Sunday

October 23, 2016
Total # Posts: 27,518

**Mathematics**

Looks like a typo problem to me the original was probably round to the nearest thousand dollars, 2000

*March 31, 2016*

**math**

Well done :)

*March 31, 2016*

**math**

Good :)

*March 31, 2016*

**math**

You are welcome, but more importantly do you know how to tell a straight line from a parabola or an exponential now?

*March 31, 2016*

**math**

You typed y = 5x, look up above. I found it correct in one of your earlier postings.

*March 31, 2016*

**AHHH!! NO**

Look, they are asking you which of these is a straight line with constant slope. In other words which looks like Y = m X + b That is B y = 19 x - 10 A, x^2 - 5 x - 14 is a QUADRATIC, parabola C, y = 5^x (NOT y = 5 x ) is exponential D, y = .03 x^2 etc is QUADRQATIC, parabola

*March 31, 2016*

**ah ha - you typed it wrong**

B)y = 19x – 10 C)y = 5^x NOT THIS. This is NOT 5 x as you wrote later.

*March 31, 2016*

**math**

There are two with constant slopes. B and C If you think there is only one answer, you have a typo.

*March 31, 2016*

**math**

B and C 19 and 5

*March 31, 2016*

**College Algebra**

the slope is -29 crimes per year In other words the number of crimes decreases by 29 every year. when t = 0, C = 31,000. That is the number of crimes when the equation started in 1998 I suspect the 31,000 is a typo as is the 29

*March 31, 2016*

**math**

the derivative is constant in B and C

*March 31, 2016*

**Physics**

334 * 2000 = 668,000 J

*March 31, 2016*

**Geometry**

x^2 + 10 x - 1200 = 0 (x-30)(x+40) = 0 x = 30 or x = -40

*March 31, 2016*

**Algebra 2**

3 150 4 150 + d 5 150 + 2 d = 180 so d = 30/2 = 15 month 1 = 150-30 = 120 An = 120 + (n-1)15 A12 = 120 + 11(15) = 285 yes correct 500 = 120 + (n-1)15 n - 1 = 25.33 n = 26.33 yes, 27

*March 31, 2016*

**math**

I guess A is at center and AB bisects CD ???? we need to know angle at center cos (T/2) = 4/5 T/2 = 36.9 degrees T = 73.7 degrees so area = (73.7/360) pi R^2

*March 31, 2016*

**physics**

horizontal speed = u = 50 cos 37 t = 80/u then how high is it at t h = (50 sin 37) t - 4.9 t^2 then d = sqrt (80^2 + h^2)

*March 31, 2016*

**math**

http://davidmlane.com/hyperstat/z_table.html

*March 31, 2016*

**math**

http://davidmlane.com/hyperstat/z_table.html

*March 31, 2016*

**Algebra 2**

d = -27 + 22 = -5 -7 - 5 = -12 -12 - 5 = -17 -17 - 5 = -22 so the last one on the list

*March 31, 2016*

**Math - Slope intercept**

sure looks like it to me. They asked the same question twice. I bet they meant to say: Write the equation in standard form 6 x - 3 y = 2 then solve for y in slope intercept form but who knows

*March 31, 2016*

**luthayi high school. science**

http://www.nlm.nih.gov/medlineplus/ency/article/002680.htm

*March 30, 2016*

**This is a trick question**

Velovicy is a VECTOR north one way, south back 31 lengths, 30 of which were back and forth so went 50 meters total north in the end How long did it take? 31 * 50 = 1550 meters back and forth at 900 m/3600 s time = 1550 *3600/900 = 6200 seconds so 50 meters/6200 seconds = 0....

*March 30, 2016*

**physics**

I will do position last a = dv/dt so a = -10*10^7 t + 3*10^5 zero at t = 3/10 * 10^-2 = .003 seconds a is positive for the whole .003s find v at t = .003 for part c x = integral v dt with x = 0 at t = 0 so x = (1/3)(-5.00×10^7)t^3 +(1/2)(3.00×10^5)t^2 length of ...

*March 30, 2016*

**physic**

k = 20 N/.098 m = 204 N/m m = 20/9.81 = 2.04 kg omega = w = sqrt (k/m) radians/second so w = sqrt (204/2.04) = 10 rad/s if x is distance from equilibrium position, positive up x = -.05 cos 10 t

*March 30, 2016*

**but**

You may not care about order withing each committee so perhaps you mean combinations, not permutations C(8,5) = 8! /[ 5! (8-5)! ] = 8*7*6/[3*2] = 56

*March 29, 2016*

**Geometry**

w L = 500 L = 5 w/4 w(5w/4) = 500 w^2 = 400 w = 20 L = (5/4)20 = 25

*March 28, 2016*

**Physics 110**

well what is the specific heat or heat capacity of water in Joules or Calories (whichever your class is using)per gram (or kilogram or oz or pound) degree centigrade (or Fahrenheit) anyway call it C Heat in = C * 500 (40-10)

*March 28, 2016*

**finance**

EUR = (1.1210/1.125 ) USD but USD = (112.15/114.25 ) JPY so EUR = (1.1210/1.125)(112.15/114.25 ) JPY so EUR/JPY =(1.1210/1.125)(112.15/114.25 )

*March 28, 2016*

**Statistics**

https://www.khanacademy.org/math/probability/statistics-inferential/normal_distribution/v/ck12-org-normal-distribution-problems-empirical-rule but if you really want the right answer to a specific problem without fooling around with 68 - 95 - 99.7 use http://davidmlane.com/...

*March 28, 2016*

**How on earth**

did you get from such a simple first problem to the problem that messes up people who write papers on this sort of thing ?

*March 27, 2016*

**probability**

LOL this is why it is so hard to do this sort of test more positives will be due to that 10% wrong accuracy problem than due to having allergy. out of 1000 people, 10 have allergy out of these 10 who have it .9 * 10 = 9 test positive correctly 1 gets missed out of the 990 who ...

*March 27, 2016*

**Hey, careful !**

first .25^5 = .0009765 NOT what you got This is a BINOMIAL DISTRIBUTION P(5,k) = C(5,k)p^k (1-p)^(5-k) here p = .25 and 1-p = .75 Now the problem 3 right: C(5,3) = 5!/[3! 2!] = 5*4/2=10 P(5,3) = 10 * .25^3*.75^2 = .0879 4 right C(5,4) = 5!/[4! 1!] = 5 P(5,4) = 5 * .25^4 * .75^...

*March 27, 2016*

**physics**

d = Vi t + (1/2) a t^2 Vi = 20 /3.6 = 5.56 m/s V final = 30/3.6 = 8.33 m/s oh well, I do not need the physics equation because Vaverage = (1/2)(5.56+8.33) = 6.94 m/s 6.94 m/s * 20 s = 139 meters

*March 27, 2016*

**Chemistry**

You are welcome.

*March 27, 2016*

**Chemistry**

note expands as pressure goes down and expands as temperature goes up so you can check if you have those ratios right.

*March 27, 2016*

**Chemistry**

T1 = 330 T2 = 380 1.02 * 2.2 / 330 = .789 * V2 /380 V2 = (1.02/.789)(380/330)(2.2) = 3.28 L

*March 27, 2016*

**typo**

P1 V1/T1 = P2 V2/T2

*March 27, 2016*

**Chemistry**

T1 = 57 + 273 T2 = 107 + 273 P1 V1/T1 = P2 V2/T1

*March 27, 2016*

**math**

you ate 4/12 = 1/3 of the area so you ate (1/3) pi (49) 2/3 of the circumference is left (2/3)(pi)(14)

*March 27, 2016*

**math**

sure, why not 12 15/24 6 9/24 5 16/24 ---------- add 23 40/24 well 40 = 24 + 16 so regroup 23 + 24/24 + 16/24 24 + 2/3 because we already know 2/3 = 16/24 24 2/3

*March 27, 2016*

**Are you there?**

It is really important that you get both of my answers.

*March 27, 2016*

**Statistics**

let me do one point so you can see if an error p of 2 right for example P(10,2)= C(10,2) (.25)^2 (.75)^8 but C(10,2) = 10![2! 8!] = 10*9/2 = 45 so P(10,2) = 45 (.0625)(.1) = .282 does that agree ?

*March 27, 2016*

**Statistics**

yes, p = .25 the mean better be ten/4 :) which is 2.5 This is a binomial distribution mean = n p = 10 * .25 = 2.5 (remarkable :) sigma^2 = n p (1-p) =10 (.25)(.75) = 1.875 sigma = sqrt 1.875 = 1.37

*March 27, 2016*

**physics**

v^2/R = 9.24^2/18.4 = 4.64 m/s^2 centripetal acceleration so apparent weight on road = m (g-4.64) = 1800 (9.81 - 4.64) = 9306 N

*March 27, 2016*

**physics**

http://www.jiskha.com/display.cgi?id=1459109572

*March 27, 2016*

**physics**

Tension holds the weight T = 1.3 * 9.81 = 12.75 N The force on the puck is the tension, 12.75 N 12.75 = m v^2/R v^2 = 1.4 * 12.75 /.24 v = 8.63 m/s

*March 27, 2016*

**Algebra**

18 / 8 = 2.25 half lives 1 unit *(1/2)^2.25 = .210 unit left

*March 27, 2016*

**science help PLS**

because the water treatment plant may be taking water from a surface reservoir or be missing contaminants at the well field that are present in your aquifer.

*March 26, 2016*

**science help PLS**

I think A

*March 26, 2016*

**chem**

200 C = 373 K 20 C = 293 K P1 / T1 = P2 / T2 or P2 = P1 T2/T1 = 5.7 (293/373)

*March 26, 2016*

**physics**

x = xo + Vo t + (1/2) a t^2 72 = (1/2) a (36) a = 4 m/s^2 v = Vo + a t at t = 6 v = 0 + 4*6 v = 24 m/s that is initial Vo at 6 if there is no friction, it continues at 24 m/s for 6 s 24 * 6 = 144 m more

*March 26, 2016*

**math**

DRAW IT !!!!! First the triangle on the left (2,3), (8,3), (8,11) base = 6, altitude = 8 so area = 3*8 = 24 Now the right side, the hard part First rectangle (8,3)(8,11)(11,11)(11,3) that area = 3*8 = 24 subtract the triangles top and bottom top is (8,11)(11,7)(11,11) area top...

*March 26, 2016*

**math**

p good = .2 p not good = .8 this is a binomial distribution problem P(n,k) = C(n,k) p^k (1-p)^(n-k) where C(n,k) = n!/[k!(n-k)!] here for example P(10,1) = C(10,1).2^1 .8^9 C(10,1)= 10!/[1(9)!] = 10 so P(10,1) = 10 * .2 * .8^9 = .268 similarly C(10,2) = 10!/[2(8!] = 10*9/2 =45...

*March 26, 2016*

**mathematics**

Oh, I see, it is math, well check physics below.

*March 26, 2016*

**not physics**

We physicists have trouble with this sort of thing. ja + jo = 25 (ja - 8) = 2 (jo-8) ---------------------- well then ja = (25-jo) so (25 - jo - 8) = 2 jo - 16 17 - jo = 2 jo -16 3 jo = 33 jo = 11 I guess ja = 25-11 = 14 then

*March 26, 2016*

**trigonometry**

height + hypotenuse height = 10 tan 60 hypotenus = 10/cos 60 = 20 so 17.32 + 20 = 37.32 meters

*March 26, 2016*

**TRIGONOMETRY**

csc = 1/sin so sin = 1/ csc sin^2 = 1/csc^2 cos^2 = 1 - sin^2 = 1 - 1/csc^2 so tan^2 = sin^2/cos^2 = (1/csc^2)/[1-1/csc^2] tan^2 = 1/[csc^2 - 1] tan = sqrt (1/[csc^2 - 1] )

*March 26, 2016*

**math**

yes = 1.31 / 31200 = 4.2*10^-5 = .000042

*March 26, 2016*

**geometry**

I have no idea what your cone looks like but to find the volume of anything with straight sides and a pointy top multiply the base area by the height and divide by 3

*March 25, 2016*

**physics**

mu * normal force = m g normal force = m omega^2 R so m g = mu m omega^2 R mu = g/(omega^2 R) or as we could have guessed mu = gravity acceleration/centripetal acceleration mu = 9.81/[ 4.2^2 * 3.1 ]

*March 25, 2016*

**wait a minute**

I am not just going to do these for you. I already took the course back in 1955. You are the one who needs the practice. I did one, the rest are really the same. Try.

*March 25, 2016*

**Algebra 2**

y-4 = 3/(x-4) maybe?

*March 24, 2016*

**Calculus Applied Integrals**

I suppose you posted that question just to find out who here was a mathematician and who an engineer :)

*March 24, 2016*

**Calculus Applied Integrals**

LOL, consider it frozen :) Well, it would be easier to siphon it out but anyway: How much work must you do to lift that amount of water from the height of its center of mass in the tank to the top of the tank? Volume of water = pi r^2(10) = 10 pi (12.2^2) m^3 center of gravity...

*March 24, 2016*

**Math**

c </= 2,000/3

*March 24, 2016*

**Physics**

It says the cross section is square Area of square = side^2 or in this case .31cm*.31 cm or 0.0031 meter * 0.0031 meter

*March 24, 2016*

**whoa**

whoa, at the instant it vanished? dA/dt = s dh/dt as s ---> 0 and dh/dt is constant dA/dt ---> 0

*March 23, 2016*

**Physics**

dA/ds = sqrt 3 (1/2) s dA/dt = .5 sqrt 3 sds/dt but altitude h = (s/2) sqrt 3 dh/dt = .5 sqrt 3 ds/dt so ds/dt = [2/sqrt 3] dh/dt dA/dt=.5 sqrt 3*[2/sqrt 3]s dh/dt = s dh/dt evaluate when s = 17 and dh/dt = -2.3

*March 23, 2016*

**Math**

assuming the driver never sleeps: 2 day = 48 hours 400 * 48/8 = 2400 miles but if the driver drives 8 hour days and the bus stops at Holiday Inn for 16 hours a day then 400 * 2= 800 miles

*March 23, 2016*

**S.S**

I would guess C since the economy certainly did boom after the war.

*March 23, 2016*

**Social Studies**

google assembly line https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&cad=rja&uact=8&ved=0ahUKEwjBpsfphNfLAhUMNSYKHcZWBZ4QFghCMAU&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FAssembly_line&usg=AFQjCNGUJZB6pZGxYuCLr23KrpIxQtld2Q&sig2=QbWU0rnUVGUiIBsPpZMxEg

*March 23, 2016*

**maths**

do you mean? (a+b)/(a-b) + (a-2b)(a+2b)/[(a-b)(a+b)] - (a-3b)/(a+b) if so then (a+b)^2/[(a-b)(a+b)] + (a-2b)(a+2b)/[(a-b)(a+b)] - (a-3b)(a-b)/[(a-b)(a+b)] [ a^2 + 2 a b + b^2 + a^2-4b^2 -a^2 -4 ab + 3b^2]/ [(a-b)(a+b)] = [a^2 -2ab ]/[(a-b)(a+b)] = a (a-2b)/(a^2-b^2) check ...

*March 23, 2016*

**Chemistry - Gas Laws**

p = pressure in tire = 14.7 + pgage CO2 = 12 + 32 = 44 grams/mol so 16 g = .364 mol of CO2 so n = .364 T = 295 deg K V = 3.15 Liters then p V = n R T get that p and subtract 1 atm to get gage

*March 22, 2016*

**PHYSICS! are these correct?**

3. A 5-kg box is sliding across a desk that has a coefficient of friction of 0.4. If the box is initially moving at 15 m/s, how far will the box slide before coming to rest. You MUST use the WORK/Kinetic Energy Theorem to solve this. ============================= initial ke...

*March 21, 2016*

**PHYSICS! are these correct?**

2. How much work must gravity do on a 2-kg falling object in order to take it from rest to 20 m/s? ============================ change in potential energy = m g h that goes into kinetic energy. so how far does it fall to go from 0 to 20 m/s? v = Vi + a t 20 = 0 + 9.81 t t = 2....

*March 21, 2016*

**UNITS !!!!!!!**

How much kinetic energy does a 700-gram baseball have that is travelling at 20 m/s? What would be the kinetic energy if you doubled the mass? What would be the kinetic energy if you doubled the velocity? V1= 20 m/s K1= ½ m v1^2 = ½ (0.7 kg)(20 m/s)^2 = ½ (...

*March 21, 2016*

**physics**

V = E times distance like work = force * distance 140 = 29 x x = 140/29

*March 21, 2016*

**physics**

LOL zero, cancels

*March 21, 2016*

**physics**

momentum change = m v force = rate of change of momentum = m v/t 55 * 28/.2 = 7700

*March 21, 2016*

**Physics helllppp**

108,000m/hr * 1hr/3600s = 30 m/s friction force= -.9 m g = -.9 * 9.81 m = -8.83 m F = m a -8.83 m = m a a = -8.83 m/s^2 v = Vi + a t 0 = 30 - 8.83 t t = 4 seconds

*March 21, 2016*

**Physics**

108,000m/hr * 1hr/3600s = 30 m/s friction force= -.9 m g = -.9 * 9.81 m = -8.83 m F = m a -8.83 m = m a a = -8.83 m/s^2 v = Vi + a t 0 = 30 - 8.83 t t = 4 seconds

*March 21, 2016*

**geometry**

(pi d/2)

*March 20, 2016*

**Physics**

one kg object 1 Newton pushing north 1 Newton pushing north east north components 1 + 1 cos 45 = 1 + .5 sqrt 2 = 1.707 Newtons north east components 1 sin 45 = .5 sqrt 2 = 0.707 Newtons east F = sqrt (1.707^ + .707^2) tan of angle clockwise from north (compass angle) = .707/1.707

*March 20, 2016*

**Math**

3 = 2*1.5 5 = 3 *1.666666 ..... 1.5 is not 1.6666.. so it is not geometric 3-2 = 1 5-3 = 2 so it is not arithmetic

*March 20, 2016*

**mathematics**

tan angle = 25/25 angle = 45 degrees

*March 20, 2016*

**Maths**

P(4) = C(10,4) .3^4 .7^6 C(10,4) = 10!/[4! 6!] = 10*9*8*7/[4*3*2] = 30*7 = 210 so P(4) = 210 * .3^4*.7^6 = 0.02 for the second part do that for 7 8 9 and 10 students and add results. Do not spend more that a week at it. If you get frustrated you can approximate a binomial ...

*March 20, 2016*

**Other**

Does it wink at you when you tap it with a hammer?

*March 20, 2016*

**asomaning**

To increase the temperature of one kilogram of Al one degree Kelvin or Centigrade, you must insert 900 Joules of heat energy.

*March 20, 2016*

**physics**

- mg = - .5 * 9.81 = - 4.9 Newtons the whole time if you ignore air frictional drag

*March 20, 2016*

**algebra**

Maybe means multiply first one by 7 and second one by 3 ? I doubt if they have gotten to Cramer or Gauss Jordan. 21 x + 35 y = 175 21 x + 18 y = 90 ---------------------subtract 0 x + 17 y = 85 y = 5 then x = 0

*March 20, 2016*

**physics**

n1 sin theta 1 = n2 sin theta 2 1.33 sin theta 1 = 1 sin 90 sin theta = 1/1.33 theta = 48.8 degrees That is actually the minimum angle for which total internal reflection occurs. Any less than that and it gets through

*March 20, 2016*

**gibberish ???**

Two charges on three vertices? anyway do E = k q/r^2 in the direction from each charge to the spot and add the vectors.

*March 20, 2016*

**physics**

a = 5 m/s^2 v = Vi + a t 25 = 10 + 5 t

*March 20, 2016*

**math**

y = k x^n maybe ? if y = x^2 then n = 2 if y = x^3 then n = 3

*March 20, 2016*

**Physics**

If there is no net force, there is no net acceleration and velocity is constant. Force = rate of change of momentum F = m A if m is constant

*March 19, 2016*

**sir damon help on these as well co-ordinate**

It is a square root sign I suspect and it is not showing up in your font. Her answer is the same as mine.

*March 19, 2016*

**sir damon help on these as well co-ordinate**

30 degrees above negative y axis y = 3 sin 30 = 1.5 x = - 3 cos 30 = -3 (sqrt 3)/2 ( -1.5 sqrt 3 , + 1.5 )

*March 19, 2016*

**math series**

1/(3*1) + 1/(4*2) + 1/(5*3) + 1/(6*4) 1/[(n+2)n] or 1/(n^2+2n)

*March 19, 2016*

**no idea**

I do not know what sort of harmonic series that is. None I know of.

*March 19, 2016*

**Math**

neither, I am not trying to fit that with a line. You might try point - slope but it will not fit very well. Your data is too bumpy to fit well with a straight line.

*March 19, 2016*