Sunday

May 29, 2016
Total # Posts: 26,282

**Math**

ln(2x+3)^4 = 11 e^ln(2x+3)^4 = (2x+3)^4 = e^11 (2x+3) = (e^11)^.25 = e^2.75 = 15.64 2x = 12.64 x = 6.32
*January 27, 2016*

**Math**

Hey Rita, read what you wrote. We have no idea what your scale ratio is hint ratio of areas = ratio of lengths^2 2/1 in length is 4/1 in area (and 8/1 in volume)
*January 27, 2016*

**history .**

I agree with A for both. However I am not a history teacher.
*January 27, 2016*

**Math**

Log2(x)+log2(x-3)=2 2^[Log2(x)+log2(x-3)] =2^2 2^log2(x^2-3x) = 2^2 (x^2-3x) = 4 x^2 - 3 x - 4 = 0 (x-4)(x+1) = 0 x = 4 or x = -1
*January 27, 2016*

**Math**

2log6(4x)=0 log6 (16 x^2) = 0 6^log6 (16 x^2) = 16x^2 = 6^0 = 1 so x^2 = 1/16 x = 1/4 normally I would say also -1/4 but try to take the log of -1 :)
*January 27, 2016*

**Physical Science**

No, I disagree. It has to transfer heat to something colder, like a vat of liquid air or the surface of Lake Superior in February.
*January 27, 2016*

**Functions 11**

well b^2-4ac = 81-24k if that is positive, two real roots if it is zero, you have one root twice (the parabola just bounces off the x axis at x = -b/2a) if it is negative there are no REAL roots (although there are two complex conjugate solutions)
*January 27, 2016*

**Chemistry 101**

.785 v = 2 v = (2/.785) cm^3
*January 27, 2016*

**Physics**

Newtons force = F = .1 kg * 9.81 x = 0.10 meter k = F/x = 9.81 N/m
*January 27, 2016*

**Math**

use -4y for x in the second equation -4 y + 5 y = 2 so y = 2 then x = -4(2) = -8 (-8 , 2)
*January 27, 2016*

**Physics**

I took this in 1955 and do not need practice. If you get stuck I will help but I will not do all the problems for you. By the way in Naval Architecture we have a term called "tons per inch" for small changes in draft. It proportional to rho g * area of water line ...
*January 27, 2016*

**Physics**

(1000 kg/m^3 * volume) - 1.8 kg
*January 27, 2016*

**Physics**

You mean 1.8 * 9.81 N
*January 27, 2016*

**PHYSICS**

weight component down slope = m g sin A = force component up slope = F 50 (1/2) = 25 N = F
*January 27, 2016*

**Physics**

The kinetic energy alone is (1/2) m v^2 = 212 * 10^12 so there is a typo somewhere either in the problem statement or in the answer.
*January 27, 2016*

**Math**

Matt had x May had 3x 4 x - 50 = 70 4 x = 120 x = 30 ======================= check Matt 30 May 3*30 = 90 sum = 120 120 = 50+70
*January 27, 2016*

**math**

tan 30 = h/a tan T = h/2a tan T = (1/2) tan 30 T = 8.21 degrees
*January 27, 2016*

**W Hist**

semiconductors sitting around staring at screens no horse messes stinking up the city streets
*January 26, 2016*

**Math**

6 * 3 * 2 = 6 * 6 = 36 ft^3
*January 26, 2016*

**math**

2*2*2*2*2*2
*January 26, 2016*

**Physics**

vertical problem: Vi = 68 sin 25 = 28.7 m/s h = Vi t -4.9 t^2 h = 28.7 t - 4.9 t^2 horizontal problem: u = 68 cos 25 = 61.6 m/s t = 50/61.6 = .811 s so h = 28.7(.811) - 4.9(.811^2)
*January 26, 2016*

**Pre Algebra**

-1 11/16 + 4 1/6 = - 27/16 + 25/6 sure enough = -81/48 + 200/48 agree = 119/48 = 2 23/48 yes ============================ NO! -1 33/48 + 4 8/48 = 3 -25/48 = 2 + 48/48 -25/48 = 2 + 23/48
*January 26, 2016*

**?**

Please proof read initially at rest maybe? 12 what? /s^2 meters, feet ? 205 what? seconds, meters ? feet? miles? after 205 what again? 12 m/s^2 for how long ? Is this after the first question using that final for this initial velocity?
*January 26, 2016*

**math**

Yours is much better. I got off on that both queens tack :)
*January 26, 2016*

**math**

Go with jolly rancher !
*January 26, 2016*

**whoops**

what I gave you was for a specific rank, like two queens multiply by 13 because there are 13 ranks
*January 26, 2016*

**math**

4/52 * 3/51
*January 26, 2016*

**Science**

storage space, closet ?
*January 26, 2016*

**typo?**

You need to specify a time before the encounter
*January 26, 2016*

**History**

Well, I suppose D, but I do not think communes were that large as a rule.
*January 26, 2016*

**math**

No, you can not copy and paste to this site.
*January 26, 2016*

**Math**

3 times as many cookies takes 3 times as many cups of sugar so 6 cups 2 cups/ 2 dozen = 6 cups/6 dozen
*January 26, 2016*

**physics**

m v^2/r 2500 (25)/50 = 1250 Newtons
*January 26, 2016*

**Physics**

dw/dt = -10 - 4.42 t so w = -10 t - 2.21 t^2 + constant find constant at t = 0 71.6 = -10(0) - 2.21(0)^2 + c so c = 71.6 w = 71.6 - 10 t - 2.21 t^2 put in t = 2.73 Now the angle dTheta/dt = w = 71.6-10 t -2.21 t^2 so Theta = 71.6 t - 5 t^2 - (2.21/3)t^3
*January 26, 2016*

**Physical sciences**

Fx = 30 cos 60 - 40 cos 60 Fy = 30 sin 60 + 40 sin 60 F = sqrt (Fx^2 + Fy^2) Tan theta = Fy/Fx Note - Theta is in quadrant 2 because 40>30 Use 180 - the magnitude of the angle
*January 26, 2016*

**Physics**

from the plus plate toward the negative plate Remember that if you integrate along them you get the voltage difference.
*January 26, 2016*

**Physics**

You do not say what the moment of inertia of the Merry Go Round is. I will call it I 39.4 revs/min (2 pi rad/rev)(1 min/60s) = 4.13 radians /s = W1 omega of John = v/r = 6.36/3.21 = 1.98 rad/s Initial angular momentum = I (4.13) + 48.4(3.21)^2 (1.98) Final I = I + 48.4(3.21)^2...
*January 26, 2016*

**Physics**

You need how far it rolled down the hill to get initial speed v = w r where w is omega, the angular velocity kinetic energy = m g h = (1/2)I w^2 + (1/2) mv^2 = (1/2)(2/3)mr^2v^2/r^2 + (1/2)mv^2 =(mv^2/2)[2/3 + 1] = (5/3) (mv^2/2) so g h = (5/3)(v^2/2) solve for v, initial ...
*January 26, 2016*

**math**

If those are matrices, no they do not necessarily commute
*January 26, 2016*

**Physics**

Your equations seem to indicate that the 70 degrees is with the FLOOR, not the Wall, which makes more sense) First, force up = force down so force up at floor = 160 Call friction force at floor Fw It is equal and opposite to horizontal force at top Now moments about ladder at ...
*January 26, 2016*

**physics**

How amusing - you happened upon a Naval Architect. increased volume of ship in the water = Area * .025 meters mass of water displaced = density of seawater * that volume density of seawater is about 1029 kg/m^3 so mass of water displaced = 1029 * .025 * A Buoyancy force due to...
*January 25, 2016*

**physics**

You mean (4/3) pi r^3 r = 0.25 meter V = .0654 meters^2 mass of water displaced-mass of foam = (1000-300)(.0654) = 45.8 kg
*January 25, 2016*

**physics**

rho fresh = 1000 kg/m^2 rho salt = 1029 kg/m^3 approx submerged draft = d vol of block = .1*.1*.1 = .001 m^3 mass of block = 700*.001 = .7 kg volume of fresh water displaced must have mass = .7 kg so .1 * .1 * d * 1000 = .7 kg d = .07 m = 7 cm so freeboard above water = 3 cm ...
*January 25, 2016*

**Physical Science**

h = 4.9 t^2 h= 4.9(1.16)^2 h = 6.19 m
*January 25, 2016*

**physics**

assume salt water density = 1029 kg/m^3 and g = 9.81 m/s^2 mass in salt water = 1029*.8*60 = 1000 * d * 60 d = .8 (1029/1000) = .8232 m
*January 25, 2016*

**History**

Knowing nothing of the subject all I can do is give you a link: http://www.geocities.ws/krysstravel/writing_kanarda.html
*January 25, 2016*

**math**

7 = 1+6 2+5 3+4 if greater than 40 and less than 50, has to start with 4 impossible to be an even number 43
*January 25, 2016*

**math**

but if being strict about significant figures you only have two 5.1 * 10^6
*January 25, 2016*

**Calc help!**

What you typed does not make sense. Perhaps you mean: [ x(x+3) - (x+1)(x-3)] / (x^2-9) ??? [ x^2 + 3x - x^2 +2 x + 3 ] /(x^2-9) (5x+3)/(x^2-9)
*January 25, 2016*

**physics**

None, no motion in the direction of the force.
*January 25, 2016*

**I did this already**

http://www.jiskha.com/display.cgi?id=1453734257
*January 25, 2016*

**Math**

7 in ( 5 miles/1 in) = 35 miles
*January 25, 2016*

**Math**

same way as the one above (15 mi)
*January 25, 2016*

**Math**

yes
*January 25, 2016*

**Math**

24 + 36 + 24 + 36
*January 25, 2016*

**Physics**

99.5-.7 = 98.8 so Reading C = Cr = .7 +(98.8/100) C if Cr = 30 30 = .7 + .988 C C = 29.7 if the same C = .7 + .988 C .012 C = .7 C = 58.3 degrees C or Cr
*January 25, 2016*

**Math**

I assume you mean a = 6.25 m/s^2 Please proof read what you type. v = a t = 6.25 m/s^2 * 7 s = 43.75 m/s Now parachute, a = - 2.5 m/s^2 and Vi = 43.75 m/s v = Vi + a t 0 = 43.75 - 2.5 t t = 17.5 seconds
*January 25, 2016*

**math**

30 km (1 hour/225 km) = .1333333... = (.1 + 1/30) hours .1 hours = .1*60 minutes = 6 minutes (1/30) hours (60 min/hour) =2 minutes 6 + 2 = 8 minutes
*January 25, 2016*

**math**

70 km/h * 1000 m/km * 1 h/3600 s = 19.44 m/s so 50 km/h = 13.89 m/s a = change in v/change in t = (13.89 - 19.44)/5 = - 1.11 m/s^2
*January 25, 2016*

**Algebra**

well, halfway between x axis intercepts is x = 12/2 = 6 so that is where the vertex must be y - b = (x-6)^2 -5 - b = (-7)^2 = 49 b = -54 y = (x-6)^2 - 54 y = x^2 - 12 x + 36 - 54 y = x^2 - 12 x - 18 ===================== check: x^2 - 12 x = y+18 x^2 - 12 x + 36 = y+ 54 (x-6)^2...
*January 25, 2016*

**math**

The top seems to be an arithmetic sequence and the bottom a geometric sequence, but I can not get that ratio to converge as i --->oo
*January 25, 2016*

**maths**

6 n + 4/n = 14 ??? 6 n^2 + 4 = 14 n 6 n^2 - 14 n + 4 = 0 3 n^2 - 7 n + 2 = 0 (3n-1)(n-2)= 0 n = 1/3 or 2 which do you think? Hint - try 2 then try 1/3 :)
*January 25, 2016*

**Physics**

work accomplished = 300*30 = 9,000 J applied work = 9,000 + 2,000 = 11,000 Joules efficiency = 100 * work accomplished/work applied = (9/11)100 = 81.8 % force applied = (300/6)/.818 = 61.1 N assuming that the 6 in your question refers to mechanical advantage
*January 24, 2016*

**Algebra**

Yes, sure :) BUT you did not need to ask because to be sure just put it back in 2 (3) + 3 (2) = ??? maybe 12 6 + 6 = YES
*January 24, 2016*

**algebra**

if + k is to the right, it is everything left of but not including -1 if k is up, it is everything below -1
*January 24, 2016*

**Algebra**

well, pick any old x, like for example x = 0 find y 0 - y = 9 y = -9 so (0 , - 9) now pick another x, like x = 1 3 - y = 9 y = 3-9 = -6 so ( 1 , - 6 ) and on and on until bored
*January 24, 2016*

**math**

I assume you mean compounded r = interest rate specified usually per year /4 because compounding every quarter year 18.5 years * 4 = 74 periods 3 = x^74 log 3 = 74 log x log x = .00644758 x =1.015 so 1.5 percent per period or r = 4*1.5 = 6%
*January 24, 2016*

**math**

Yes, I think that is the intent of the fuzzy phrasing :)
*January 24, 2016*

**Pre-Calculus**

50 72 45 (the 72 is 60+12) 32 45 12 ----------subtract 18 27 33 now being a navigator, I happen to know that a minute of latitude is a nautical mile so I would say 18*60 + 27 + 33/60 = 1107.5 NAUTICAL miles However do not tell you math teacher that :) we have 1107.5 Minutes of...
*January 24, 2016*

**trig**

one goes 60 km one goes 20 km the third side is 50 km divide them all by ten (similar triangle then law of cosines 5^2 = 6^2 + 2^2 - 2*6*2 * cos A 25 = 36 + 4 - 24 cos A -15 = -24 cos A cos A = .625 A = 51.3
*January 24, 2016*

**English I guess**

Specify subject so you do not get a physicist.
*January 24, 2016*

**Physics**

2 * 6.41 = 12.82 :)
*January 24, 2016*

**Physics- HELP PLEASE!**

F = k q1 q2 /r^2 all on 6 F = (k q6) q/r^2 let d = .15 meter due to 3 Fx = - (k q6) 3/d^2 negative direction because +and - Fy = 0 due to -2.4 repelling Fx = +(kq6) 2.4 /[ 2*(d)^2](sqrt2/2) Fy = same up due to -1 only up Fy = (kq6)9/ d^2 so Fx = -c 3/d^2 + c 2.4 *sqrt 2/4d^2 ...
*January 24, 2016*

**math**

26 miles ( 1 hour/175 miles) = .14897 hours .14897 hours * (60 minutes/1 hour) = 8.915 minutes .915 minutes ( 60 seconds/1 minute) = 55 seconds so 8 minutes and 55 seconds approximately.
*January 24, 2016*

**Algebra**

x and 83-x x - (83-x) = 24 2 x - 83 = 24 2 x = 107 x = 53.5 83.0 - 53.5 = 29.5
*January 24, 2016*

**Grade 11 Functions**

parabola, complete square -x^2/100 + 3x - 50 = p x^2 - 300 x + 5000 = -100 p x^2 - 300 x = - 100 p - 5000 x^2 - 600 x + (150)^2 = -100p+22,500 (x-150)^2 = -100 (p-225) vertex at x = 150, p = 225 so once again, 150
*January 24, 2016*

**typo at start 150 is right**

profit = 6x - x^2/100 - 50 - 3x p = -x^2/100 + 3 x - 50 dp/dx = 0 at max = -x/50 + 3 x = 50 * 3 = 150
*January 24, 2016*

**Grade 11 Functions**

whooops, you have not have derivatives parabola, complete square -x^2/100 + 6x - 50 = p x^2 - 600 x + 5000 = -100 p x^2 - 600 x = - 100 p - 5000 x^2 - 600 x + (300)^2 = -100p+85,000 (x-300)^2 = -100 (p-850) vertex at x = 300, p = 850 so once again, 300
*January 24, 2016*

**Grade 11 Functions**

profit = 6x - x^2/100 - 50 - 3x p = -x^2/100 + 6 x - 50 dp/dx = 0 at max = -x/50 + 6 x = 50 * 6 = 300
*January 24, 2016*

**math**

1.5 * 8/5 = .3*8 = 2.4 2.5 * 8/5 = .5*8 = 4.0
*January 24, 2016*

**Calculus**

1/3 - 1/sqrt 2 = (sqrt 2 -3)/3 sqrt 2 divide that by 2-9 = -7 (sqrt 2 -3)/-21
*January 24, 2016*

**Physics II**

m = mass of electron q = electron charge F = m a q E = m a a = q E/m v = Vi - a t v = 0 at stop so t = 5 * 10^5/a d = Vi t - .5 a t^2 .09 = 5*10^5 t - .5 a t^2 now the easy way :) if acceleration is constant the average speed during this stop is 2.5 *10^5 m/s it stops in .09 m...
*January 24, 2016*

**Math**

He works two and then four more for a total of six ? 2 + 4*4 = 18
*January 24, 2016*

**math**

a + ar + ar^2 ...+ ar^(n-1) a = 3 a r+ a r^2 = 60 r + r^2 = 20 r^2 + r - 20 = 0 (r-4)(r+5) = 0 r = 4 3 + 12 + 48 .... 3 * 4^(n-1)
*January 24, 2016*

**physics**

m (g - v^2/r)
*January 24, 2016*

**physic**

change of momentum = Force * time = 200 * s If you really mean for s seconds.
*January 24, 2016*

**phy 231**

horizontal problem: u = 338 cos 40.5 = 257 m/s x = u t = 257 * 8.09 h = Vi t - 4.9 t^2 where Vi = 338 sin 40.5 = 220 m/s h = 220(8.09) - 4.9 (8.09)^2
*January 24, 2016*

**Been there - use d = v t **

http://www.jiskha.com/display.cgi?id=1453604915 Yes, it is composite, so use d = v t which was part B and please use one name so I do not have to search all over for your earlier attempt.
*January 24, 2016*

**Physics**

initial momentum east = 3 * 1 - 5 * 2 = -7 so final momentum east = -7 I guess left is west, negative -7 = 3 * -1 + 5 * u 5 u = -4 u = 4/5 west
*January 24, 2016*

**physics**

Vertical problem: v = a t x = Xi + Vi t + .5 a t^2 here x = 0 Xi = 49.1 Vi = 0 a = -9.8 0 = 49.1 + 0 t - 4.9 t^2 so a) t = sqrt ( 49.1 m/4.9 m/s^2 ) b) v = -9.8 t c) u = 12.6 v = -9.8 t tan(angle from horizontal) =(v/u)
*January 24, 2016*

**Physics**

a = 710/275 v = a t t = 20 * 275 / 710 seconds
*January 24, 2016*

**physics**

assuming the weight is evenly distributed radius = .04 meters pi r^2 * P = 400 P = 400/.00503 = 79577 Pascals or about .8 atmospheres
*January 23, 2016*

**Math**

Area = 2400 ft^2 2400 (2/12) = 400 cubic feet(7.48 gal/ft^3 = 2992 gallons
*January 23, 2016*

**Math**

for P sin T is -1 when T = (3/4)(2pi) = 3 pi /2 for Q sin T then becomes +1 when sin T = 2 pi + pi/2 = 5 pi/2 ------------------------ now x for P 6 pi x = 3 pi/2 x = 1/4 and y is -2 now x for Q 6 pi x = 5 pi/2 x = (5/12) and y is +2 slope = 4/(5/12 - 1/4) = 4 / (5/12 - 3/12...
*January 23, 2016*

**physics**

F = 800 - 300 = 500 N a = F/m = 500/400 = 1.25 m/s^2
*January 23, 2016*

**Math**

so d = v t
*January 23, 2016*

**Math**

what is that x about? d is distance traveled k is hypotenuse 10 , not (10+x) is original distance leg k = sqrt (d^2 + 100) d = v t Is c) the dot product of k and d? I do not understand your notation If it is k dot d then that is |k| |d| cos angle between path direction and ...
*January 23, 2016*

**Physics**

m = mass of proton q = charge of proton During first E1 F = q E1 north a = q E1/m north vfinal = (q/m) E1 (14.8) x final = 0 + 0 +.5 (q/m)E1 (14.8)^2 = (q/m)E1 (110) Now during E2 x initial = 110 (q/m)E1 Vinitial = (q/m)E1 (14.8) x final = 0 so 0 = 110(q/m)E1 + 14.8(q/m)E1(8....
*January 23, 2016*

**Math**

Well, anyway, the Physics was easy :)
*January 23, 2016*

**Physics**

cool :)
*January 23, 2016*