Saturday

January 31, 2015

January 31, 2015

Total # Posts: 20,444

**Algebra**

The function is undefined for x = 1 (x = 1 is not in the domain but otherwise all real x) but the range is from -oo < y < +oo
*October 26, 2014*

**calculus**

f(x) = 5/x f(x+h) = 5/(x+h) [f(x+h) - f(x) ]/h = [5/x+h) -5/x]/h = 5[ x - (x+h) ] /(x^2+xh)h =5 [ -h ] /(x^2+xh)h = -5/(x^2+xh) ======================== if you want use x = 4 right now = -5/(4^2 + 4 h) let h-->0 = -5/16 ======================== I would continue though let h...
*October 26, 2014*

**calculus**

here for example: http://www.sosmath.com/calculus/diff/der00/der00.html
*October 26, 2014*

**typo**

f(x) is a function of x or is it 5/8? the slope for part b is the definition of the derivative in beginning calculus. You are not supposed to know that but google it and the result will show you how to find the limit of: [ f(x+h) - f(x) ] / h as h -->0
*October 26, 2014*

**differential equation**

dA/dt = k(M-A) - cA dA/dt = k M - kA - c A but k - c is a constant call it r dA/dt = kM- rA dA/dt +rA = k M = constant for this case of M
*October 26, 2014*

**math/limits**

(x-3) / [ (x-3)(x+3) = 1/(x+3) so when x is 3 1/6
*October 25, 2014*

**math/limits**

You mean???? [(14+y)^2 - 14^2 ] / y ???? if so [ (14+y-14)(1+y+14)] / y y (y+15)/ y y+15 as y--->0 15
*October 25, 2014*

**College Algebra**

First draw the picture, center at (0,0) and radius = 5 You see that point in Quadrant 3? It makes a triangle with the -y axis hypotenuse 5, bottom 3, 4 down y axis Now draw the tangent to the circle at that point. It will be perpendicular to the radius at that point. Since the...
*October 25, 2014*

**Math**

v^3 = 5.8^3 vsphere = (4/3) pi (5.3/2)^3 subtract
*October 25, 2014*

**physics**

G Me m /r^2
*October 25, 2014*

**csci 120**

f(t) = initial balance ( 1 + r )^t where r = yearly interest percent/100 t is time in integer years after deposit
*October 25, 2014*

**physics**

m g = 690 mg - 1000 kg/m^3 * V g = 36 1000 V g = 690 - 36 = 654 N g = 9.8 so V = 654/(1000 g) rho = m/V = 690/g / .654/g = 1055 kg/m^3 just a little heavier than water (must have expelled all air from her lungs to sink like that)
*October 25, 2014*

**Physics**

friction work done = ke lost = (1/2)m v^2 force = .7 m g = .7 * 66 * 9.8 mass = 66 a = F/m = .7 * 9.8 = 6.86 m/s^2 0 = 4.4 - 6.86 t t = .641 seconds average speed = 2.2 m/s distance = 2.2 * .641 = 1.41 meters
*October 25, 2014*

**Calculus**

and also draw a picture of the circle of radius 2 and center at (0,0) so you get an idea what is going on.
*October 24, 2014*

**physics**

(1/2) m v^2 = m g h so v1 = -sqrt(2 g h) (coming down) h = -2.87 coming down going up same deal but opposite sign for v v2 = + sqrt(2*9.8*1.12) change in momentum = m (v2-v1)
*October 24, 2014*

**Algebra Slope Intercept Form**

slope is -6/2 = -3 look at what Steve did He did it exactly the way I did your last problem!
*October 23, 2014*

**Algebra Slope Intercept Form**

see http://www.jiskha.com/display.cgi?id=1414100500 exactly the same procedure.
*October 23, 2014*

**math**

2 pints then is 4 cups that is 8 half cups
*October 23, 2014*

**Trig/Pre-Calc...**

That is interesting, a little tailwind helps but a lot hurts. also zero has an effect? why? anyway if s = 3 t = .0119 *9 - .308 *3 - .0003 t = +.1071 - .924 - .0003 t = -.8172 seconds (wow, a lot) = -.82 rounded
*October 23, 2014*

**Math**

No if x = 2 y = 20 * 2 + 10 y = 40 + 10 y = 50
*October 23, 2014*

**Math**

to graph it use any two points on the line if x = 0 , y = 10 so (0,10) is one point if x = 532, y = 20*532 + 10 = 10,650 so (532 , 10,650) is a point take ruler and draw line through those points (I would suggest using like x = 5 instead of 532)
*October 23, 2014*

**Algebra Slope Intercept Form**

change in y/change in x = -2/7 = m so y = (-2/7) x + b put in any old point from the table, like (0,-1) -1 = 0 + b so y = (-2/7) x - 1 or 7 y = -2 x - 7
*October 23, 2014*

**Language Arts HELP!**

B is the only sentence in the list that is English.
*October 23, 2014*

**math**

1 gallon = 231 in^3 37 * 10 * 12.5 in^3 (1 gal / 231 in^3) = 20 gallons
*October 23, 2014*

**Finance**

You mean you just deposited once with 27K ? If so amount = 27,000 (1.11)^20 = 27,000 (8.062311536) = 217,682.41 gee, I agree to the penny wonderful stuff that compounding :)
*October 23, 2014*

**Math 12**

The main reason for giving appreciated stock to a charity is to avoid paying any capital gains tax on the stock. but anyway paid 100 * 10 = 1,000 for stock it was worth 100 N when granted got a refund of .6(100 N) = 60 N paid tax = .2(100 N - 1,000) so 60 N - (20 N - 200) = 1...
*October 23, 2014*

**ap calc**

oh, the first one is always v positive v = 2t+2 so when does the second one go - ?
*October 23, 2014*

**ap calc more on d**

they both start out + find out when the first one hits v = 0 then it goes - while the first is + find when the second hits v=0 then they both are - etc
*October 23, 2014*

**ap calc**

b) find the velocity of each when t = 8 v1 = 2(8)+2 = + 18 v2 = -2t +8 = -8 so in opposite directions d ) they start out both moving + when is v1-v2 = 0 ?
*October 23, 2014*

**physics**

vertical problem first Vi = 9.52 sin 24.4 v = Vi - 9.8 t at top v = 0 so 9.8 t = 9.52 sin 24.4 = 3.93 t = .401 seconds to top height h = Vi t - 4.9 t^2 h = 3.93 (.401) - 4.9(.401)^2 h = .789 m now the horizontal problem time = 2 t = .802 seconds because he spends as much time ...
*October 23, 2014*

**Algebra**

slope = -1/5 y = -(1/5)x + b -1 = -3/5 + b b = -2/5 so y = -(1/5)x -2/5 or 5 y = -x -2
*October 22, 2014*

**PHYSICS**

I initial = 4.90 + .565^2 (4.95*2) = 4.90 + .319 (9.9) = 4.90 + 3.16 = 8.06 Ifinal = 4.90 + .2^2(9.9) = 4.90 + .396 = 5.30 oh my, much less angular momentum initial and final = I omega1 = 8.06*2.8 = 22.6 (answer A) = 5.3*omega2 so omega 2 = 22.6/5.3 = 4.26 rad/s Now I think ...
*October 22, 2014*

**physics Damon???**

(i) half (it arrives randomly polarized. The first filter takes out all the x components for example) (ii) None. The x components go out with the first filter. The second filter takes out the y components :) (iii)http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polcross.html ...
*October 22, 2014*

**Algebra 2**

Assuming you are not allowed to use calculus, complete the square to find the vertex. x^2 + 12 x = y-5 x^2 + 12 x + 6^2 = y - 5 + 6^2 (x+6)^2 = y + 31 vertex at x = -6, y = -31 so never smaller than -31
*October 22, 2014*

**physics**

You are welcome.
*October 22, 2014*

**physics**

it would have to be 8 times as thick 64 cm (pretty thick :)
*October 22, 2014*

**science**

http://curious.astro.cornell.edu/question.php?number=300 Note that if you are in the Northern hemisphere You can see stars that are high in the northern sky all year. It is the ones close to the plane of the earth's orbit that you can only see when you are facing them at ...
*October 22, 2014*

**Sorry**

mistook / for times (bottom*derivative of top - top * derivative of bottom)/bottom^2 (x)/(2+x)= = [(x+2)(1)-x(1) /(x+2)^2 = [ x+2 -x ] /(x+2)2 = 2/(x+2)^2
*October 22, 2014*

**Derivatives**

(2+x) d/dx (x) + x d/dx (2+x) (2 + x) + x(1) 2 + 2 x 2(x+1)
*October 22, 2014*

**physics**

6.3 * sin 34 = 3.52 meters up m g h = 13.2 * 9.81 * 3.52 = 456 so I get -456 J I think you reversed digits
*October 22, 2014*

**physics...damon please help**

delta L/L = k delta T delta L/L = 6.7*10^-6 inches/inch degree F (117 deg F) L = 10 meters =394 inches so delta L = 394 * 6.7 *10^-6 * 117 = .309 inches = .784 cm = .0784 meters
*October 22, 2014*

**COLLEGE PHYSICS**

|v1| = 2|v2| v1^2 = 4 v2^2 so Ke v1 = 4 Ke v2 so 5 Ke2 = 7500 J so Ke2 = 1500 J and Ke1 = 6000 J
*October 22, 2014*

**Trig ?**

I do not understand your symbols, do not have whatever font you are using. I assume this is some kind of a phase shift in the argument of the cotangent.
*October 22, 2014*

**physics**

17 = (1/2) 1.9 t^2 solve for t v = 1.9 t
*October 22, 2014*

**Math**

II. When calculating interest , you need too ______? (1pt) a. Change the percentage to a decimal then multiply the decimal against the amount. --------------------------------- VIII .0450 - .0375 = .0075 .0075 * 1000 = 7.50 or answer c BUT this assumes interest PER MONTH (very...
*October 22, 2014*

**Algebra 1 ?????**

do you mean -2/5x - 9 < 9/10 ??? or (-2/5)(x - 9) < 9/10 or -2/(5x - 9 )< 9/10 or -2/[5(x - 9)] < 9/10 or what ?
*October 22, 2014*

**math**

You mean he checked 25 wheels ????? If so then what times 4 + what times 3 = 25? 1*4 + 21 sure 1*4 + 7*3 2*4 + 17 nope 3*4 + 13 nope 4*4 + 9 sure 4*4 + 3*3 4*5 + 5 nope 4*6 + 1 nope
*October 21, 2014*

**math**

1.65 * 100 = 165
*October 21, 2014*

**Precalculus**

PLEASE PROOF READ - I assume you mean 2.2ft. / HOUR a) A = pi r^2 b) 2.2 t c) 2.2 * 2 2.2 * 2.5 d)A(2) = pi * (4.4)*2 A(2.5) = pi (5.5)^2 e) A(t) = pi (2.2 t)^2 f) A(2) = pi (2.2*2)^2 A(2.5) = pi (2.2*2.5)^2 g) [ A(2.5) -A(2) ]/.5 h) [A(7.)-A(5)]/ 2.5 i) you will see that it ...
*October 21, 2014*

**precalculus**

A = Ai e^kt 2 = e^3k 3 k = ln 2 k = .231 30 = e^(.231 t) ln 30 = .231 t t = 14.7 hours
*October 21, 2014*

**precalculus**

already did this above
*October 21, 2014*

**algebra**

note a may not be 10 or -4 or the problem is undefined (zero denominator)
*October 21, 2014*

**algebra**

(a+10)/(a+4) by dividing top and bottom by (a-10)
*October 21, 2014*

**Physics**

4.63 * 4.10 work "in" in x direction -3.79 * 3.86 work "out" in y direction (braking) 2.16 * -2.45 work "out" in z direction now just add the results (not a vector) 4.63 * 4.10 - 3.79 * 3.86 - 2.16 * 2.45 Joules (scalar but does have sign)
*October 21, 2014*

**math**

Oh, I misunderstood question I think.
*October 21, 2014*

**math**

72 min and 75 min (common multiple?) 72 = 12 * 6 = 3*3 *2*2*2 75 = 5*5*3 so we need 25 * 3 * 8 = 600 minutes = 10 hours
*October 21, 2014*

**Physics**

just did it below kinetic energy at bottom = potential energy at top
*October 21, 2014*

**Physics**

m g h = (1/2) m v^2 v^2 = 2 g h v = sqrt (2 g h) h = 3.27 (1 - cos 35.6) meters g = 9.8 m/s^2
*October 21, 2014*

**Physics**

work done by gravity = change in potential energy in falling from initial height h = L (1-cos theta) h = .8 (1 - cos 40.4) so work=m g h = 2.63*9.8*.8 (1 - cos 40.4)
*October 21, 2014*

**Physics**

2275 = 207.8 t +(1/2)15.6 t^2 solve quadratic for t then v = 207.8 + 15.6 t
*October 21, 2014*

**Physics**

d = Vo t + (1/2) a t^2 219 = 0 + 2.3 t^2 solve for t v = Vo + a t v = 0 + 4.6 t
*October 21, 2014*

**Physics**

here for example Vo = 14.8 a = 11.3 ft/s^2 t = 6.6 sd v = Vo + a t so v = 14.8 + 11.3 * 6.6 = 89.4 feet/second
*October 21, 2014*

**Physics**

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html You want the box labeled: "Description of motion in one dimension"
*October 21, 2014*

**inequalities**

10x -20 + 11 < 10 -12 + 2x 8 x < 7 x < 7/8
*October 21, 2014*

**Note**

0 = Vo - a t ignore that line, final speed is not zero and is not needed
*October 21, 2014*

**please help! physics**

95 km/hr *1000/3600 = 26.4 m/s in .5 seconds moves 13.2 meters then a = F/m = -2400/1200 = - 2 m/s^2 0 = Vo - a t d = 13.2 + 26.4 t - (1/2) (2) t^2 t = 2 so d = 13.2 + 48.8 = 62 meters d = 13.2 +
*October 21, 2014*

**Physics**

at bottom v = sqrt (2 g h)
*October 21, 2014*

**science**

They are all gasses but the most dense of the bunch is ethyne (Acetylene)
*October 21, 2014*

**physic**

Vi = 28 sin 60 =24.25 m/s v = Vi - 9.8 t h = Vi t - 4.9 t^2 = 24.25(3.3) - 4.9(3.3)^2 = 26.7 meters high not 29.2 when does v = 0 (top of arc)? 0 = 24.25 - 9.8 t t = 2.47 seconds so h at top = 24.25(2.47) - 4.9 (2.47)^2
*October 21, 2014*

**Physics**

m v^2/r = 1630 * (13.9)^2/385 b) I do not know what is holding your vehicle in the circle !!!! If on a string - tension force If held in a big cylinderical rotating can (carnival ride) - normal force If on a flat curving road - friction on the tires If in orbit in space - ...
*October 21, 2014*

**math**

I (we) did not use any resources such as the web or books other than our textbook. ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; and you want me to work on it ?
*October 19, 2014*

**Physics**

Total momentum = 0 Mb*2.2 - Me*2.9 = 0 Mb/Me = +2.9/2.2 the ratio of masses is unlikely to be negative and I think you have it upside down.
*October 19, 2014*

**Physics!**

force = rate of change of momentum (only m a if m is constant :) change of momentum = 79.1 (0 - -6.27) change of time = 2.06*10^-3 s so force = + 6.27*79.1 / .00206 Newtons
*October 19, 2014*

**Calculus**

You are welcome.
*October 19, 2014*

**Calculus**

In other words you got the slope right but did not get b right in y = m x + b
*October 19, 2014*

**Calculus**

you say y = 5 x + 2.5 then if x = pi/4, y = 5 pi/4 + 2.5 You say but what is f(x) at x = pi/4 ??? sin^2 = 1/2 2 tan = 2 so f(x) = 2.5 You better take 5 pi/4 off your answer
*October 19, 2014*

**Physics**

Keb = (1/2)55 (7.5^2) = 1545 J = 1.55kJ I agree with your book :)
*October 19, 2014*

**Physics**

55*22 = 1210 = 55 Vb cos 76+ 45 Vs cos 12 45*26 = 1260 = 55 Vb sin 76 - 45 Vs sin 12 1210 = 13.3 Vb + 44.0 Vs 1260 = 53.4 Vb - 9.36 Vs times 44/9.36 1210 = 13.3 Vb + 44.0 Vs 5922 = 251 Vb - 44.0 Vs ------------------------- 7132 = 264.3 Vb Vb = 27 km/hr = 7.5 m/s humm, we ...
*October 19, 2014*

**physics**

force in = mg sin 12 = m v^2/r v^2/r = g sin 12 turns 180 deg = pi radians r = v^2/(g sin 12) pi r = distance flown = pi v^2/(g sin 12) time = distance / speed = pi r/v = pi v/(g sin 12 here v = 450/3.6 and g = 9.81
*October 19, 2014*

**physics**

m = 5 t^.8 - 3 t + 20 dm/dt = (5*.8) t^-.2 - 3 so at max or min 0 = 4/T^.2 - 3 T^.2 = 4/3 = 1.333 .2 log T = log 1.333 Log T = .6246 T = 10^.6246 = 4.21 seconds find m at t = 4.21 and find dm/dt at 3 and 5 (signs better be different)
*October 19, 2014*

**Math**

You are welcome.
*October 18, 2014*

**Math**

A + B = 500,000 .14 A + .10 B = 59,000 multiply second equation by 10 1.0 A + B = 500,000 1.4 A + B = 590,000 --------------------- subtract -.4 A = -90,000 A = 225,000 etc
*October 18, 2014*

**Math**

A + B = 260,000 A = B + 77,000 so B + 77,000 + B = 260,000 or 2 B = 183,000 etc
*October 18, 2014*

**Math**

Yes, it is a quadratic equation
*October 18, 2014*

**Math**

been there, done that see below
*October 18, 2014*

**math**

LOL, well if 6 is the answer then -16(36) + 96 (6) better be zero is it ? :)
*October 18, 2014*

**Math**

yes, quadratic equation
*October 18, 2014*

**Math**

speed before Forde = (s-5) speed after Forde = s time to Forde = (t+96) note 4 days = 96 hr time after Forde = t then distance = speed * time 70 = (s-5)(t+96) 60 = s t or t = 60/s 70 = (s-5)(60/s + 96) 70 = 60 + 96 s - 300/s - 480 490 = 96 s - 300/s 245 = 48 s - 150/s or 48 s^...
*October 18, 2014*

**Physics**

v = Vi + a t d = Vi t + (1/2) a t^2 2240 = 199 t + 4.25 t^2 solve quadratic for t then v = 199 + 8.5 t
*October 18, 2014*

**Physics**

d = (1/2) a t^2 176 = 4.35 t^2 solve for t v = a t = 8.7 t
*October 18, 2014*

**PHYSICS 1301 (COLLEGE)**

When people do not even go back and see if what they typed makes any sense, I just skip to the next question.
*October 18, 2014*

**Calculus**

v = -gt = -9.8*2 100 = 4.9 t^2 t = 4.52 seconds to ground so v = -9.8*4.52
*October 18, 2014*

**chemistry**

.55 a = .45(715)
*October 18, 2014*

**calculus**

4 cos anything is never bigger than 4 so there is no overlap here
*October 18, 2014*

**Calculus**

2 * L + 4*w = 800 so L = 400 - 2w A = L * w A = (400 -2w)w A = -2 w^2 +400 w dA/dw = 0 at max = -2 w + 200 w = 100 L = 400 -200 = 200
*October 18, 2014*

**Calculus**

decrease at start due to startup costs and paying off initial investment in staff and equipment increase over long term because we have to go deeper and deeper into the vein as we mine out the easy stuff near ground level.
*October 18, 2014*

**Chemistry**

heat it 25 deg C 550 * 2.03 * 25 Joules
*October 18, 2014*

**physics please help**

In the end torque = F*.4 - friction torque = I d omega/dt
*October 18, 2014*

**physics please help**

Not knowing that frictional torque, I am at a loss. Please proof read question.
*October 18, 2014*

**Physics**

69 v1 = 47 v2 so v2 = (69/47)v1 19 = (v1+v2)t 19 (v1 + 1.47 v1) t so v1 t = distance of first = 19/2.47 = 7.69 meters
*October 18, 2014*

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