Tuesday

March 31, 2015

March 31, 2015

Total # Posts: 20,745

**maths**

A work/day = 1/16 B work/day = 1/12 2(1/16) + x(1/16 +1/12) = 1
*February 13, 2015*

**Physics Dynamics**

work done against friction = Initial PE - final Ke = Force of friction * distance down slide = mu * normal force * distance Now where is your work?
*February 12, 2015*

**Math**

ds/dt = v = -4 (2t) /t^4 = -8/t^3 if t = a v = -8/a^3 if t = 1 v = -8/1 = -8 if t = 2 v = -8/8 = -1 if t = 3 v = -8/27
*February 12, 2015*

**Algebra 1**

6 a + 2 s = 40 5 a + 12 s = 85 multiply first one by 6 36 a + 12 s = 240 5 a + 12 s = 85 ------------------ subtract 31 a = 155 a = $5 now find s
*February 12, 2015*

**Physics**

so the magnitude is 3.31 kg m/s
*February 12, 2015*

**Physics**

-29 m/s - (+29 m/s) = -58 m/s velocity change change in momentum = -.057(58) kg m/s
*February 12, 2015*

**Physics**

Ac = v^2/R = 100/20 = 5 m/s^2 so F = m g + 5 m = (9.81+5)m = 14.81 m so F = 14.81*85 = 1258 Newtons f = change in momentum / time change in momentum = 85*10 = 850 kg m/s time = distance/average speed = 1.5/5 = .3 seconds so f = 850 kg m/s / .3 s = 2833 kg m/s^2 = 2833 Newtons
*February 12, 2015*

**math**

(1/2) x = 3 x = 6
*February 12, 2015*

**Math**

HE HAD 280 dollars. That is the answer. It is not a fraction.
*February 12, 2015*

**Math**

he started with x he spent 3/7 of x then he had 4/7 of x left he then gave his sister 1/4 of 4/7 of x so he was left with 3/4 of 4/7 of x that is 120 so 3/4 * 4/7 * x = 120 3/7 * x = 120 x = 40*7 = 280 =========================== NOW spend 3/7 * 280 = 120 280 - 120 = 160 left ...
*February 12, 2015*

**Math**

Look at this simpler problem: http://www.jiskha.com/display.cgi?id=1423783632
*February 12, 2015*

**Math**

He spent 3/7 which left him with 4/7 !!! 7/7 - 3/7 = 4/7 then he gave away 1/4 of his 4/7 1 - 1/4 = 3/4 so he is left with 4/7 * 3/4
*February 12, 2015*

**math**

HUH ? 120.25 + (7/28)^(3/7) ??? but 7/28 = 1/4 = .25 120.25 + .25^.42857 ??? 120.25 + 0.55 = 120.80
*February 12, 2015*

**Physics**

You have to tell us the angles from vertical or horizontal or what. if from horizontal T1 cos 45 = T2 cos 30 .707 T1 = .866 T2 T2 = .816 T1, so T1 is the one that will break m g = T1 sin 45 + T2 cos 30 m g = .866 T1 + .866 (.816) T1 m g = .866 T1 + .707 T1 = 1.57 T1 m g = 1.57...
*February 12, 2015*

**Math**

I am getting a bit bored with this question as well.
*February 12, 2015*

**Math**

he was left with 4/7 he gave 1/4 of the 4/7 to his sister so he was left with 3/4 of the 4/7 so (3/4)(4/7)x = 120 (3/7) x = 120 x/7 = 40 x = 280
*February 12, 2015*

**math**

1 - 3/7 = 4/7 is remainder (4/7) x - 3 = 120
*February 12, 2015*

**TYPO ???**

5 seconds up, 5 seconds down do vertical problem first with Vi = initial velocity component up v = Vi - g t at top v = 0 0 = Vi - 10 t but t = 5 seconds so Vi = 50 m/s IF YOU LAUNCH AT 10 m/s, YOU CAN NOT STAY IN THE AIR TEN SECONDS!
*February 12, 2015*

**Science**

If there are no mutations, there are no changes with generations, no adaptation of beak lengths or anything.
*February 12, 2015*

**Science**

You are welcome.
*February 12, 2015*

**Science**

n = 2 , 3, 4, 5, 6 ,7 ..... n^2 = 4 , 9 , 16, 25,...... 1/n^2 = 1/4 , 1/9 , 1/16 .....
*February 12, 2015*

**Science**

so did I :)
*February 12, 2015*

**Science**

1/n^2
*February 12, 2015*

**math**

100(3/4) = 300/4 = 75
*February 12, 2015*

**math**

sorry, slope intercept form is y = -(1/2) x + 1/2
*February 12, 2015*

**math**

2 y = -x + 8 y = -(1/2) x + 4 so slope = -1/2 -2 = -(1/2) 5 + b -2 = -2.5 + b b = .5 = 1/2 so y = -(1/2) x + 1/2 2 y + x = 1
*February 12, 2015*

**Math**

x in 1900 y in 2015 x = .64 y so y = 1.56 x so y now is 156 percent of x which is an increase of 56 percent
*February 12, 2015*

**Math**

(4/7)(3/4) x = 120
*February 12, 2015*

**Physics**

d = (1/2) g t^2 6.9 = 4.9 t^2
*February 12, 2015*

**Physics**

horizontal x = 100 + 35 cos 35 vertical y = 35 sin 35 or sqrt(x^2+y^2) at tan^-1(y/x) above x axis
*February 12, 2015*

**Physics**

There is no force on the ball after it its release so it continues at 4,000 km/s while the ship accelerates away from it. Therefore: -6 m/s^2 Think, how does your head move when you accelerate a car as opposed to driving straight at constant speed.
*February 12, 2015*

**math**

3/2 is bigger than 1 4/9 = 13/9
*February 12, 2015*

**linear algebra**

well, we agree this time :)
*February 12, 2015*

**linear algebra**

I am going to call c1 = a c2 = b c2 = c +3a + 1b -11c = 0 +3a + 2b -13c = 0 -1a - 3b + 9c = 0 -b+2c = 0 from first 2 so b = 2 c +3a + 2b -13c = 0 -3a - 9b +27c = 0 from second 2 ---> -7 b + 14 c = 0 so -7(2c) = -14 c c = anything at all In your linear algebra using Cramer&#...
*February 12, 2015*

**Physics**

You must know the acceleration A of the shuttle F = m A + m g
*February 12, 2015*

**physics**

work = change in potential energy = m g h No work is done in the horizontal motion because the force, gravity, is perpendicular to the direction of motion during horizontal displacement.
*February 12, 2015*

**math**

A = x y cost = 75 x + 60 y = 1750 60 A = x (1750 - 75 x) maximize 60 A = z z = 1750 x -75 x^2 75 x^2 -1750 x = -z complete square to find vertex o parabola x^2 - 23.3 x = -z/75 x^2 - 23.3 x + 11.7^2 = - z/75 + 136 (x -11.7)^2 = -(1/75)(z-10,200) x = length = 11.7 A = z/60 = ...
*February 12, 2015*

**Physical sciences**

v = 0 + a t 12 = a (5) a = 12/5 Is that the whole question? Surely they ask about the deceleration as well? d = Vi t + (1/2) a t^2 so 24 = 12 t + (1/2) a t^2 v = Vi + a t so 0 = 12 + a t
*February 12, 2015*

**algebra**

if b is base 8 b + b = 36 9 b = 36 b = 4 then 4 b = 16 so 16 , 16 and 4
*February 12, 2015*

**physics**

(1/2) m v^2 = m g h v = sqrt (2 g h) = sqrt(2*9.81*4.03) average speed = v/2 5.00 = (v/2) t t = 10/v
*February 12, 2015*

**Physics**

tension - T inward force = T cos 30 so T cos 30 = m v^2/.6 downward force = mg = T sin 30 = T/2 T = 2 m g so 2 m g = m v^2/.6 v^2 = 1.2 g if you want angular velocity v = omega r omega = v/.6
*February 12, 2015*

**Physics**

power = V^2/Req = 9.8 = 26^2/Req so Req = 69 ohms so R = 2 Req = 138 Ohms new Req: 1/Req = 1/138 + 1/276 Req = 92 Ohms power = 26^2/92 = 7.35 Watts
*February 12, 2015*

**physics**

correct
*February 12, 2015*

**GEO**

7.5 * 5*10^4 cm (1 km/10^5 cm) = 37.5 * 10^-1 = 3.75 km
*February 11, 2015*

**Physics**

If there were a potential difference current would flow. i = V/R where V is the potential difference and R is the resistance In a perfect conductor the resistance is zero and and V ( potential difference) would lead to infinite current to wipe out the potential difference. ...
*February 11, 2015*

**math(5th grade)**

.375 = 3/8 (3/8)120 = 3*15 = 45
*February 11, 2015*

**physics**

m1 = 3 m2 = 4 F = m1 a1 = 3 a1 F = m2 a2 = 4 a2 so 3 a1 = 4 a2 3 * 6 = 4 * a2 a2 = 18/4 = 9/2 = 4.5 m/s^2
*February 11, 2015*

**physics**

40 = (1/2)(9.81) t^2 t = 2.86 seconds v = g t = 9.81 * 2.86
*February 11, 2015*

**physics**

13400 cm (1.72*10^-6) ohm cm /2.07*10^-2 cm^2 = [(1.34)(1.72)/(2.07)] * 10^(4-6+2) = 1.11 Ohms
*February 11, 2015*

**physics**

multiply
*February 11, 2015*

**MATH**

18 qt * 1 gal/4 qt = 2.5 gal 7 half gal * 1 gal/2 half gal = 3.5 gal 2.5 + 3.5 = 6 gal
*February 5, 2015*

**Math Ms Sue please help me**

We can not see your picture so have no idea how to help.
*February 5, 2015*

**physic**

420 rev/min = 420 * 2 pi /60 = 44 radians/second 120 rpm = 12.6 rad/s alpha = (12.6 - 44 ) /4 = -7.86 radians/s^2 (part A) angle = initial omega t + (1/2) alpha t^2 angle = 44 (4) - (1/2)(7.86)(16) = 113 radians 113/2pi = 18 revolutions 18 * 2 pi r =18 *2 pi * .2 = 22.6 cm but...
*February 5, 2015*

**physics**

d = u t for car d = 20.1 t d = (1/2) a t^2 for cruiser d = (1/2) 2.65 t^2 = 1.325 t^2 so 1.325 t^2 - 20.1 t = 0 t (1.325 t - 20.1) = 0 they are together at t = 0 and again at t = 20.1/1.325 = 15.2 seconds
*February 5, 2015*

**physics**

v = Vi - 9.81 t at top v = 0 t = Vi/9.81 at top Vi = 20.1 sin 33.4 = 11.1 so t = 11.1/9.81 = 1.13 seconds at top how far horizontal in 1.13 seconds? u = 20.1 cos 33.4 = 16.8 m/s 16.8 m/s * 1.13 s = 19 meters
*February 5, 2015*

**algebra**

2.5 pounds/month is a constant slope of weight as a function of age. Therefore we are dealing with a straight line, linear function. (eventually it will hopefully turn exponential as it approaches full grown weight but that is a later problem :) six weeks = 1.5 months weight...
*February 5, 2015*

**chemistry**

They are liquid.
*February 5, 2015*

**Math**

You are welcome.
*February 5, 2015*

**Math**

7/8 = .875 14/14 = 1 1 4/9 = 1.444444444...... 3.2 = 1.5 which is the winner
*February 5, 2015*

**math**

Hey, you are supposed to know how to solve simultaneous linear equations or would not have been given this problem: http://www.jiskha.com/display.cgi?id=1423159248#1423159248.1423161824
*February 5, 2015*

**math**

2.4 * 2.8 * 3 = 20.16 m^3 1 m^3 = 1000 liters so V tank = 20,160 liters we have 3,600 liters so must deliver 16,560 liters 16,560 liters / .5 liters/s = 33,120 seconds /3600 = 9.2 hours 9 hours and 12 minutes
*February 5, 2015*

**math**

p = 1040 - c L = c - 880 but p = 3 L 3 L = 1040 - c so 3 (c-880) = 1040 - c 3 c - 2640 = 1040 - c 4 c = 3680 c = 920
*February 5, 2015*

**Latin I**

Well, I took Latin in 1951 and 1952 but your work looks ok to me for what that is worth.
*February 5, 2015*

**statistic**

This might help: http://davidmlane.com/hyperstat/z_table.html
*February 5, 2015*

**chemistry**

Mg ion +2 Cl ion - 1 so I suspect MgCl2
*February 5, 2015*

**math/algebra**

s = a + 70 s+a=770 so (a+70) + a = 770 2 a + 70 = 770 2 a = 700 a = 350
*February 5, 2015*

**physics**

v = Vi - 9.81 t if it came down at 28 m/s, it started up at 28 m/s (symmetry) v = 28 - 9.81 t at the top, v = 0 0 = 28-9.81 t so t = 28/9.81 at the top, same time falling h = (1/2)(9.81) t^2 h = (4.9)(28/9.81)^2 h = 4.9
*February 5, 2015*

**math - We do not do magic**

If you do not give some sort of distribution of customer arrival times I have no idea when Tony's staff should be there.
*February 5, 2015*

**math Huh ?**

I do not have the slightest idea what you are talking about. Arithmetic sequences or series maybe ?
*February 5, 2015*

**physics**

Now you should be able to do them.
*February 5, 2015*

**physics**

in general if speed is s at A degrees above horizontal u = s cos A Vi = s sin A
*February 5, 2015*

**physics**

Now divide your problems up into vertical problems and horizontal problems if horizontal speed is u at start it is u to the finish so x = u t if initial vertical speed is Vi then v = Vi - 9.81 t (remember v=0 at top) h = Hi + Vi t - 4.9 t^2
*February 5, 2015*

**physics**

just a vertical problem Vi = 17.9 sin 36.0 h = 0 + Vi t - 4.9 t^2 0 = 0 + t (Vi - 4.9 t) so it is at ground level at t = 0 of course and at t = Vi/4.9
*February 5, 2015*

**Algebra typo ???**

a (b+c) = db a (b+c) = dc If you did not make a mistake than of course b = c but CHECK FOR TYPOS
*February 5, 2015*

**physics**

first do the horizontal problem with constant horizontal speed of 13 for 21.4 m to get time in air distance = speed * time t = 21.4 /13.0 now we know it fell for time t h = (1/2) g t^2 h = (1/2) (9.81) (21.4/13.0)^2 meters
*February 5, 2015*

**Math**

Not that I know of.
*February 5, 2015*

**Math**

(5/12) s = 10 c + 20 (7/12) s = 15 c well we have two linear equations with two unknowns. We can do it. 7 s = 180 c so c = (7/180) s (5/12) s = (7/18)s + 20 (5/4) s = (7/6) s + 60 (15/12)s = (14/12) s + 60 s/12 = 60 s = 720 kg
*February 5, 2015*

**physics**

speed east = 1.65 cos 18.8 t = 21.0 / ( 1.65 cos 18.8 )
*February 5, 2015*

**Maths**

5 f + 10 t = 500 f + t = 90 f + 2 t = 100 f + 1 t = 90 ------------- subtract t = 10 then f = 80 check 400 +100 = 500 80 + 10 = 90 ok
*February 5, 2015*

**chm**

water is 1 g/mL so we have 0.105 g of H2SO4 per mL How many mols is that H2 = 2 g/mol S = 32 g/mol 4 O = 64 g/mol total = 98 g/mol .105/98 = .0011 mol /mL which is 1.1 mol/Liter
*February 2, 2015*

**Math**

4 * (13/5) = 10.4 which is more than 10
*February 2, 2015*

**math**

Oh, sorry, assumed that if you were deciding what was a function you also were doing inverses of functions. You will be :)
*February 2, 2015*

**math**

Just watch out if you are asked for the inverse next.
*February 2, 2015*

**math**

It IS a function ! (I said that)
*February 2, 2015*

**math**

there is only one value of y for every x so it is a function (a parabola in fact) The inverse of this however is NOT a function
*February 2, 2015*

**Physics**

Gravity is not in this because it says in a HORIZONTAL circle (it must be on a table top or something but 625 is much bigger than g anyway))
*February 2, 2015*

**Physics**

Your problem only has two significant figures accuracy to two significant figures w = 25 so r w^2 = 625 F = m a F = 2 kg * 625 m/s^2 = 1250 N
*February 2, 2015*

**money**

1 small + 1 medium = $20 so $60 left we can only buy one large for $32 60 = 32 = $28 left (3 bought so far) One medium for 16 leaves 12 (4 so far) finally 3 smalls leaves $0 (7 total)
*February 2, 2015*

**4th grade math fraction**

4 * 6 = 24 juice boxes gave away 2 * 3 = 6 24 - 6 = 18 are left
*February 2, 2015*

**5th grade math**

to divide fractions invert the one on the bottom and multiply (8/11) / (4/1) = (8/11)(1/4) = 2/11 (3/8)(10/3) = 10/8 = 5/4 = 1 1/4 (6/1)(4/3) = 24/3 = 8 (1/6)(6/5) = 1/5 etc etc etc
*February 2, 2015*

**algebra**

5 b (6b+5) + 6 (6b+5) = (5b+6)(6b+5)
*February 2, 2015*

**algebra**

no (s + 7)(s - 4 )
*February 2, 2015*

**algebra**

********2 b + 8 (b-3) | 2 b^2 + 2 b - 9 ********2 b^2 - 6 b ------------------------ ************** + 8 b - 9 ************** + 8 b - 24 ------------------------- *****************R = + 15 In other words I suspect you have a typo or you have a remainder of 15
*February 2, 2015*

**maths please help**

x^2 = 2 x = +/- sqrt 2 2. not sure what is denominator 3. x^2/2 + 6 x + 10 = 0 x^2 + 12 x + 20 = 0 (x+10)(x+2) = 0 x = -10 or x = -2 4. x^2 - 7 x = 0 x (x-7) = 0 x = 0 or x = 7
*February 2, 2015*

**Science**

The jellyfish, which I suspect is actually a squid (I gather your teacher is not a fisherman), moves forward as the water moves aft. I suppose you could say that this is due to the third law (action---> equal magnitude opposite direction reaction). However I consider this a...
*February 1, 2015*

**Physics**

F = k Q1 Q2 /d^2 due to right one, +force because opposite sign due to middle one, -force because of same sign F = k(10^-6)(10^-6)(65) [ 95/1.2^2 -48/.6^2 ]
*January 31, 2015*

**Physics**

force up = 25 sin 25 force down = 5(9.81) normal force = 5(9.81) - 25 sin 25 pull force = 25 cos 25 back friction force = 4 so 25 cos 25 - 4 = 5 * a
*January 31, 2015*

**Physics**

force up from left wire = 106 sin 5 force up from right wire also 106 sin 5 total force up = 212 sin 5 = weight down = m g = 9.81 m m = (212/9.81)sin 5
*January 31, 2015*

**physics**

u = speed * cos 27.1 7.72 = speed *cos 27.1 * t where t is time in air h = Hi + Vi t - (1/2) g t^2 0 = 0 + speed*sin 27.1 * t - 4.9 t^2 but we know t = 7.72/(speed*cos 27.1) call speed s 0 = 0 + 7.72 tan 27.1 - 4.9*7.72^2/(s^2cos^227.1) solve for s do it again for s = 1.05 s
*January 31, 2015*

**math**

j = dan + 30 dav = 3 j j + dan + dav = 400 (dan+30) + dan + 3 (dan+30) = 400 5 dan + 120 = 400 5 dan = 280 dan = 56 j = dan + 30 = 86 dav = 3 j = 258
*January 31, 2015*

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