Wednesday

April 1, 2015

April 1, 2015

Total # Posts: 20,759

**Math**

Yes, it is a quadratic equation
*October 18, 2014*

**Math**

been there, done that see below
*October 18, 2014*

**math**

LOL, well if 6 is the answer then -16(36) + 96 (6) better be zero is it ? :)
*October 18, 2014*

**Math**

yes, quadratic equation
*October 18, 2014*

**Math**

speed before Forde = (s-5) speed after Forde = s time to Forde = (t+96) note 4 days = 96 hr time after Forde = t then distance = speed * time 70 = (s-5)(t+96) 60 = s t or t = 60/s 70 = (s-5)(60/s + 96) 70 = 60 + 96 s - 300/s - 480 490 = 96 s - 300/s 245 = 48 s - 150/s or 48 s^...
*October 18, 2014*

**Physics**

v = Vi + a t d = Vi t + (1/2) a t^2 2240 = 199 t + 4.25 t^2 solve quadratic for t then v = 199 + 8.5 t
*October 18, 2014*

**Physics**

d = (1/2) a t^2 176 = 4.35 t^2 solve for t v = a t = 8.7 t
*October 18, 2014*

**PHYSICS 1301 (COLLEGE)**

When people do not even go back and see if what they typed makes any sense, I just skip to the next question.
*October 18, 2014*

**Calculus**

v = -gt = -9.8*2 100 = 4.9 t^2 t = 4.52 seconds to ground so v = -9.8*4.52
*October 18, 2014*

**chemistry**

.55 a = .45(715)
*October 18, 2014*

**calculus**

4 cos anything is never bigger than 4 so there is no overlap here
*October 18, 2014*

**Calculus**

2 * L + 4*w = 800 so L = 400 - 2w A = L * w A = (400 -2w)w A = -2 w^2 +400 w dA/dw = 0 at max = -2 w + 200 w = 100 L = 400 -200 = 200
*October 18, 2014*

**Calculus**

decrease at start due to startup costs and paying off initial investment in staff and equipment increase over long term because we have to go deeper and deeper into the vein as we mine out the easy stuff near ground level.
*October 18, 2014*

**Chemistry**

heat it 25 deg C 550 * 2.03 * 25 Joules
*October 18, 2014*

**physics please help**

In the end torque = F*.4 - friction torque = I d omega/dt
*October 18, 2014*

**physics please help**

Not knowing that frictional torque, I am at a loss. Please proof read question.
*October 18, 2014*

**Physics**

69 v1 = 47 v2 so v2 = (69/47)v1 19 = (v1+v2)t 19 (v1 + 1.47 v1) t so v1 t = distance of first = 19/2.47 = 7.69 meters
*October 18, 2014*

**Physics**

same old conservation of momentum at first p = 2 *1 + m * 4.4 at the end p = (2+m) * 1.5 so (2+m)(1.5) = 2 + 4.4 m 3 + 1.5 m = 2 + 4.4 m etc
*October 18, 2014*

**DIFF CALCULUS**

now the volume is another story because now we will have to make some assumption, for example constant slope up from 3 m to 1 m. In that case we have a trapezoid and V = 6 * 12 * average depth = 6 * 12 * 2 - 144 m^3
*October 18, 2014*

**DIFF CALCULUS**

Part b is easy and requires no assumptions about the shape of the pool bottom. The rate of change of height = volume added per second / surface area of water. (draw a picture) or dh/dt = .25 m^3/min / (12m*6m) = .00347 m/min or 21 cm/hour
*October 18, 2014*

**Algebra typo**

22x maybe? what is the question anyway? factor it? put in an equal sign and solve it find zeros of parabola ? by the way x^2 + 22 x + 121 = (x+11)^2 (x+11)^2 -y^2 = (x+11-y)(x+11+y) because a^2-b^2 = (a-b)(a+b)
*October 18, 2014*

**physics**

if v = 0, p = 0 v ground = sqrt (2 g h) momentum at ground = m v = 4 sqrt (2 g h)
*October 18, 2014*

**physics**

falls from top for 6.65/2 = 3.325 seconds h = (1/2) (9.8) (3.325)^2
*October 17, 2014*

**MATH HELP PLEASE**

no sqrt 36 = 6 but what is sqrt 21 ?
*October 17, 2014*

**Last ratio/equation**

m/w = 10/12 so w = 12 m/10 then w/c = 10/15 12 m /10 / c = 10/15 12 m/c = 100/15 = 20/3 m/c = 20/36 = 5/9
*October 17, 2014*

**equations**

a/b = 5 so b = a/5 use that in the second a/5 / c = 2/3 multiply by 5 a/c = 10/3
*October 17, 2014*

**equation**

divide both sides by 3b 3a/3b = 5b/3b simplify a/b = 5/3
*October 17, 2014*

**Science-Physics**

y velocity = 8.9 x velocity = 19.3 tan theta = (8.9/19.3)
*October 16, 2014*

**Science (Physics)**

acceleration = change in velocity / change in time same as speed on the way up, but down not up or in other words it falls from 3 meters up m g h = (1/2) m v^2 so v = -sqrt(2 g h) = -sqrt ( 2 * 9.81 * 3)
*October 16, 2014*

**Algebra**

LOL - It happens
*October 16, 2014*

**Algebra**

for heaven's sake subtract: 1.10 - 1.05 = ?????? 1.15 - 1.10 = ??????
*October 16, 2014*

**physics**

I hope and pray we are talking about buoys and not boys. Anyway, here is a solution for totally immersed spheres. If the sphere waterlines are halfway down the sphere, then double the resistance. (you have half as much conductance) http://www.df.unipi.it/~macchi/TEACHING/...
*October 16, 2014*

**normal distribution**

http://davidmlane.com/hyperstat/z_table.html
*October 16, 2014*

**Algebra 2**

Now that point (5,6) 5 is beyond x = 3.5 so on the way down y = (7/2)(5)- (1/2)(25) = 10/2 = 5 will not make it over wall
*October 15, 2014*

**Algebra 2**

y = (7/2) x - (1/2)x^2 reasonable domain is above ground, y>0 so solve for zeros of x x^2 -7x = 0 x = 0 to x = 7 range (vertex is when x = half 7 or3.5 ymax = (7/2)(7/2) -(1/2)(7/2)^2 = (1/2)(49/4) = 49/8 = 6.125 so range is 0 to 6.125
*October 15, 2014*

**Algebra 2 well, maybe**

well, maybe y = a + b x + c x^2 3 = a + b(1) + c(1) 5 = a + b(2) + c(4) 6 = a + b(3) + c(9) so 2 = b + 3 c 1 = b + 5 c ---------- 1 = -2c c = -1/2 1 = b -5/2 b = 7/2 3 = a + 7/2 -1/2 = a + 3 so a = 0 y = 0 + 7 x/2 - x^2/2 y = (7/2) x - (1/2)x^2 or 2 y = 7 x - x^2
*October 15, 2014*

**Algebra 2**

y = a + b x^2 3 = a + b (1)^2 = a + b 5 = a + b(2)^2 = a + 4 b b = 3 - a 5 = a + 4(3-a) 5 = a + 12 - 4a 3 a = 7 a = 7/3 b = 9/3 - 7/3 = 2/3 so y =7/3 +2/3 x^2 check last point y = 7/3 + (2/3)(9) = 7 1/3 I do not think your table is a parabola
*October 15, 2014*

**Pre-Calc/Trig...**

v = Vi - g t at top v = 0 0 = 4 -32 t t = 1/8 second (4 ft /sec is really slow. I bet you mean 40) h = 10 + 4 (1/8) - 16 (1/64) = 10 + .5 - .25 = 10.25 ft
*October 15, 2014*

**science**

both the same ( Newton #3 action equal and opposite to reaction) F = G * 6 * 7.3 / .65^2 where G = 6.67 * 10^-11
*October 15, 2014*

**Math**

maybe ln y = ln 35.15 + 1.04 x
*October 15, 2014*

**Algebra 2**

332/2 = 166 = v+c 318/2 = 159 = v-c 2 v = 166+159 = 325 etc
*October 15, 2014*

**Math 112**

(86 + 88 + z)/3 = 90 86 + 88 + z = 270 etc
*October 15, 2014*

**Pre-Calc/Trig...**

(2x-1)(x+3) = 0 --> 1/2 and -3 x^2 - x - 1 = 0 x = [1 +/- sqrt (1+4) ]/2 = 1/2 +/- (1/2) sqrt 5
*October 15, 2014*

**Physics Help Please!!!!**

the mass of the shuttle is decreasing because fuel is being expelled F = m a same F less m more a Same happens with your car, but so slowly it does not matter :)
*October 15, 2014*

**physics... PLEASE HELP**

F = m a = 9 * 10^13 * 1.7 * 10^-27 = 15.3 * 10^-14 = 1.5 * 10^-13 Newtons
*October 15, 2014*

**physics**

R = 1.3 * 10^7 * 8 * 10^-9 ohms/m^2 i = V/R b. Same V, double R, half i
*October 15, 2014*

**physics... PLEASE HELP**

The relation between the energy and the momentum is E = p c where p is momentum so our momentum density here is our energy density/c
*October 15, 2014*

**physics... PLEASE HELP**

well, it is late but I might be able to point a direction average intensity = average power / (pi r^2) Peak = twice average power if a sine wave [ average of sin^2 = (1/2) ] energy density usually in Joules per square centimeter where it hits I assume per second. That would be...
*October 15, 2014*

**Physics - Second Law Question**

same as the other question (3-1.96)10^5 Newtons up
*October 15, 2014*

**Physics - Second Law Question**

4.9 - 4.5 = 0.4 0.4*10^-3 = 4 * 10^-4 down because weight is bigger than drag up
*October 15, 2014*

**Physics**

NO!! Usually in a given problem the SUM of kinetic and potential energy is more or less constant. I say more or less because energy is easily lost from your nice neat problem through friction or escape of heat or whatever messy thing happens to make your car not coast forever ...
*October 15, 2014*

**Math**

(7/8)hr/(3/4) puzzle = 7/6 hours/puzzle or 6/7 puzzles/hour
*October 15, 2014*

**Pre-Calc/Trig...**

b^2-4ac 5^2 -4(2)(-3) that will be positive and therefore its square root is a real number. Thus two real roots, nothing imaginary about it 1^2 - 4(-1)(1) everything positive again, two real roots
*October 15, 2014*

**Physics**

x momentum after = x momentum before after p = 38 (6.06 cos 10.5) + 2.37 v before p = (38+2.37)(5.07) set equal, solve for v which will probably be negative (kid goes forward, board goes backward)
*October 15, 2014*

**Pre-Calc/Trig...**

4 sqrt -1 = 4i 7 sqrt -1 = 7i
*October 15, 2014*

**Math**

do you mean??? [(-3) ^-4 ] [ 5 ^-2 ]=? if so [1/(-3^4 )] (1/25) (1/81)(1/81)(1/25)
*October 15, 2014*

**Algebra 2**

1+6+11 = 18 = 2/3 of cast so cast = 27 27 - 18 = 9 which is indeed 1/3 of the cast
*October 15, 2014*

**Algebra 2**

cost = c Briana + 73 = c Molly + 65 = c Briana + Molly = c so molly = (c - Briana) now we have Briana + 73 = c (c - Briana) + 65 = c or Briana = 65 Briana + 73 = c 73 + 65 = c = 138
*October 15, 2014*

**Science**

been there, done that http://www.jiskha.com/display.cgi?id=1413411282
*October 15, 2014*

**Mechanics**

x velocity = 15 across y velocity = .9 downstream note order of answers c. time to cross = 500/15 = 33.33 seconds b. downstream y = .9 * 33.33 = 30 meters a. sqrt (.9^2+15^2) angle to x axis = A = tan^-1(.9/15) d. sin^-1 (.9/15) toward upstream e. component of speed across = ...
*October 15, 2014*

**math**

n-2, n, n+2 4(n-2) - n = 12 + 2(n+2) 4 n - 8 - n - 12 - 2n - 4 = 0 n - 24 = 0 n = 24 22 , 24, 26
*October 15, 2014*

**Math**

let z = cos A 4 z^2 - 3 z = 0 z(4z-3) = 0 z = 0 or z = 3/4 so cos A = 0 when z = 90 degrees cos A = 3/4 when z = 41.4 degrees but that is not between 90 and 180
*October 15, 2014*

**Algebra Help**

(2x+5)^.5 - (x+6)^.5 = 1 you mean maybe, parentheses missing if so (2x+5)^.5 = 1 +(x+6)^.5 2x+5 = 1 + 2 (x+6)^.5 + x+6 (x - 2) = 2 (x+6)^.5 x^2 - 4 x + 4 = 4 x + 24 x^2 - 8 x - 20 = 0 (x-10)(x+2) = 0 x = 10 or x = -2 go back and see what works in the original now.
*October 15, 2014*

**algebra**

1. let z = x^2 then yo have 4 z^2 - 4 z + 1 (2z-1)(2z-1) which is (2x^2-1)(2x^2-1) 2. a^2 (a^2+1)
*October 15, 2014*

**algebra**

(x+3)(x-2) 3*5 = 15 which differs from 113 by 2 so (3x-1)(2x+5)
*October 15, 2014*

**Linear Algebra**

.866 is sin 60 by the way
*October 13, 2014*

**Linear Algebra**

L1 = 1 L2 = -.5 - .866 i L3 = -.5 + .866 i eigenvectors messy see http://www.wolframalpha.com/widgets/view.jsp?id=9aa01caf50c9307e9dabe159c9068c41
*October 13, 2014*

**Algebra 2 Answer Check**

just put in a bunch of cost values from 0 to 10 if the cost were $2 change = 8 cost - change = 2 - 8 = -6 and 10 - 2(8) = -6 It is only non-negative if the cost is between 5 and 10
*October 13, 2014*

**Algebra 2 Answer Check**

I misread it cost = 10 - change cost - change = 10 - 2 change
*October 13, 2014*

**Algebra 2 Answer Check**

Huh f(x) = 10 - x
*October 13, 2014*

**Algebra 2**

You are right, there is no solution to #1
*October 13, 2014*

**Algebra 2**

Oh, I see, you are right. no positive divided by -5 could be + 6
*October 13, 2014*

**Algebra 2**

-26/3 is negative 34/3 is positive but they could both be negative anyway
*October 13, 2014*

**Algebra 2**

By the way to graph that parabola y = 5 x^2 - 8 x + 11/3 http://www.wolframalpha.com/input/?i=5+x^2+-+8x+%2B+11%2F3
*October 13, 2014*

**Algebra 2**

1. (|3x -4|)/(-5) = 6 A: no solution ========================= multiply both sides by - 5 , no arrow business because just an equal sign 3 x - 4 = -30 or -3x + 4 = -30 3 x = -26 or -3x = -34 so there are two solutions
*October 13, 2014*

**Algebra 2**

6. |12 - 4x| - 4 > 20 A: x < -3 and x > 9 ============================ 12 -4 x > 24 or -12 + 4x >24 -4x > 12 or 4x > 36 x < -3 or x > 9
*October 13, 2014*

**Algebra 2 #1 sorry**

1. (|3x -4|)/(-5) = 6 (|3x -4|) = -30 or 3x-4 = -30 or -3x+4 = -30 so x = -26/3 or x = 34/3
*October 13, 2014*

**Algebra 2**

5. (-3) x (|5x - 8|)- 5 greater than or equal to 6 A: no solution? ============================ 3 (-3) x (|5x - 8|)- 5 >/= 6 (-3) x (|5x - 8|) >/= 11 x (|5x-8)| </= -11/3 x(5x-8) </= -11/3 x(-5x+8) </= -11/33 5 x^2 - 8 x </= -11/3 -5 x^2 + 8 x </= -11/3 ...
*October 13, 2014*

**Algebra 2**

4. [(|x - 2|)/ (4) greater than or equal to 5 A: -18 < x < 22 =========================== x-2 >/=20 or -x+2 >/= 20 x >/= 22 x </= -18
*October 13, 2014*

**Algebra 2**

3. |5x + 15| > 20 A: x > 1 and x < -7 5x +15 > 20 or -5x - 15 >20 x > 1 yes -5x > -35 divide both sides by -5 and reverse the arrow x < 7 why the sign ?????
*October 13, 2014*

**Algebra 2**

2. (|2x -5|) = x + 3 A: x = 8 or x = 2/3 ========================= 2x-5 = x+3 or -2x+5 = x+3 x = 8 or x = 2/3 you got that right. Why did you miss #1 ????
*October 13, 2014*

**Algebra 2**

1. (|3x -4|)/(-5) = 6 (|3x -4|) = -30 or 3x-6 = -30 or -3x+6 = -30 so x = -24/3 = -8 or x = 12
*October 13, 2014*

**Math**

2 sin T cos T cos T /sinT = 2 cos^2 T
*October 13, 2014*

**algebra**

a circle with radius 10 and center at (7,6) (x-7)^2 + (y-6)^2 = 10^2 = 100 where does it hit the y axis (y=0) ? (x-7)^2 + 36 = 100 (x-7)^2 = 64 (x-7) = +/- 8 x = 7 +/- 8 if x = 7+8 then x = 15 if x = 7-8 then x = -1
*October 12, 2014*

**Physics**

The lighter one had to be going faster. To be at 45 degrees, momentum north = momentum east 2000 v = 3000 (2v/3)
*October 12, 2014*

**Physics 12**

u = 12 cos 42 = 8.92 m/s the whole time Vi = 12 sin 42 = 8.03 v = Vi - 9.81 t h = 9.5 + Vi t - 4.9 t^2 0 = 9.5 + 8.03 t - 4.9 t^2 t = [ -8.03 +/- sqrt (8.03^2-4*-4.9*9.5)]/-9.81 t = [-8.03+/-sqrt(251) ]/-9.81 t = 2.43 s
*October 12, 2014*

**calculus**

a = area = 6 x^2 da/dx = 12 x so da = 12 x dx
*October 12, 2014*

**calculus**

Again? http://www.jiskha.com/display.cgi?id=1413057054
*October 12, 2014*

**Nonsense**

It is true that in obsolete English units H(t) = Hi + Vi t - (1/2)(32) t^2 if Hi = 72 ft and Vi = 0 then H(t) = 72 - 16 t^2 however your next line makes no sense at all. perhaps you mean when is H = -184 ??? then 72 - 16 t^2 = -184 16 t^2 = 256 t^2 = 16 t = 4 seconds
*October 12, 2014*

**Calculus**

dv = surface area * dh so dv/dt = surface area * dh/dt width at surface = 40 + (80-40)(30/40) = 40 + 30 = 70 cm = 0.70 m so surface area = 9 * .7 = 6.3 m^2 so .3 m^3/min = 6.3 m^2 * dh/dt and dh/dt = .047 meters/min or 4.7 cm/min
*October 12, 2014*

**physics**

time 2 = 1.81 * time 1 h = (1/2)gt^2 so h/t^2 = constant = H1/t1^2 = H2/t2^2 H2/H1 = t2^2/t1^2 = 1.81^2 = 3.27
*October 12, 2014*

**College Algebra**

Yes, but it is trivial p = 0 and p = 10 give R = 0
*October 12, 2014*

**physics**

m g down ramp = m g sin 28 mg normal = m g cos 28 m g sin 28 - mu m g cos 28 = m (.98) sin 28 - mu cos 28 = .98/9.81
*October 12, 2014*

**physics**

potential energy proportional to height, m g H so Hn = (1-.0534)Hn-1 Hfinal = .9466^n Hinitial 2.79 = .9466^n 5.89 .9466^n = .4737 n log .9466 = log .4737 n = 13.6 so 13
*October 12, 2014*

**calculus**

well Diagonal = D = x sqrt 2 so dD = sqrt 2 dx a = x^2 so da = 2 x dx etc
*October 11, 2014*

**physics**

horizontal force exerted = 75 cos 22 +75 cos 32 vertical force exerted = 75 (sin 32 -sin22) normal force on ground = m g - 75(sin32-sin22) friction force = mu (normal force) acceleration = 0 so friction force = horizontal force exerted work done = 13 * friction force You did ...
*October 11, 2014*

**Math**

.12/52 = .002308 so for n weeks 3 = 1.002308^n log 3 = n log 1.002308 n = 476 weeks or 9.16 years
*October 11, 2014*

**Calculus**

v = x^3 729 = 9^3 so x = 9 dv = 3 x^2 dx = 243 dx = 849.285 - 729 etc
*October 11, 2014*

**Calculus**

Hint: http://www.jiskha.com/display.cgi?id=1413053733
*October 11, 2014*

**Trigonometry**

Hint: sin 2a = 2 sin a cos a so what is 3 sin 2a?
*October 11, 2014*

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