Thursday

July 24, 2014

July 24, 2014

Total # Posts: 17,501

**algebra 2**

7^(1/5) = fifth root of 7

**Science**

It did not result. The force caused the acceleration ! F = m a F = 3000*20 = 60,000 N

**Algebra**

total females = 80+80+70 = 230 80 of those in humanities first 80/230 = .348

**algebra**

1/sin (7pi/6) 7 pi/6 is pi/6 or 30 degrees below -y axis in quadrant 3 so sin 7 pi/6 = -1/2 inverse is csc = -2

**math**

sketch it in quadrant 2 y = 1 x = -1 tan theta = y/x = -1

**geometry**

1 meter by 0.2 meter boards? area board = .2 m^2 area floor = 50 m^2 50/.2 = 500/2 = 250

**Math**

when does 4 x = 2 pi , a full circle ? x = 2 pi/4 = pi/2 the single amplitude is given, 3

**Algebra**

d^2 = (y2-y1)^2 + (x2-x1)^2 d^2 = 5^2 + 12^2 = 13^2 (5 12 13 right triangle) 13

**math help**

x = -7717/335 y = -36738/335 z= 16662/335

**typo?**

-3 ??? x ???? +13y+29z=81 and what do you want? A solution using Gauss Jordan ? If so go to: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

**algebra 1**

multiply both sides by -1 This changes the direction of the arrow 2 x/7 + 8 < 9 2 x/7 < 1 2 x < 7 x < 3.5

**math help**

Just told you, 4/5

**math help**

sin R = 3/5 = r/s t^2 = 5^2 - 3^2 = 25 - 9 = 16 so t = 4 tan T = t/s = 4/5

**Algebra 2**

ten scores in order: 81 81 82 84 85 86 89 93 94 95 sum = 870 mean = 870/10 = 87 median 85.5 (5 above, 5 under) mode = 81 (there are two of them)

**algebra1**

LOL - Yes, I suppose.

**algebra1**

******** -6 x^2 + 4 x + 3 ********____________________________ (3x+2) | -18 x^3 + 0 x^2 + 17 x + 6 *********-18 x^3 -12 x^2 *****************12 x^2 + 17 x + 6 *****************12 x^2 + 8 x *************************9 x + 6 *************************9 x + 6 remainder = 0 ========...

**Algebra 2**

area bulls eye = pi (6^2) total area = pi (16)^2 6^2/16^2 = 2^2*3^2 /2^8 = 9/2^6 = 9/64 = .140625

**Algebra/can you explain how this is set up**

150 * 150 * 150 ------ P1 times or log x = P1 log 150 using for example ten base then x = 10^log x

**Math help**

1, 2, 3, 4, 5, 6 p greater than 2 = 4/6 = 2/3 p 2 or smaller and prime = 2/6 = 1/3 2/3 + 1/3 = 1 every single side of the cube is either greater than 2 or prime (two and one) so you simply will get the desired result in 100% of your rolls :)

**Algebra**

combinations of 10 three at a time = 10!/ [ 3! (10-3)! ] = 10 * 9 * 8 *7* .......1 /[3*2(7*....1) ] = 10 * 3 * 4 = 120 or look up binary coefficient (10,3) = 120 or look at Pascal's triangle row 10, element 3 (the eleventh row, fourth element really since zero counts) agai...

**Science**

Note from wikioedia: Uranium-238 - Wikipedia, the free encyclopedia en.wikipedia.org/wiki/Uranium-238 Wikipedia Jump to Radioactivity and decay - [edit]. 238U radiates alpha-particles and decays **** (by way of thorium-234 ****** and protactinium-234) into uranium-234. 234U ...

**Science**

U238 for example has 92 protons and therefore 146 neutrons It loses an alpha particle which is a helium nucleus, 2 protons and 2 neutrons so NOW we have an element with 92 - 2 = 90 protons 238 - 4 = 234 atomic mass ignoring E = mc^2 What has 90 protons? Th Thorium 232 is usual...

**math**

33/77 , 21/77 , 7/11 Now which is big and which is small ?

**Social Studies**

Gee, remarkable. You are welcome :)

**Social Studies**

1 yes 2 yes 3 yes !!! 4 yes 5 I think d because of shape of Mexico city valley between mountains 6 yes 7 yes 8 yes 9 yes

**Algebra 2**

I still do not have any idea what the question is.

**Algebra 2**

| 3 2 5 | |B11 B12 B13| |5| | 2 -1 8|+|B21 B22 B23|=|0| | 4 7 -3| |B31 B32 B33| |3| Is that what you mean ??? or what ?

**Algebra 2**

Are they matrices? is A square and C a column?

**Algebra 2**

Sorry, I don't get it. Is A a square matrix and B and C columns or what? do you mean A * B or A + B ???

**Algebra 2**

|+3 +2 -5 | |x| |20| |+5 -10 8 |*|y|=|17| |-8 +11 -2| |z| |29| to solve: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

**CHEMISTRY**

22.4 liters/mole for perfect gas at STP C + O2 --> CO2 46 g of C which is 12 grams/mole so we need 46/12 = 3.83 moles of carbon and 3.83 moles of O2 3.83 * 22.4 = 86 liters

**Math Help? Check my answers.**

4. Write the expression using a single exponent. 2^2 • 2^8 Oh my ! a^x * a^y = a^(x+y) !!!!! 2^2 * 2^8 = 2^(2+8) which is 2^10

**Math Help? Check my answers.**

3. ***** -7x***** - 5x^2 + 5, what is the coefficient of x -7x so -7 is coef of x !!!

**Math Help? Check my answers.**

2. 6x^4 + 3x^3 - 2x^2 + 15x - 14 one ,,, two,,,three,,four, five

**Math Help? Check my answers.**

2. disagree 3. disagree 4. disagree 5. disagree sort of. A works but 10*10^7 is usually written 1 * 10^8 which is C

**Algebra 2**

For heavens sake look up the definition of these terms instead of guessing !!!

**Physics**

b) w = 2 pi/T = sqrt (k/m) (2 pi)^2 /T^2 = k/m k = 2 (2 pi)^2/ 1.8^2 = 24.4 n/m (I do not know how you got the same answer) !!!! c) when it is stopped at x = .08 meters, all the energy is potential (1/2) k x^2 (1/2)(24.4) (.08)^2 = .0781 Joules d) Ke = .0781 Joules = (1/2)mv^2...

**algebra/have no idea how to do this one**

That is what I did Mary Ann I use r = interest rate/number of payments in a year so your n t is my n and your r/n is my r and my n is 40 *12, your n t

**algebra/have no idea how to do this one**

are we doing amount of an annuity here? here? If so: A = payment [ (1+r)^n -1 ]/r r is interest PER PERIOD = .08/12 = .006667 n is number of periods = 40*12 = 480 months payment = 150 A = 150 [1.006667^480 -1 ] / .006667 A = 150(23.27339)/.006667 A = $ 523,624.99

**math**

Try tessellating a circle.

**Calculus**

Ah hah, a real calculus problem :) dV/dt = - k V dV/V = -k dt ln V = -k t + C e^ln V = V = e^(-kt+C) = c e^-kt so here at t = 0 e^0 = 1 so V = c = 70,000 and V = 70,000 e^-kt now when t = 10/12 years, V = 22,222 22,222 = 70,000 e^-k(.83333) ln (22,222/70,000) = - .83333 k -1.1...

**Calc**

total lift if you release water to the atmosphere .5 meter up at the pump = .5 + .75 meters = 1.25 meters total mass = 1000kg/m^3 * 50*25*1.5 = 1.875 *10^6 kg total weight = m g = 9.81*1.875*10^6 = 1.84*10^7 Newtons total work done = weight * distance = 1.84*10^7 *1.25 = 2.23 ...

**Calc**

The same. You have to lift the center of mass of the water up to the pump either way. Of course if you attach a tight hose to the outlet you save lifting a half meter by releasing the water at ground level rather than .5 meter up.

**Calculus**

See the reply to the question above this one from "Jen". Find the mass and m*g, the center of mass, and how far you lift the center of mass. In all these questions if you lower the outlet with a tight outlet hose you can siphon the water out without doing any work at...

**Calculus**

In your earlier question you did not "factor out" the log function. you (or Reiny) said if log x = log y then x = y

**Calculus**

Not calculus. algebra .08/4 = .02 per quarter n = number of quarter years 15825 = 4500 (1.02)^n log (15825/4500) = n log 1.02 .546131 = n log 1.02 n = 63.5 quarter years so 63.5 /4 = 15.88

**Need help with pre cal hw**

LOL, physicist / mathematician again.

**Need help with pre cal hw**

note 160 = south - 20 North distance = 8 cos 30 - 6 cos 20 East distance = 8 sin 20 + 6 sin 20 distance = sqrt (North^2 + east^2) 2. call it 132 = south - 48 north distance=-61.1cos 48 +76.5sin 36.5 east distance =61.1sin48 +76.5cos 36.5 tan theta = north/east

**Physics**

How far does it go in 30 seconds (1/2 minute)? (1/2)min * 1 hr/60 min *120,000 m/hr = 1000 meters 360/20 = 18 the circumference of the turn = 18,000 meters 2 pi r = 18,000 r = 2865 meters Ac = v^2/r = (120,000/3600)^2 / 2865 = .388 m/s Check my arithmetic !! usually plane bank...

**HELP PLEASE !!**

b a

**algebra/I appreciate your help**

.055/4 = .01375 per quarter 3 * 4 = 12 quarters 625 * 1.01375^12 = 736.29

**Calculus**

You are NOT factoring it out You are saying that if log x = log y then x = y

**Physics**

Max force at bottom = 20 = m v^2/r + m g 20/.25 = v^2/1.2 + 9.81 v^2 = 70.19 *1.2 = 84.2 v = 9.17 m/s

**Physics**

ax = 5 m/s^2 ay = 8 m/s^2 vx = Vi + ax t = 5 + 5 t vy = 0 + ay t = 8 t x = Xi + Vi t +(1/2) ax t^2 x = 0 + 5 t + (1/2)5 t^2 y = 0 + 0 t + (1/2)8 t^2 at t = 2 x = 0 + 10 +2.5(4) = 20 y = 0 + 0 + 4 (4) = 16

**Algebra 1**

when x = 0, undefined (very big to the right, very negative to the left of origin) when x = 1, y = 2 when x = -1, y = -2 when x = 2, y = 1 when x = -2, y = -1 when x = 5218575, y is very close to x axis above it when x = -5218575, y is very close to the x axis under it.

**Physics**

You are welcome.

**Physics**

Maybe in c you mean what is the velocity angle to the x direction at t = 1 ??? dx/dt = 4 dy/dt = 3-2 = 1 tan^-1 (1/4) = 14.03 degrees

**error - sorry**

|V| = sqrt (16+4) = 4.47 m/s theta = tan^-1 -2/4 = -26.6 degrees

**Physics**

dx/dt = 4 dy/dt = 3 t -2 so at t = 0 V = 4 i - 2 j |V| = sqrt (16+9) = 5 m/s theta = tan^-1 -2/4 = -26.6 degrees b) Ax = 0 Ay = 3 m/s^2 c) is a typo I think.

**physics**

angular velocity constant = omega = 2 pi radians/3600 seconds tip speed is also constant = omega * r

**physics**

d is diameter of circle :) C = pi d t = pi d/v

**physics**

v = sqrt (2 g h) u = 14.4 so |V|= sqrt(u^2+v^2) = sqrt (2 g h + 14.4^2)

**physic**

twice

**pre algebra**

SKETCH A GRAPH !! Always note symmetric about x = 0 note hits bottom at (0,-20) (so that is the vertex) so (y +20) = a x^2 at (10,80) 100 = a(100) a = 1 so I claim x^2 = y + 20

**algebra**

(m-1) or (1-m) 7 - m + 1 = 3 or 7 -1 + m = 3 m = 5 or m = -3 check if m = 5 |m-1| = 4 4 = 4 good if m = -3 |m-1| = 4 good

**Math - Int Trig**

for every a+bi there is a a-bi solution (x-2i)(x+2i)(x-3i)(x+3i) = 0 (x^2 +4)(x^2+9) = 0 x^4 + 13 x^2 + 36 = 0 ================= (x-i)(x+i)(x+10) = 0 etc

**Precalculus**

2 x^2 + 4 x = y^2 - 4 y +4 2 x^2 + 4 x = (y-2)^2 x^2 + 2 x - (1/2) (y-2)^2 = 0 x^2 + 2 x + 1 - (1/2) (y-2)^2 = 1 (x+1)^2 - (y-2)^2/(sqrt 2)^2 = 1 Hyperbola, you look up the stuff for it.

**math**

when x = 0, y = 3 so make a point (0,3) when y = 0, x = -1 so make a point (-1,0) the line goes through those two points and off into the distance left and right

**Science**

sorry, typo 1.5 * 63 Joules

**Science**

4.5 * 63 Joules

**Algebra 1 Answer Evaluation**

Yes, the only solutions to the first problem are the imaginary numbers +/- i sqrt 7 where i is the sqrt of NEGATVE one

**Algebra 1 Answer Evaluation**

x^2 + 0 x + 7 = 0 x = [ 0 +/- sqrt(-28) ]/2 x = +/- 2 sqrt -7/2 x = +/- sqrt(-7) or +/- i sqrt 7 ================================= x^2 - 0 x -16 = 0 x = [ 0 + /- sqrt (0 +4*16) ]/2 x = +/- 2*4/2 x = +/- 4 what else is new :)

**math**

position = speed * time + initial position or d = s t + di 90 = s (2) + di 135 = s(3) +di subtract 45 = 3 s - 2 s = s and 135 = 45(3) + di or di = 0 so in the end d = 45 t check that for example does 270 = 45(6) ???

**math**

-4 = (1/4) (0) + b b = -4 y = (1/4) x -4 or 4 y = x - 16 graph that line

**math**

(-2 - 2) / (4-6) -4 / -2 2

**trig**

sin 15 = 135/ hypotenuse tan 15 = 135/d along ground

**Calculus answer check**

s = .7 t^2 ds/dt = v = 1.4 t 4 = 1.4 t t = 4/1.4 = 2.86 seconds It asked for t given v = 4 m/s

**maths**

(p-7)(p+6)

**Pre - Cal**

start at the middle up 4 end up 20 (16 above middle y = kx^2 16 = k(40)^2 16 = k (4)^2(10^2) 1 = 100 k k = .01 so , adding the 4 at the middle y = 4 + .01 x^2 at x = 20 y = 4 + .01 (4)(100) = 8

**Pre - Cal**

(x+3)^2 = a(y-1) (4+3)^2 = a (-5-1) 49 = -6 a a = -49/6 x^2 + 6 x + 9 = -49/6 (y-1) x^2 + 6 x + 54/6 = - 49 y/6 + 49/6 x^2 + 6 x - 5/6 = -49 y/6 6 x^2 + 36 - 5 = -49 y

**Pre - Cal**

4y^2+4y-16x=0 y^2 + y = 4 x y^2 + y + 1/4 = 4 (x + 1/16) (y+1/2)^2 = 4 (x+16) vertex at (-16 , -1/2)

**Calculus**

integral f(t) dt from 5 to 7 sin 7 - sin 5 = .657 - .959 = -.30 if angle is in radians or .121 - .087 = .0347 if angle is in degrees I suspect you left a pi out or something.

**English**

d cautionary is correct

**Help math only 1 question easy!**

3 * 2 * 1 = 6

**physics**

distance = speed * time wavelength = speed * period = speed / frequency = 3 * 10^8 m/s / 76 = 3.95 * 10^6 meters = 3950 km Either you have a typo or you are talking about a super low frequency system that is used to communicate with submarines underwater. It is 360 degrees aro...

**Physics**

7 m/s * 3600 s/hr /1000 m/km = 25.2 km/hr which is about 15 miles/hour 10m/s = 36 km/hr which is about 22 miles/hour If you are really, really fast you might run 100 meters in around 10 seconds :) I am sure you can think of something :)

**Algebra**

-10 (x^3 - 3 x + 2) /[-5(x-1)] 2 (x^3 - 3 x + 2) / (x-1) 2 (x^2+x-2)(x-1)/(x-1) 2 x^2 + 2 x -4

**pre calculus**

Vx = 1200 cos 6 Vy = 1200 sin 6 |V| = sqrt (Vx^2 + Vy^2)

**maths**

18 - 24 j 9 - 8 j

**maths**

That is not an equation.

**maths**

[(x+a)(x+b)] / [(x+a)(x+a)] = (x+b)/(x+a)

**math?**

82 - 5(15) = 82 - 75 = 7

**AP Physics B**

4.2 * heat of fusion of water 4.2 * 79.5 kcal/kg = 334 kCal

**maths**

I have no idea. You did not say how much he spent total.

**physics**

The net horizontal force is 35 -2.9 = 32.1 N

**Calculus**

if dP/dx = x e^-x^2 then p = -(1/2)e^-x^2 + C 15,000 = -.5 /e + C agree so C = 15,000 + .1839 = 15,000.1839 so p = -(1/2)e^-x^2 + 15,000.18

**Physics/Calculus**

f = -k x 21714 = k (3) W = (1/2) k x^2 do for x = 1/2 inch then for 1 inch and subtract the half inch result

**Calculus**

work = integral w dz w = 4.5 (180-z) so work = integral from z = 0 to z = 180 of 4.5 (180-z)dz = 4.5*180 (180) - 4.5 (1/2)(180)^2 = (1/2)(4.5)(180)(180) or half the weight over the whole distance of course :)

**Physics**

half the weight times the distance

**Physics**

twice as far (1/2) k x^2 where k = force/distance

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