=5r^2-25r Rearranging and combining you had an error. I got 5r^2-81r+235=0 Solve using the quadratic formula. One of the solutions of the quadratic will be less than 5 which would imply that the canoe would change ...
*February 9, 2008*

**chem**

Google is your friend. google: solute solution Read all of the first hit (Wikipedia).
*February 8, 2008*

**chemistry**

google for: solute solution Read all of the first hit.
*February 8, 2008*

**Physics II**

power = (voltage^2)/resistance
*February 8, 2008*

**Physics**

The formula for resistance is: total resistance = (resistivity) * (length)/(area) filling in from the problem statement: 0.32 ohms = (resistivity) * 1m/(area) looking up the resistivity for tungsten: at 20 C restivity = 5.28 * 10^-8 ohms*m leaving just the area unknown: 0.32 ...
*February 8, 2008*

**math**

For the first problem: x=length of side before the addition x+4=the length of the longer side after the the additional 4 feet is purchased. The new area of the property is: x(x+4)=9600 x^2 + 4x =9600 x^2 + 4x - 9600=0 which is: (x-96)(x+100)=0 Ignore the negative solution for ...
*January 29, 2008*

**Physics**

Also, it is assumed there is no friction with the ramp.
*January 28, 2008*

**Physics**

The initial potential energy of the watermelon is mass*acceleration*height or mgh for Earth. For practical purposes assume that total energy of the watermelon is constant. So the potential energy + the kinetic energy is constant. The kinetic energy is given at the point of ...
*January 28, 2008*

**science**

assuming there are no other forces (ignoring friction between the skates and the ice) force=mass*acceleration so... acceleration=force/mass acceleration=234n/45kg=??
*January 24, 2008*

**C++ Programming**

Nope, for some reason the last lines which are not posting. Anyway, the last lines are the same as your source.
*January 23, 2008*

**C++ Programming**

I guess the double blank lines caused the posting applet to clip the last two lines. Here they are finally, I hope. -------------------- return 0; }
*January 23, 2008*

**C++ Programming**

missed the end ------------------------- #include <tchar.h> #include <iostream> using namespace std; int _tmain(int argc, _TCHAR* argv[]) { cout << "-Calculate the area of a circle-" << endl; cout << endl; cout << "Radius = &...
*January 23, 2008*

**C++ Programming**

I did a quick check using MS VS2005. The following compiles correctly. Changing the line: cout<<endl; to: cout<<end; gives a compile error -------------------------------- Below compiles using VS2005 -------------------------------- #include <tchar.h> #...
*January 23, 2008*

**C++ Programming**

Did you intend for the line: cout<<end; to really be: cout<<endl;
*January 23, 2008*

**math**

x=first number y=second number 3*x + y = 18 2*x - y = 12 You need to eliminate one of the variables and solve for the remaining one. In this problem, you can see that if the two equations are added together, the y variable is eliminated: adding: 3x + y = 18 2x - y = 12...
*January 23, 2008*

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