# Posts by Quidditch

Total # Posts: 512

**C++**

First problem I see is in your first case 'Yes'. You output "Does the hard drive spin?", then you have a "break;". The break takes you out of the first enclosing switch. So, the cin does not get executed. This same error is in the first case 'No...

**Algebra**

when you write: (sqrt5/15) Which do you mean: sqrt(5/15) or (sqrt(5))/15

**Algebra**

If you let y=x^2 then: 25*y^2 - 20*y +3 = 0 You can use the quadratic formula to solve. Your solution for y will really be for x^2. Take the square root of y values to get the x values. I did not get any imaginary values. One answer (that I got) IS sqrt(3/5), but as a real ...

**Algebra**

That's the answer I got.

**math**

Calculate the total cost of each game including tax--remember that 8.125% = 0.08125 total cost = game + tax = $3.95 + 0.08125($3.95) Now divide the money he has ($20.92) by the total cost per game. Disregard any remainder/fraction/decimal of a game since you cannot normally ...

**Algebra**

For the equation of a line in the form: y = m*x + b m=slope b=y intercept The lines will be parallel if the slopes (m value) are the same. Rearrange the equations to the above form and check for m being the same value for each.

**Algebra 1**

Correct!

**Algebra**

Generally, the rational expression will be undefined when the denominator is 0. Find all values for z which would result in the denominator being 0.

**College Algebra**

For: Assume all values are in inches. L=length of rectangle W=width of rectangle P=perimeter of rectangle P=L + W + L + W The problem states: P=78 L=W + 3 Substitute the above values into the perimeter equation.

**math**

Generally, a fraction would be considered undefined if the denominator is 0. For the problem, what values of x would make the denominator 0?

**algebra**

p1 = -2,0 p2 = 0,1 slope = (1 - 0)/(0 - (-2) ) = 1/2 =0.5

**math final tmrw!!!!!!!**

(x + a)(x + a) =x^2 + 2ax + a^2 In the problem, 2ax=22x 2a=22 a=11 substituting back (x + 11)(x + 11) x^2 + 22x + 121

**Math**

The exponent is evaluated first. So, you would evaluate 4^x first, then multiply that by 3. If the expression had been written as: (3*4)^x, then you would multiply 3*4 first and raise that product to the x power.

**Math question - plz correct**

Glad to help!

**Math question - plz correct**

Actually, you would have to multiply them by 4 because they are squared. You have the right idea. For the legs use: b=2*450km=900km c=2*355km=710km This would then give you 800^2=900^2 + 710^2 - 2(900)(710)cos(A) Solve for A using the same method you did before.

**Math question - plz correct**

A quick look finds that you missed the fact that the time the airplanes are 800km apart is 2 hours not 1 hour. The legs should be 2*355km and 2*450km.

**Algebra**

A good way to be sure is to put the lines into the familiear y=mx+b form. The second equation is already there. If the value of m is the same, the lines are parallel.

**Algebra**

Since it says "factor completely", I think you are correct with D.

**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=?

**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 =?

**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))

**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 ...

**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.

**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 ...

**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 ...

**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

**physics- is this correct?**

That looks good. Remmeber there will be two phases that have a positive cosine value.

**physics**

looks OK. Don't forget the units.

**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.

**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.

**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.

**simple math check**

For the formula posted, I got about 1.72 also.

**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%

**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...

**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.

**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.

**5TH GRADE MATH NEED HELP ASAP**

^ means "raised to the power of" X ^ 2 = X * X X ^ 3 = X * X * X etc.

**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.

**C++ Programming**

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

**C++ Programming**

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

**C++ Programming**

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

**C++ Programming**

Does it compile OK? If so, what is the test input that is failing?

**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...

**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 ...

**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 ...

**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 ...

**physics**

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

**physics**

I think the horizontal velocity is 10.0m/s not 50.0m/s.

**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 ...

**physics(ignore previous)**

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

**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 ...

**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 ...

**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.

**Physics**

Your answer looks good! I got the same answer.

**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.

**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...

**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 ...

**physics**

I think both answers are correct.

**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.

**Physics(check)**

Yes! 2.52 m/s^2 is what I got also.

**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...

**Physics(check)**

That's the answer I got also.

**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.

**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?

**Physics Cont'd - Quidditch**

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

**Physics**

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

**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?

**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...

**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.

**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 ...

**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...

**Physics**

Check my reply to this yesterday at 12:27pm

**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...

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

I'm not sure what you mean by that.

**Algebra II**

Factoring x^2 - 2x - 24 =(x + 4)(x - 6) Roots are -4 and 6.

**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.

**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.

**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 ...

**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.

**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

**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

**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....

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

Good! That is what I got, also.

**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

**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 ...

**Math**

Part 1 I don't think you set up the equation correctly. See me first response

**Math**

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

**Math**

OK, I think you are correct 5 - |-2| =5-2 =3

**Math**

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

**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

**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.

**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 ...

**Math**

That is what I got, also.

**Math**

Yes, that is what I got.

**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.

**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

**math**

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

**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 =??

**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.

**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....