Friday

September 4, 2015
Total # Posts: 21,884

**chemistry**

Use what Dr Bob said
*July 31, 2015*

**chemistry**

CaCO3 + HCOOH = H + CaCOOH + CO3 of course that needs H2 on the right so double everything if writing a balanced equation
*July 31, 2015*

**College Algebra**

Are you sure? The air resistance is more like proportional to velocity square. However it does not matter since you do not need to know. v(0) = 180 * (1- e^0) but anything to the 0 is 1 so v(0) = 180 (1 - 1) = 0 v(5) = 180(1 - e^-1.15) = 180(1-.317) = 123 v(10) = 180 (1 - e^-2...
*July 31, 2015*

**Math 157**

11^5 * 3^5 * 3^1 * 13^1 * (5^2)^12 3^6 * 5^24 * 11 * 13
*July 31, 2015*

**problem solving**

.71 J/g C * 2 g * 100 C = 142 J
*July 31, 2015*

**Science**

I agree, low density gas due to high temp and low compressing gravitational force.
*July 30, 2015*

**Statistics**

Since you probably want to run equipment that will only work withing a specified voltage range, you want range. For example if your motor requires a voltage between 90 and 110 it does not do much good to say the median is 100 if it burns out at 80 or 120
*July 30, 2015*

**algebra**

time to get there = d/60 time to get back = d/30 so d/60 + d/30 = 9 3 d/60 = 9 d = 3 * 60
*July 30, 2015*

**alg**

just did that above
*July 30, 2015*

**whoops, sorry**

carol drove for 3.5 hours d = 30*3.5 = 105 miles mother drove 105 miles in 3 hours 105 = r(3) r == 35 so yes a
*July 30, 2015*

**Easy Math??**

carol drives for 3 hours d = r t d = 30 * 3 = 90 miles mother drove 90 miles in 2.5 hours 90 = r (2.5) t = 36 miles/hour
*July 30, 2015*

**Maths**

3 is also a typo mess
*July 30, 2015*

**Maths**

I do not understand the first question I assume question 2 is a typo. If u = v then the angle is 0 u dot v = |u| |v| cos theta u dot v = 2*2 + 2*2 = 8 |u| = sqrt(4+4) = 2 sqrt 2 |v| = 2 sqrt 2 also (you put in the right values) 8 = 8 cos theta = 8 * 1 cos zero = 1
*July 30, 2015*

**witbank high**

the base is square so 5 * 5
*July 30, 2015*

**algebra**

I guess you probably mean y = -7/ (x^2-4) if x = 0 , y = -7/-4 if x = 1 , y = -7/-3 if x = -1, y = -7/-3 if x = 2, y = -7/0 if x = -2 y = -7/0 if x = 3, y = -7/5 if x = -3, y = -7/5 same left and right of x = 0 :)
*July 30, 2015*

**Science**

1000*9.81 + 1000*1 = 10810 N total mass = 1000 kg F = 1000*10 + 1000 * 1 = 11,000 N Ke on collision = m g h = m g (5) so (1/2) m v^2 = 5 m g v^2 = 10 g on rebound v^2 = 3.6 g sqrt (3.6/10) = sqrt .36 = sqrt (36/100) = 6/10 = 3/5
*July 30, 2015*

**physics**

v = Vi - 9.81 t 0 = Vi - 9.81*3 so Vi = 29.4 m/s m g h = (1/2) m v^2 so v^2 = 2(9.81) h 29.4^2 = 19.6 h h = 44.2 m
*July 30, 2015*

**Easy Math??**

He bought it for $80 As far ads I know the "par value" is of only academic interest. 100 * 6/80 = 7.5 %
*July 29, 2015*

**Math**

1) biggest side <12 because must stretch 6+6 so perimeter < 24 2) sum has to be 180 40+40 + 70 = 150 no but 40 + 70 + 70 = 180 ok 45+45 + 90 = 180 ok 100 + 100 too big so C is no good 60+60 + 90 = 180 ok 3) make a table 1 to 10 across 1 to 10 down start to fill it in ...
*July 29, 2015*

**Maths**

4! = 4*3*2*1 = 24 actually the order matters so it is permutations not combinations (google permutations and combinations) n! where the ! means "factorial" n! = n (n-1)(n-2) ....... (1) 5! = 5*4! = 5 * 24 = 120 or if you already know the 24 for four of them, then if ...
*July 29, 2015*

**Science**

Vi = 72,000 meters/3600 seconds = 20 m/s v = Vi + a t what is a ? a = F/m F = -m g (.5) so a = -.5 g = -9.81/2 = -4.9 close enough v = 20 - 4.9 t when v = 0 0 = 20 - 4.9 t so t is about 4 seconds average speed during stop = 20/2 = 10 m/s so d = 10 m/s * 4 s = 40 meters ( ...
*July 29, 2015*

**Calculus**

700 sin 23 = 273.51 = x speed (east) 700 cos 23 = 644.35 = y speed (north) If x = 333.51 and y = 644.35 then resultant = sqrt (x^2+y^2) and for angle east of north tan angle = x/y
*July 29, 2015*

**College Algebra**

well, I suppose I will try although really I am sure you could try first. 9 e^-(.012*49) = 9/(e^.588) = 9/1.8 = 5
*July 29, 2015*

**physics**

push force forward = 300 cos 30 = 260 N push force down = 300 sin 30 = 150 N normal force = 100*9.81 + 150 = 1131 N static maximum friction = 1131*.5 = 565 N so it does not budge and we need an older stronger woman or you have typos. It is unlikely that u static and u kinetic ...
*July 29, 2015*

**physics**

u = 11.3 cos 60 forever Vi = 11.3 sin 60 v = 11.3 sin 60 - 9.8 t v = 0 at top so at top t = 11.3 sin 60 /9.8 and also at top h = 49 + Vi t - 4.9 t^2 now at h = 0 0 = 49 + Vi t - 4.9 t^2 solve quadratic for t when h = 0 then that is time in air range = u t
*July 28, 2015*

**physics**

If it is thrown horizontal, the horizontal speed u remains 5 m/s until it stops. u = 5 v = - g t h = 19.6 - .5 g t^2 or 0 = 19.6 - 4.9 t^2 t^2 = 19.6/4.9 t = 2 seconds in air u = 5 v = -9.8 * 2 = -19.6 m/s so speed at ground = sqrt (25 +19.6^2 ) tan angle below horizontal = 19...
*July 28, 2015*

**Physics**

http://www.jiskha.com/display.cgi?id=1438108230
*July 28, 2015*

**Physics**

I just answered this below and you mean 30 degrees
*July 28, 2015*

**Math**

I would say that y = x - 8 then I would say -3x - 2(x-8) = 16 I would solve that for x and use that x to go back and find y
*July 28, 2015*

**Physics**

3.9/2 = 1.95 left of pivot --- right of pivot 50(1.95) = 27 x + 35(1.95) 27 x = 15(1.95) x = 15(1.95)/27 to the right of pivot
*July 28, 2015*

**Physics**

Original ke = (1/2)I omega^2 + (1/2)m v^2 omega = v/r = 35/r ke=(1/2)(2/5)mr^2(35^2)/r^2+(1/2)m (35)^2 = (1/2)m [(2/5)+1](35)^2 = (1/2) m (1.4)(35)^2 distance up = h = L sin 20 = L/2 potential energy stopped at top = original ke m g h = m g L/2 = (1/2)m(1.4)(35)^2 note m ...
*July 28, 2015*

**Physics**

.28 F = 98
*July 28, 2015*

**Algebra**

maybe you mean 5^4 - (8x)^2 = 0 ?????? that would be 25^2 - (8x)^2 =0 (25 - 8x)(25+8x) = 0 8x = 25 or 8x = -25
*July 27, 2015*

**pre-calculus**

112.5 = 90 + 22.5 ah ha, 22.5 is half of 45 we will be able to do this with half angle formulas first: sin 122.5 = sin(90+22.5) = sin 90 cos 22.5 + cos 90 sin 22.5 = cos 22.5 + 0 so we want cos (45/2) cos 45/2 = sqrt[ (1+cos 45)/2] = sqrt [ (1 +1/sqrt2)/2 } = sqrt [ (1 + sqrt ...
*July 27, 2015*

**Integral calculus**

h^2 = 2 x^2 2 h dh/dx = 4 x dh/dx = 2 x/h dh/dt = dh/dx * dx/dt = 2 (x/h) dx/dt but x/h = 1/sqrt 2 so dh/dt = sqrt 2 * dx/dt ---------------------- but more simply h = x sqrt 2 dh/dx = sqrt 2
*July 25, 2015*

**Physics**

total contact area = 4 * .217 * .179 = .155 meters^2 total force = gage pressure * area = 176357*.155 = 27,401 Newtons weight It might be interesting to also find the mass in kilograms assuming this all happens on earth 27,401/9.81 = 2793 Kg mass (heavy car)
*July 24, 2015*

**Math**

whoops (x+5)(x-2) (y^3+5)(y^3-2)
*July 23, 2015*

**Math**

I personally would say x = y^3 then it is x^2 + 3 x - 10 (x-5)(x+2) now go back (x^3-5)(x^3+2)
*July 23, 2015*

**calculus**

obviously 2 r </= 12 volume = pi r^2 h =1000 then pi r^2 = 1000/h but h</= 11 so pi r^2 >/= 1000/11
*July 22, 2015*

**Physics**

http://www.mathpages.com/home/kmath114/kmath114.htm u + 2 U
*July 22, 2015*

**Math**

find how many minutes until one is left. That one leaves immediately if the first ones left at t = 0 a(n+1) = (1/2)A(n) a(1) = 256 a(N) = 1 an = a1 r^n-1 1 = 256 *.5^(n-1) .5^(n-1) = 1/256 (n-1) log .5 = log (1/256) n-1 = 8 n = 9 so 9 * 5 min = 45 min if the first ones left at...
*July 22, 2015*

**Physics**

suspect you mean 30 south of west momentum east 7900*5 - 1650*20 cos 30 momentum south 1650*20 sin 30 new mass = 7900 + 1650 new mass*speed east = old momentum east new mass* speed south = old momentum south
*July 22, 2015*

**Math Check**

x^2+ x(6)^1/2+3=x^2 +3 makes no sense at all. 6^(1/2) is not zero
*July 22, 2015*

**Physics**

T = our unknown angle theta angular acceleration = alpha = d omega/dt I = (1/2) M R^2 FF = friction force up slope FF * R = torque so FF * R = (1/2) M R^2 * alpha FF = (1/2)M R alpha R omega = no slip velocity R alpha = no slip acceleration = a so FF = (1/2) M a F = m a F = M ...
*July 21, 2015*

**Algebra1**

of course we could have short cut that by saying 9 (x-8)^2 = y-8 then (x-8)^2 = (1/9)(y-8) but best do the whole completing the square process.
*July 20, 2015*

**Algebra1**

If it is indeed y = 9 (x-8)^2 + 8 then y = 9 (x^2 - 16 x + 64) + 8 y = 9 x^2 - 144 x + 576 + 8 y = 9 x^2 - 144 x + 584 then x^2 - 16 x + 64 8/9 = y/9 x^2 - 16 x = y/9 -64 -8/9 x^2 - 16 x + 64 = y/9 -64 -8/9 + 64 (x-8)^2 = y/9 -8/9 = (1/9)(y-8) vertex at (8,8)
*July 20, 2015*

**not a parabola**

Need an x^2 somewhere perhaps y = 9 (x-8)^2 + 8 or something?
*July 20, 2015*

**math**

3^2 * 6 = 9*6 = 54
*July 20, 2015*

**math**

area is proportional to scale squared volume is proportional to scale cubed
*July 20, 2015*

**physics**

10 = (1/2)(.4) t^2 solve for t v = 0 + .4 t
*July 20, 2015*

**physics**

v = 30 - a (75) assume you mean 1.5 km or 1500 meters 1500 = 30 (75) - .5 a (75)^2 solve for a then go back and get v at 75 seconds
*July 20, 2015*

**physics(kinematics)**

You are welcome. Actually that was fun :)
*July 19, 2015*

**physics(kinematics)**

me too :)
*July 19, 2015*

**physics(kinematics)**

we agree !
*July 19, 2015*

**physics(kinematics)**

v = Vi - 8 t x = Xi + Vi t - 4 t^2 20 = Xi + Vi (4) - 4(16) 20 = Xi + 4 Vi - 64 Xi + 4 Vi = 84 so Vi = (84-Xi)/4 16 = Vi - 8 t 16 = (84-Xi)/4 - 8 t (eqn 1) 4 = Xi + [(84-Xi) /4] t - 4 t^2 (eqn 2) from eqn 1 64 = 84-Xi - 32 t so Xi + 32 t = 20 Xi = (20 - 32 t) substitute that ...
*July 19, 2015*

**college chem**

Na ---- 23 g/mol Cl ---- 35.5 g/mol so NaCl 58.5 g/mol so 42.6/58.5 = .728 mols .728/1.58 = .461 mol/Liter
*July 19, 2015*

**how exasperating !**

Oh my :(
*July 19, 2015*

**calculus**

sketch it of course where does y = 2x hit y=6/x? 2 x = 6/x x^2 = 3 x = sqrt 3 then y = 2 sqrt 3 so from x = 0 to x = sqrt 3 integrate y = (2 x-.5 x) = 1.5 x 1.5 x^2/2 .75 ( 3-0) = 2.25 then where does y = .5 x hit y = 6/x? .5 = 6/x^2 x^2 = 12 x = 2 sqrt 3 so integrate y = (6/x...
*July 19, 2015*

**Physics**

.... and if you have a physics question type PHYSICS, not Blue bells....,
*July 19, 2015*

**blue bells model school**

The momentum does not change but the balls are together during the collision so initial ke = (1/2) m V1^2 = total energy initial momentum = m V1 final momentum = (n+1)m V2 so V2 = V1/(n+1) so ke during collision = (1/2) (n+1)m V2^2 = (1/2) (n+1)m V1^2/(n+1)^2 = (1/2)m V1^2/(n+...
*July 19, 2015*

**Thokozwayo secondary school**

So I can talk to my neighbors.
*July 19, 2015*

**Physics**

Jai showed you how to do this type of problem below. All your problems are really the same and are solved the same way. In other words if you do what Jai said, you can do them all. We are not here to do them all for you. http://www.jiskha.com/display.cgi?id=1437267116
*July 19, 2015*

**math**

DRAW IT ON A GRAPH ! square in quadant 3 or 4 diagonals hit at (2,-2) or (-2,-2) that is a and d NOT B or C
*July 18, 2015*

**math**

+1 -2 +3 -4 +5 .....+99 -100 = +1 +3 +5...... +99 -2 -4 -6...... -100 --------------------- ADD -1 -1 -1 ..... -1 :) LOOKS LIKE AROUND -50
*July 18, 2015*

**math**

you left the lower limit out so I assume 0 sum of all integers from -9 to +9 looks like zero to me so 4 pm
*July 18, 2015*

**Algebra**

LOL - thanks
*July 18, 2015*

**Algebra**

The key is "rounded to 82" :)
*July 18, 2015*

**Algebra**

Well, try 49 :) 49 +4*90 = 409 409/5 = 81.8 rounds to 82 so ok try 48 48 +4*90 = 408 408/5 = 81.6 rounds to 82 so ok try 47 47 + 4*90 = 407 407/5 = 81.4 no good
*July 18, 2015*

**Algebra**

( a+b+c+d+e)/5 = 82 a+b+c+d+e = 410 but 90*4 = 360 so he got at least 50
*July 18, 2015*

**math**

500/9 = 55.5 so there are 55 from 9 to 495 j = 55 500/7 = 71.4 so there are 71 from 7 to 497 k = 71
*July 18, 2015*

**Math 115 Statistics**

draw a box or matrix with your 36 possibilities make an x for win, an o for lose(3,4,6) 0 1 2 3 4 5 6 1 x o o x o x 2 o o x o x x 3 o x o x x x 4 x o x x x x 5 o x the rest x 6 number of o s = 10 so 10/36
*July 18, 2015*

**Math URGENT!!!**

10/30 * 9/29 * 8/28
*July 18, 2015*

**calculus**

dM/dt = -k M separate variables, M left, t right dM/M = -k dt integrate ln M = -k t + c note e^log a = a ao e^ln M = M = e^(-kt+c) = e^-kt e^c or since e^c is some constant call it C M = C e^-kt note that when t = 0 e^-kt = 1 so C = initial amount so here M = 100 e^-kt
*July 17, 2015*

**physics**

good answer.
*July 17, 2015*

**maths**

f-5 = 3(s-5) f + 10 = 2(s+10) ---------------- -15 = 3(s-5) - 2(s+10) -15 = 3s - 15 -2s -20 20 = s then f-5 = 60 -15 f = 50
*July 16, 2015*

**geometry**

(8/6)(12+12+6)
*July 16, 2015*

**geometry**

Pure luck ! You are welcome :)
*July 16, 2015*

**geometry**

ah, I think 155 / C
*July 16, 2015*

**calculus**

dT/dt = -k (T-38) try dT/(T-38) = -k dt ln(T-38) = -k t T-38 = ce^-kt T = 38 + ce^-kt when t = 0, T =75 75 = 38 + c c = 37 so T = 38 + 37 e^-kt at t = 30 , T = 60 so 60 = 38 + 37 e^-30 k 22/37 = .595 = e^-30 k ln .595 = -.52 = -30 k so k = .0173 and T = 38 + 37 e^-.0173 t you ...
*July 16, 2015*

**Alebra 2**

law of cosines c^2 = a^2 + b^2 - 2 a b cos angle
*July 16, 2015*

**Stats**

first pick p anchovies = 1/12 p not anchovies = 11/12 second pick you now only have 11 toppings one of which is anchovies p anchovies = 1/11 p not anchovies = 10/11 11*10/(11*12) = 10/12 = 5/6 p neither anchovies =
*July 16, 2015*

**Math**

We want y = m x + b ? solve for y 3 y = 2 x - 6 so y = (2/3) x -2 so m = 2/3 , b = -2
*July 16, 2015*

**Math**

slope = 4162 but when t = 10, n = 45400 so n - 45400 = 4162 (t-10) ----------------------------- n at t = 19 n = 45400 + 4162(9)
*July 16, 2015*

**Math**

slope = slope (y-9)/(x-1) = (19-9)/(3-1) = 5 you can do the rest
*July 16, 2015*

**Math**

12 y = 15 x - 1 y = (15/12) x - 1/12 y = (5/4)x - 1/12
*July 16, 2015*

**Geometry**

180-45
*July 16, 2015*

**Physics**

If Fx is on a graph, you need the area under the curve from x = 0 to x = 3
*July 15, 2015*

**Physics**

work done = integral Fx dx from 0 to 3 calculate that then Ke final = Ke initial + work done
*July 15, 2015*

**Physics**

assume you mean dot or scalar product A dot B = |A| |B| cos theta so here cos theta = .5 cos^-1 0.5 = 60 degrees
*July 15, 2015*

**Calculus Partial Derivatives**

x y z = k sum = s = x + y + z ds = ds/dx dx + ds/dy dy + ds/dz dz = 0 for max or min but ds/dx = ds/dy = ds/dz = 1 so ds = dx + dy + dz = 0 so max when dx = -dy - dz but x = k/(yz) dx/dy = -kz /(yz)^2 dx/dz = -ky /(yz)^2 so dx = -[-kz/((yz)^2 ]dy - [-ky/(yz)^2]dz so k y = k z ...
*July 15, 2015*

**precalc**

well, if you are not allowed to use calculus you must complete the square to find the vertex of that parabola. 16 t^2 - 48 t -20 = -h t^2 - 3 t - 1.25 = -h/16 t^2 - 3 t = -h/16 + 1.25 t^2 - 3 t + 9/4 = -h/16 + 3.5 (t - 1.5)^2 = - 1/16 ( h - 56) so at peak t = 1.5 and h = 56...
*July 12, 2015*

**sjrkps**

By the way, if you post a physics question say physics, not sjrkps so you get a physicist.
*July 12, 2015*

**sjrkps**

v = Vi + a t v = 0 + .8 m/s^2 * 20 s = 16 m/s
*July 12, 2015*

**math**

but much more fun my way :)
*July 12, 2015*

**math**

x^2 = tan^2 +2 tan sin + sin^2 y^2 = tan^2 -2 tan sin + sin^2 x^2-y^2 = 4 tan sin = 4 sin^2/cos ____________________________________ x y = tan^2 -sin^2 = sin^2/cos^2 -sin^2 = sin^2/cos^2 - sin^2 cos^2/cos^2 = [sin^2 / cos^2] [1-cos^2] but 1 - cos^2 = sin^2 so = sin^4/cos^2 so ...
*July 12, 2015*

**physics**

weight = 320+ 4*15 + 500 + 750 + 1000 = total force up moments around one end Ff=force up from floor at far end Fn = force up at near end 320(1.2)-Fn*.3+30*.3+500*.4+750*1.2 +1000*2 +30*2.1-Ff*2.1 = 0 but Fn = (total force up - Ff)
*July 12, 2015*

**Physics**

V = Vi e^-(t/RC) .1 = e^-(10/10C) = e^-(1/C) ln .1 = -1/C
*July 11, 2015*

**Physics**

but the boat is also being pushed by the propeller m = 60 + 500 = 560 kg
*July 11, 2015*

**physics**

You are welcome.
*July 11, 2015*

**physics**

6000 Pascals is 6000 Newtons per square meter. Multiply that by the area of the wall in square meters to get the force in Newtons. The density of water is about 1000 kg /m^3 so the weight of a cubic meter of water is about m g = 1000 * 9.81 = 9810 Newtons so a column of water ...
*July 11, 2015*