Tuesday

October 13, 2015
Total # Posts: 22,114

**MATHS**

http://www.jiskha.com/display.cgi?id=1422706001
*January 31, 2015*

**Calculus**

Do it Reiny's way (but mine is more fun)
*January 31, 2015*

**Calculus**

2.44 + .0053535353 ....... 2.44 + (1/100)(.53535353 .... ) .53535353... = .53 + .53/100 + .53/100^2... = a + a r + a r^2 + a r^3..... where a = .53 and r = .01 sum of that infinite series = a/(1-r) so .53 /(.99) = 53/99 so we have 2.44 + (1/100)(53/99) (1/100)( 244 + 53/99) I ...
*January 31, 2015*

**Statiatics**

4/5 is chance wrong problems are independent so (4/5)(4/5) = 16/25
*January 31, 2015*

**math**

independent probabilities (1/42)(1/47)
*January 31, 2015*

**Trigo**

you went a total of 160 straight north you went 40 east draw that on a graph, 40 right, 160 up the angle between north and east is 90 deg so you want the hypotenuse of that right triangle.
*January 31, 2015*

**Trigo**

y distance = 90 + 70 = 160 x distance = 40 want hypotenuse d = sqrt (40^2 + 160^2)
*January 31, 2015*

**Calculus**

e^.2x = 75/(30*16) .2x ln e = ln (75/480) but ln e = 1 .2 x = ln (75/480) ======================= 8^tan x * ln 8 * (1/sec^2 x)
*January 31, 2015*

**Physics - must know all masses**

insufficient information
*January 30, 2015*

**Physics**

F = G Mearth Moject / r^2 F1 = G Mearth m/6400^2 F2 = G Mearth m/6780^2 so F2/F1 = (6400/6780)^2 = 0.891
*January 27, 2015*

**math**

150 C (95-0) = 21 (heat of fusion of ice) where C is heat capacity of iron about 0.450 J/ g deg c
*January 27, 2015*

**Math - Problem solving**

last year = x this year = x+20 x = 0.75(x+20)
*January 27, 2015*

**COMBINED OPERATION ON FRACTIONS**

(3/5)j +(3/4)(2/5)j + 3 = j
*January 27, 2015*

**physics**

both angular and linear momentum are conserved. The center of gravity of the system is height: h (m+m) = m x + m (L/2) and it is moving at speed Ucg (m+m) = u m then of course you have the rotation about the cg
*January 27, 2015*

**Math**

the area of the base is easy w*L if the height is h then two of the slope triangles have base w and height along the slope of sqrt(h^2 + L^2/4) so the sum of those two triangle areas is w *sqrt(h^2 + L^2/4) similarly the sum of the other two triangle areas is L * sqrt (h^2 + w...
*January 27, 2015*

**Math / Algebra**

I agree
*January 27, 2015*

**algebra**

-y = 18 * 7 y = -18 * 7
*January 27, 2015*

**Math word problems**

100 - 25 = 75 so you took 25 off every hundred In other words, 25% discount another way to do it: 160 - 120 = 40 dollars discount 100 (40/160) = 25 %
*January 27, 2015*

**Science**

If you look at the inside of a tree branch or tap maple sap you see straw like tubes that transport fluid around.
*January 27, 2015*

**Physics - Vectors**

|A| |B| cos theta = A dot B 2.2 * 10 * cos theta = 14 theta = cos^-1 (14/22) theta = 50.5 degrees
*January 26, 2015*

**Math**

x = 2log4 y log4 y = (x/2) 4^log4 y = 4^(x/2) y = 4^.5x
*January 21, 2015*

**math**

1.75 inches ( 45 miles/2.25 inches) =
*January 21, 2015*

**physical**

(1/2)m v^2 = m g h so v = sqrt (2 g h) remember that
*January 21, 2015*

**physics**

Google Millikan oil drop experiment for example: http://ffden-2.phys.uaf.edu/212_fall2003.web.dir/Ryan_McAllister/Slide3.htm
*January 21, 2015*

**math**

the absolute value of (sec x)^2 is just sec^2 x because when you square a number it is positive so we have (1/cos^2) sin / [1/sin] or tan^2 x dx = tan x - x + c
*January 19, 2015*

**physics**

calculate F = Q E do a = F/m to get a, the acceleration then plain old constant acceleration x = (1/2) a t^2
*January 19, 2015*

**Civics**

YES !!!!!
*January 19, 2015*

**Civics**

Google Marbury vs Madison
*January 19, 2015*

**Physics**

x = 2 + 3.2 t - 3.85 t^2 when is v = 0? dx/dt = v = 3.2 - 7.7 t v = 0 when t = 3.2/7.7 = .416 seconds then x = 2 + 1.33 - .665 = 2.66 If you do not know calculus, find the vertex of the original equation.
*January 19, 2015*

**Physics**

d = Vi t + (1/2) a t^2 98.3 = 29 t - 1.8 t^2 1.8 t^2 -29 t + 98.3 = 0 solve quadratic for t [ 29 +/- sqrt ( 841 - 708) ] /3.6 [ 29 +/- 11.5 ] / 3.6 = 4.86 (ignore the big time, it is on the way back :)
*January 19, 2015*

**Civics**

14 says they have the right to go to school just as other citizens do. 13 just says they are not slaves any more.
*January 19, 2015*

**Civics**

I would say 14 which does not allow sates to deny privileges to any citizens.
*January 19, 2015*

**ap calc**

(z^2)^(1/2) = z you have (x^2 - 4) dx x^3/3 - 4 x + C
*January 19, 2015*

**Math**

I am an engineering teacher not a civics teacher jeje.
*January 19, 2015*

**Math**

(4/5)(7/8) = 7/10
*January 19, 2015*

**Algebra**

s = side length x = long length 2 s + x = 1600 so x = (1600-2s) 2 s (5) + x (20) = 13,000 10 s + 20 (1600-2s) = 13,000
*January 19, 2015*

**physics**

By the way to do the second one you must know the density of the fluid. These are very strange questions. Are you sure you have copied them correctly?
*January 19, 2015*

**physics**

I do not know. All I know is the ratio of areas. 80 (pi r^2) = 1300 (pi R^2)
*January 19, 2015*

**SCIENCE**

3. An electric engineer might (1 point) Build a dam to provide a source of hydroelectric power.*** Develop new fuels to use in the next generation of fuel-efficient cars. Design a remote-controlled toy race car. Design an airplane wing that changes shape to increase stability...
*January 19, 2015*

**Physics**

at the same acceleration as (one g about 9.8 m/s^2)
*January 19, 2015*

**physical science**

------>------> = ------------> <----- ------> = 0
*January 19, 2015*

**physics**

F * d = work done by friction = initial Ke = (1/2) m v^2 F * 50 = (1/2)(750)(900) 100 F = 750 * 900 F = 750 * 9 There are harder ways of course using F = change of momentum/ time = 750*30/t then 50 = (1/2)a t^2 where a = 30/t so 100 = 30 t t = 10/3 seconds to stop a = 9 m/s^2...
*January 19, 2015*

**Physics**

56.1 mi/h = 25.1 m/s v = Vi - a t 0 = 25.1 - a t so t = 25.1/a d = (1/2) a t^2 206 = a t^2 206 = a(630/a^2) = 630/a a = 3.06 m/s^2 (negative of course)
*January 19, 2015*

**Physics**

h = (1/2) g t^2 = 4.9 (3.12)^2
*January 19, 2015*

**Physics**

d = (1/2) a t^2 194 = 6.60 t^2 solve for t then v = Vi - 6.60 t = 0 at end Vi = 6.60 t
*January 19, 2015*

**Physics**

v = Vi + a t 104/3.6 = 74/3.6 + 1.52 t
*January 19, 2015*

**ap calc**

d/dx (2x+7)) = 2 so the 1/2 gets rid of the 2
*January 19, 2015*

**Physics**

but you forgot the work done against gravity m g * 500 = .05 * 9.81 * 500 = 245 J 1450 - 245 = 1204 J so 1204 = .05 * a * 500 =============================== The second one looks ok
*January 18, 2015*

**physics**

Aaaarrrgghhhhh !!!!! Joey is right. At the top v = 0, no velocity except that of mother earth spinning on her axis and making her way around the sun. The acceleration is constant, -9.81 m/s^2 the whole time. Charlie could not possibly, any way, be more wrong. Send him to his ...
*January 18, 2015*

**Physics**

t = 4.2 T = 8 a = 5 I guess it started from a dead stop. x = (1/2) a t^2 during acceleration Vi = speed at end of acceleration = a t X = Vi T during constant speed so distance = (1/2) a t^2 + Vi T
*January 17, 2015*

**Math**

32 = 8 * 4 56 = 8 * 7 so 32 + 56 = 8 (4+7)
*January 15, 2015*

**MATH**

how much of the 40 hours he works at each job
*January 15, 2015*

**Torque**

Huh? 30 * 1.5 = 45 BUT 60 * 1 = 60
*January 15, 2015*

**physicsss**

You are welcome.
*January 15, 2015*

**physicsss**

If the velocity is constant you are correct.
*January 15, 2015*

**Physics**

work done in stopping = brake force * 75 meters (1/2) m v^2 = work done by brakes (1/2) m (93)^2 = F * 75 F = m (57.66) but F = ma so a = 57.66 that is the acceleration the tires can give you. Use it for centripetal acceleration a = 57.66 = v^2/r r = (93)^2/57.66
*January 15, 2015*

**Physics**

mu m g = m v^2/r = m omega^2 r mu g = omega^2 r omega = 2 *2pi radians/s = 4 pi rad/s mu = .4 g = 9.81 so r = .4 * 9.81 /16 pi^2 = .0248 meters = 2.5 cm
*January 15, 2015*

**Physics**

m v2/r
*January 15, 2015*

**No picture**

If you drew a picture, we can not see it.
*January 15, 2015*

**Alegebra**

Huh? how can it be even and 9 ????
*January 15, 2015*

**physics**

Without friction, the angular velocity will not change. In real fife of course it will. Even in the vacuum there is loss of kinetic energy to the bearings. However as far as the information given here, the answer is constant angular velocity forever. or at least for a very ...
*January 15, 2015*

**Physics**

Vi = 15 sin 30 = 7.5 v = Vi - gt v = 0 at top so g t = 7.5 t = 7.5/9.8 = about .77 s
*January 15, 2015*

**Calculus POwer RUle**

dy/dx = -3 k x^2 3 = -3 k (4) k = -1/4
*January 14, 2015*

**Math**

b = 2 r b - r = 24 2 r - r = 24 r = 24 b = 48
*January 14, 2015*

**math**

16 x = x^2 x^2 -16 x = 0 x(x-16) = 0 x = 0 or x = 16 y = 0 or y = 256 (0,0) and (16 , 256)
*January 14, 2015*

**calculus**

If 5x2 + y4 = -9 then evaluate d^2y/dx^2 when x = 2 and y = 1 5 x^2 + y^4 = -9 10 x + 4 y^3 dy/dx = 0 so dy/dx = -2.5 x/y^3 d^2y/dx^2=-2.5[ y^3 - 3 x y^2 dy/dx]/y^6 at (2,1) dy/dx = -2.5(2/1) = -5 so d^2y/dx^2=-2.5[1-6(-5)]/1 = -2.5 [ 31] = - 77.5
*January 14, 2015*

**calculus**

Find the slope of the graph of the relation x2y + 4y = 8 at the point (2, 1) (x^2+4)y = 8 (x^2+4)dy/dx + y (2x) = 0 8 dy/dx + 4 = 0 dy/dx = -1/2
*January 14, 2015*

**calculus**

f(x)=(x+1)^2x d/dx u^v = v u^(v-1)du/dx + u^v lnu dv/dx so f'=2x(x+1)^(2x-1) + 2(x+1)^2x ln(x+1)
*January 14, 2015*

**calculus**

x^2 + y^2 = 2 x y 2 x + 2 y dy/dx = 2 x dy/dx + 2 y dy/dx (y-x) = y-x dy/dx = 1 easier way without using calculus: x^2-2xy+y^2 = 0 (x-y)(x-y) = 0 x = y slope = 1
*January 14, 2015*

**calculus**

y = 7 x^6 cos^-1 x f' = 7[ x^6 (1/(1-x^2)) + 6x^5 cos^-1 x ] = 7 x^6/(1-x^2) + 42x^5 cos^-1 x
*January 14, 2015*

**history**

LOL - well, to start with which ? TR or FDR ? You better Google which ever president you mean because it would take all night to summarize.
*January 14, 2015*

**science**

NaCl = 23 + 35 = 58 g/mol 22/58 = .379 mol .379 mol/ .2 liter = 1.9 M
*January 14, 2015*

**Math**

Get it now. When you subtract you reverse the sign of the bottom one and add.
*January 14, 2015*

**Math**

y=-x+4 y=-2x-4 ------------- SUBTRACT these two equations 0 = -x -(-2x) + 4 - (-4) 0 = -x + 2x + 4 + 4 0 = x + 8 x = -8
*January 14, 2015*

**Math**

y = -x + 4 y = -(-8) + 4
*January 14, 2015*

**Math**

for elimination subtract as is y=-x+4 y=-2x-4 ------------- 0 = 1 x + 8 so x = -8 then get y = 8+4 = 12
*January 14, 2015*

**Math**

I did the earlier ones with elimination here is substitution y = (-x+4) so (-x+4) = -2x - 4 x = -8 then go back and get y y = -x + 4 = 8+4 = 12
*January 14, 2015*

**Math**

You are welcome.
*January 14, 2015*

**Math**

10 x - 8 y = 20 -10 x + 2 y = -20 ----------------- add 6 y = 0 y = 0 if y = 0 5 x = 10 x = 2 so the two lines intersect at (2 , 0)
*January 14, 2015*

**calculus**

1. If f(x) = ln(1 - 2x), then find f '(x) f' = [ 1/(1-2x) ]( -2) = -2/(1-2x) = 2/(2x-1) I do not understand your second function
*January 14, 2015*

**Math**

check it then 2 (9) - 5 (-7) = 53 ? yes -9 +5(-7) = -44 ? yes so it is right, why ask ? :)
*January 14, 2015*

**Calculus**

x = 15 t h = 5 t + 5 = 5(t+1) y^2=x^2+h^2 = 225t^2 + 25t^2 + 50 t + 25 y^2 = 250 t^2 + 50 t + 25 2 y dy/dt = 500 t + 50 now at t = 3 y^2 = 250(9)+150 +25 = 2425 y = 49.2 so 2(49.2) dy/dt = 1500 + 50 = 1550 dy/dt = 15.8 m/s
*January 14, 2015*

**Math**

add the two equations as is x = -10 5 y = 50 y = 10
*January 14, 2015*

**physics**

easier way work done = final (1/2) m v^2 4310 = (1/2)2520 v^2 v = 1.85 m/s ;)
*January 14, 2015*

**physics**

change of momentum = 2520 v average force = change of momentum/time F = 2520 v/t work = force * distance 4310 = 2520 v/t * 22.4 or v/t = .0764 or t = v/.0764 now if force were constant d = (1/2) a t^2 44.8 = a t^2 v = a t so a = v/t so 44.8 = v t = v^2/.0764 v^2 = 3.42 v = 1....
*January 14, 2015*

**Sixth Grade Math**

m = p - 3 = 51 - 3 = 48 yes, you did it right. Either the question was meant to be confusing or the person who wrote the question never asked the more complicated question he/she had in mind :)
*January 14, 2015*

**Socal Studies**

sorry: http://science.kqed.org/quest/2007/12/12/snakes-are-not-poisonous/
*January 14, 2015*

**physics**

mass has nothing to do with this work = force * distance in direction of force = 4.67 * 10^5 * 528 Joules
*January 14, 2015*

**Math**

multiply the second equation by 2 5 x - 4 y = 10 -20x +4 y = -40 ------------------ mow add -15 x = -30 x = 2 etc
*January 14, 2015*

**history**

You are welcome.
*January 14, 2015*

**history**

here: http://en.wikipedia.org/wiki/Iron_Brigade and Texas was very, very important to the Confederacy.
*January 14, 2015*

**history**

Oh my, oh my, you better Google Minnesota and Michigan Gettysburg
*January 14, 2015*

**Math**

24.8 / 0.6 = 41.3
*January 14, 2015*

**Math**

100 * 1382.5/27650 = 5 %
*January 14, 2015*

**english**

http://www.jiskha.com/display.cgi?id=1421282361
*January 14, 2015*

**english**

I did, scroll down to where you first posted this question.
*January 14, 2015*

**english**

( It is a real marine biology subject )
*January 14, 2015*

**english**

I gave you a reference for the lobster and the cod. Try. Read the article and see if you can come up with something.
*January 14, 2015*

**us history**

http://www.bostonmassacre.net/timeline.htm
*January 14, 2015*

**English**

might help: http://www.gma.org/lobsters/allaboutlobsters/society.html
*January 14, 2015*