Friday

May 27, 2016
Total # Posts: 11,830

**Math**

I chose b also.
*February 16, 2016*

**Math**

1. Correct. 2. Correct.
*February 16, 2016*

**Basic electricity**

The given units are incorrect.
*February 16, 2016*

**science**

20 minutes per hour makes no sense. So let's assume: Vo = 20m/s[53o]. Xo = 20*Cos53 = 12.0 m/s. Yo = 20*sin53 = 16.0 m/s. a. Y = Yo + g*Tr = 0, 16 - 9.8*Tr = 0, -9.8Tr = -16, Tr = 1.63 s. = Rise time. Tf = Tr = 1.63 s. = Fall time. T = Tr+Tf = 1.63 + 1.63 = 3.27 s. = Time ...
*February 16, 2016*

**MATH HELP**

My answer is b.
*February 15, 2016*

**Math College Algebra (urgent before 2hours)**

The zero of a function is the value of X when Y or F(x) = 0. In other words it is the solution to the function. Example: Y = 2x + 4 = 0, 2x = -4, X = -2. So x = -2 when y = 0: (-2,0). 2. g(y) = y^2-3y+2 = 0, C = 2 = (-1)*(-2), Sum = B = -3. (y-1)(y-2) = 0, y-1 = 0, Y = 1, y-2...
*February 15, 2016*

**Science**

10o W. of N. = 100o CCW. 10o E. of N. = 80o CCW. Fr = 1.8*10^6[100o] + 1.8*10^6[80o]=-3.126*10^5+1.773*10^5i + 3.126^10^5+1.773*10^5i = 3.546*10^5i = 3.546*10^5[90o] = 3.546*10^5N., Due North. = Resultant force. Work = Fr*d = 3.546*10^5 * 500 =
*February 15, 2016*

**science**

F1 = 15N[0o], F2 = 20N[135o]. F1+F2 = 15 + -14.1+14.1i = 0.9 + 14.1i = 14.2 N.[86.4o] = 0.89 + 14.18i F3 must be equal to F1+F2 in magnitude and 180o out of phase: F3 = 14.2N[86.4+180] = 14.2[266.4o].= 14.2[86.4o] S. of W. = -0.89 - 14.18i
*February 14, 2016*

**Physics**

a. Work = Change in kinetic energy: Work = 0.5M*V^2-0.5M*Vo^2, Vo = 0, Work = 0.5M*V^2 = 0.5*845*25^2 = 264,063 J. Po = W/t = 264063/60 = 4401 J/s. = 4401 Watts = Power out. Pi = 4401/0.95 = 4633 Watts = Power in. Pi = 12*I = 4633, I = 4633/12 = 386 Amps.
*February 14, 2016*

**Maths**

r = 35/4 = 8 3/4% = 8.75% = 0.0875. I = Po*r*t = 3000*(0.0875/12)*14 = 90.0875*
*February 14, 2016*

**Physics**

1. V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 80 m, V = ?. 2. 0.5g*t^2 = 80. g = 9.8 m/s^2, t = ?.
*February 13, 2016*

**Physics**

Wc = M*g = 35*9.8 = 343 N. Fp = 343*sin26 = 150.4 N. = Force parallel to the incline. Fn = 343*Cos26 = 308.3 N. = Normal. Fp-Fk = M*a. 150.4-0.33*308.3 = 35*a. 35a = 48.66. a = 1.39 m/s^2. V = Vo + a*t = 0 + 1.39*2.5 = 3.48 m/s.
*February 13, 2016*

**physics**

V^2 = Vo^2 + 2g*h, V = 0, Vo = 5m/s, g = -10m/s^2, h = ?.
*February 13, 2016*

**physics**

177 + 0.00506(T-28.7)177 = 116. 177 + 0.8956(T-28.7) = 116. 177 + 0.8956T-25.7 = 116. 0.8956T + 151.3 = 116. 0.8956T = 116-151.3 = -35.3. T = -39.4oC.
*February 13, 2016*

**physics**

F = M*a, F = 37.5 N, a = 2.80 m/s^2, M = ?.
*February 13, 2016*

**Cal 2**

The region bounded by y=3/(1+x^2), y=0, x=0 and x=3 is rotated about the line x=3. Using cylindrical shells, set up an integral for the volume of the resulting solid. The limits of integration are:
*February 12, 2016*

**Biology**

How are the following terms related to the process of transcription and one another? Eukaryotes, Prokaryotes, membrane-bound nucleus, introns, exons, precursor RNA (pre-mRNA), mature RNA, heterochromatin, euchromatin, and structural gene.
*February 12, 2016*

**geometry**

Or 7.5/7 = x/18, X = ?.
*February 11, 2016*

**physics**

Vector V1: Vx = -6.5 units. Vy = 0. Vector V2: V = 8.3[60o] Vx = 8.3*Cos60 = Vy = 8.3in60 = V1 + V2 = -6.5 + 8.3[60] = -6.5 + 4.15+7.19i = -2.35 + 7.19i = 7.56[71.9o] N. of W. = 108o CCW
*February 11, 2016*

**physics**

Do you mean constant velocity? If so, d = V*t, t = d/V.
*February 11, 2016*

**physical science**

d = 0.5a*t^2. d = 10 m, t = 45 s, a = ?.
*February 10, 2016*

**mathematics**

David: X years old. John: x+4 years old. Joe: x-2 years old. x + x+4 + x-2 = 41. Solve for x and add 8 years to John's age.
*February 10, 2016*

**Pysics**

D = -400 + 200[135o] + 100i = -400 - 141.4+141.4i + 100i = -541.4 + 241.4i = 593 km[24o] N. of W.
*February 10, 2016*

**Science**

1. V = Vo + a*t. V = 3.2 m/s, Vo = 1.2 m/s, t = 2.5 s, a = ?. 2. V = Vo + a*t. V = 34 m/s, Vo = 0, t = 2.5 s, a = ?. d = 0.5a*t^2. 3. r = 30m/4.2s. = 4. d = r*t, t = ?.
*February 10, 2016*

**Math**

The simplification process starts inside the expression: ((3x)^4/(6x)^3)^2 = (81x^4/216x^3)^2 = (3x/8)^2 = 9x^2/64.
*February 9, 2016*

**Math**

Y = 3x. 3 = The number of wheels per tricycle. X = The number of tricycles. Y = The total number of wheels to be ordered.
*February 9, 2016*

**math**

15, 5, 3, 0.8, -5.
*February 9, 2016*

**Math**

Po*e^(rt) = 2Po. Divide both sides by Po: e^rt = 2. rt = 0.08t. e^0.08t = 2. 0.08t*Log e = Log 2. 0.08t = Log2/Log e = 0.693 t = 8.7 yrs.
*February 9, 2016*

**math**

a. Tan 24 = h/500, h = ?. b. Tan 27 = (h+L)/500. L = Length of flag = ? h = Ht. of bldg. calculated in part a.
*February 9, 2016*

**physics**

See previous post: Mon, 2-8-16, 7:33 PM.
*February 9, 2016*

**science**

30kW = 30 kJ/s. Work = 30kJ/s * 15s = 450 kJ.
*February 9, 2016*

**Calculus 2**

Sketch the region in the first quadrant enclosed by y=7/x, y=2x, and y=x/2. Decide whether to integrate with respect to x or y. Then find the area of the region.
*February 9, 2016*

**physics**

a. Vr = -100i + 70km[330o] = -100i + 60.6-35i = 60.6 - 135i = 148km[65.8o] S. of E = 24.2o E. of S. = Resultant velocity of the goose. Do you mean 20o W. of S.? If so, Direction = 24.2 + 20 = 44.2o W. of S. b. d = V*t = 1000, t = 1000/V = 1000/148 = 6.76 h.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 5000*0.04*2 = $400. (25/4)% = 6.25%. I = 5000*0.0625*2 = $625. Gain = (625-400)/2yrs. = $112.50 per year.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 0.125Po. Po*0.1*t = 0.125Po. t = 1.25 yrs. = 15 Months.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 1344-1200. Po = 1200, r = 0.06, t = ?.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 1770. r = 0.08 t = (15/2)yrs. = 7.5 yrs. Po = ?.
*February 8, 2016*

**SIMPLE INTEREST**

There are errors in your problem; please make corrections.
*February 8, 2016*

**science**

Sketch: From a point, draw 2 arrows pointing upward and one arrow pointing downward.
*February 8, 2016*

**science**

Fr = 5.3 + 2.2 - 10.7 = -3.2 N. = 3.2 N. downward.
*February 8, 2016*

**Engineering**

Mass = 0.00072kg/cm^3 * 37854cm^3/gal * 5gal. = 136.3 kg.
*February 7, 2016*

**physics**

h = V*t + 0.5g*t^2 = 2 m. V*0.1 + 4.9*0.1^2 = 2. 0.1V = 2-0.049 = 1.95. V = 19.5 m/s at top of the window. V^2 = Vo^2 + 2g*h. V = 19.5 m/s, Vo = 0, g = 9.8 m/s^2, h = ?.
*February 7, 2016*

**Finance**

P = Po(1+r)^n. P = $16,860. r = 5.75%/100 = 0.0575 n = 1Comp./yr. * 6yrs. = 6 Compounding periods. Po = ?.
*February 7, 2016*

**science**

a. h = 0.5g*t^2. h = 20 m., g = 9.8 m/s^2. t = ?. b. d = 10m/s * t. t = Value calculated in part "a".
*February 7, 2016*

**physics**

a. V^2 = Vo^2 + 2g*h = 0. Vo^2 = -2g*h = -(-19.6)*0.18 = 3.53, Vo = 1.88 m/s. b. KE = 0.5M*Vo^2 = 0.5*0.54*10^-6*1.88^2 = 9.54*10^-7 Joules.
*February 7, 2016*

**Physics**

Xo = 2.9 m/s. Tan 43.6 = Y/Xo = Y/2.9. Y = 2.9*Tan 43.6 = 2.76 m/s. V^2 = Vo^2 + 2g*d = 2.76^2. 0 + 19.6d = 7.63, d = 0.389 m. h = 1.14 - 0.389 = 0.751 m. Above the floor.
*February 6, 2016*

**physics**

See previous post: Thu,2-4-16, 11:20 AM.
*February 5, 2016*

**Physics**

Oops! The name is Henry.
*February 4, 2016*

**Math**

A semicircular shape is in a (fills it) rectangle. If the perimeter of the rectangle is 30 cm, fine the perimeter of the semicircular shape.
*January 11, 2016*

**Calculus**

A water tank is made by rotating f(x)=2^x-1 between [0,2] about the y-axis. The water tank is initially full, when a hole is opened at the bottom tip so that the water drains at a rate of 4 unit^3/second. How fast is the height of the water decreasing when the height has ...
*January 8, 2016*

**Calculus**

Okay I didn't know you could do that haha
*January 2, 2016*

**Calculus**

Wait, but how can you let u=x^3 when there is not an x^3 in the equation
*January 2, 2016*

**Calculus**

Last one for me. Evaluate (3x^2)/sqrt(1-x^6) dx I changed sqrt (1-x^6) into 1^(1/2)- x^(3) and let u=1^(1/2)- x^(3) then I got du/dx=3x^2 and subbed dx=du/3x^2 into the equation and got x-(x^4/4) but its not right. Options 7x^3/(3sqrt(1-x^7) + C cos-1(3x) + C sin-1(x^3) + C ...
*January 2, 2016*

**Calculus**

Im think im going to go with B then
*January 2, 2016*

**Calculus**

Okay Thanks. That one confused me.
*January 2, 2016*

**Calculus**

Okay will do. So I'm presuming your saying the answer is B. I think D works.
*January 2, 2016*

**Calculus**

Wait so A is right or is it wrong?
*January 2, 2016*

**Calculus**

Which of the following integrals cannot be evaluated using a simple substitution? I think it is A because if you would substitute there would be nothing left in the equation? Is that right? Options ∫√(x-1) ∫1/√(1-x^2) ∫x/√(1-x^2) ∫&#...
*January 2, 2016*

**Calculus**

Oh shoot I misread my answer. Thanks
*January 2, 2016*

**Calculus**

∫(x^3-x^2)/x^2 I got (x^2)/(2)-x + C but that's not one of the answers Options x - 1 + C (x^2/2)-(x^3/3) + C (x^4-x^3)/4x^2 + C (x^2/2) -x + C
*January 2, 2016*

**Calculus**

Yes Thank you
*January 2, 2016*

**Calculus**

∫((cos^3(x)/(1-sin^(2)) What is the derivative of that integral? I have been trying to use trig identities but can't find one to simplify this equation. I can't find one for (cos^3(x) or (1-sin^(2)) My options -sin(x) + C sin(x) + C (1/4)cos^(4)(x) + C None of ...
*January 2, 2016*

**Calculus**

I tried to use partial fractions to do it, but it didn't work.
*December 27, 2015*

**Calculus**

Find the integral of 2/((2-x)(x+2)^2)dx.
*December 27, 2015*

**Calculus**

Solve the following problem by separation of variables. T^2y'-t=ty+y+1 Y(1)=0 Y is a function of t.
*December 25, 2015*

**Calculus**

Wait, why is it the integral from 3-5 instead of 0-2?
*December 23, 2015*

**Calculus**

Sorry I forgot the dx after the integral.
*December 23, 2015*

**Calculus**

Find the integral from 0 to 2 of xsqrt(5-sqrt4-x^2)). The hint said to use substitution of u=sqrt(4-x^2), and that I needed one more substitution, but I don't know how to do it.
*December 23, 2015*

**Math**

(Y+1)+4
*December 16, 2015*

**Mathematics**

Show that the root equation (x-a)(x-b)=k^2 are always real numbers
*December 16, 2015*

**Mathematics**

Factorice 16(a-2b)^2-(a+b)^2 leaving your answer in the simplest form
*December 16, 2015*

**math-unit rates**

Jeff hikes 1/4 mile every 2/5 hour. John hikes 1/2 mile every 1/7 hour. A) Who hikes farther in one hour. Show work. B) Who walks faster, Jeff or John. Explain your reasoning
*November 28, 2015*

**science**

I did a physics experiment and now i need help with the matlab fit. can someone tell me how to do this fit in matlab? a1*[sin(a2(x-a3))/(a2(x-a3)]^2*[cos(a4(x-a3))]^2 i have a program that fits a graph to a function i just ask how to write this function.
*November 27, 2015*

**Calculus**

Find the Absolute Maximum and Absolute Minimum of f on (0,3]. f(x)=(x^3-4x^2+7x)/x Multiple choice question I know the minimum is (2,3) but the maximum is either nothing or (0,7) but I can't tell which one
*November 7, 2015*

**math**

1. 6x4 in., P = 2*6 + 2*4 = 20 in. 2. 6*2 in., P = 2*6 + 2*2 = 16 in. 3. 4x2 in., P = 2*4 + 2*2 = 12 in. =
*November 5, 2015*

**Physics**

Units of time are seconds, minutes, and hours. Vf = Vo + a*t = 0. t = -Vo/a = -22/-6.1 = 3.61 s.
*November 5, 2015*

**Precalc**

a. P = Po(1+r)^n Po = $60,000. r = (4%/4)/100% = 0.01 = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 10yrs. = 40 Compounding periods. P = ? c. Same procedure as a.
*November 5, 2015*

**physics **

Vf = Vo + a*t = 3 + 1*5 = 8 m/s. d = Vo*t + 0.5a*t^2.
*November 5, 2015*

**universitu**

See Related Questions: Sun, 2-17-13, 5:09 AM.
*November 5, 2015*

**Calculus**

use tangent line approximation (linear approximation) to estimate The cube root of 1234 to 3 decimal places. Hint: the equation should be y=f'(x0)(x-x0)+f(x0) 11^3=1331 can be easily computed using binomial theorem. I used linear approximation and got 10.733, but it is not...
*November 4, 2015*

**Physics**

a. 0.5g*t^2 = 4780. 4.9t^2 = 4780. t^2 = 975.5 t = 31.2 s. Dx = 175m/s * 31.2s = 5465.8 m.
*November 3, 2015*

**math**

2. X = -30 km/h = velocity of the wind. Y = -295 km/h.=velocity of the aircraft. Q3. tan A = y/x = -295/-30 = 9.83333. A = 84.2o S. of W. = direction. Vr = y/sin A=-295/-sin84.2 = 296.5 km/h.
*November 3, 2015*

**physics**

Vf^2 = Vo^2 + 2g*h. g = 9.8 m/s^2.
*November 3, 2015*

**Science**

10cm3 * 10.5g/cm^3 = wt. in grams. mass = g/1000 = kilograms.
*November 2, 2015*

**physics**

140rev/60s * 20s =
*November 2, 2015*

**science**

T1*Cos26.5 - T2*Cos49.5 = 0 T1*Cos26.5 = T2*co49.5. T1 = 0.73T2. T1*sin26.5 + T2*sin49.5 = 155. Replace T1 with 0.73T2: 0.73T2*sin26.5 + T2*sin49.5 = 155. 0.324T2 + 0.760T2 = 155. 1.084T2 = 155. T2 = 143 N. T1 = 0.73*143 = 104 N.
*November 1, 2015*

**Physics**

Work = F*d = 23*Cos39 * 8 =
*November 1, 2015*

**physics**

Fs = M*g = 14 * 9.8 = 137 N. Fp = 137*sin 0 = 0 = Force parallel to the gnd. Fn = 137*Cos 0 - 25*sin30 = 124.5 N. = Normal Force. Fk = u*Fn = Force of kinetic friction. a = (25*Cos30-Fp-Fk)/M. Vf^2 = Vo^2 + 2a*d Vo = 1 m/s. d = 18 m. Vf = ?
*November 1, 2015*

**Physics**

See Relate Questions: Tue, 2-7-12, 1:11 AM.
*November 1, 2015*

**science**

Vf = Vo + g*Tr = 0. 24.5 + (-9.8)Tr = 0. -9.8Tr = 24.5 Tr = 2.5 s. = Rise time. Tf1 = Tr = 2.5 s = Time to fall back to roof. h = Vo*t + 4.9*t^2 = 24 m. 24.5*t + 4.9t*2 = 24. 4.9t^2 + 24.5t -24 = 0. Use Quad. Formula. t = 0.839 s. = Time to fall from top of roof to gnd. T = Tr...
*November 1, 2015*

**physics**

Fb = M*g = 13 * 9.8 = 127.4 N. Fp = 127.4*sin 0 = 0. = Force parallel to the floor. Fn = 127.4 - 86.4*sin64. = Normal force. Fk = u*Fn = Force of kinetic friction. a = (F*Cos64-Fp-Fk)/M.
*November 1, 2015*

**physics**

F1 = 2123 N. F2 = 1930 N. M = 1245 kg. a = (F1-F2)/M.
*November 1, 2015*

**Physics**

Vo = 90,000m/3600s = 25 m/s. d1 = Vo*t = 25m/s * 0.75s. = 18.75 m. d2 = 40 - 18.75 = 21.25 m. = Required stopping distance. Vf^2 = Vo^2 + 2a*d. Vf = 0. Vo = 25 m/s. a = -10 m/s^2. If d is => 21.25 m, the car will hit the barrier.
*October 31, 2015*

**physics**

4. Incomplete. Vo = 35m/s[33o]. Xo = 35*Cos33 = 29.4 m/s. Yo = 35*sin33 = 19.1 m/s. 5A. Yf^2 = Yo^2 + 2g*h. Yf = 0. g = -9.8 m/s^2. h = ? B. Dx = Vo^2*sin(2A)/g. A = 33o. g = +9.8 m/s^2. Dx = ?
*October 31, 2015*

**SPR,PHYSICS**

Vs = Db/Dw * Vb = 2/3 * Vb = 2Vb/3. Db/Dw * Vb = 2Vb/3. Db/Dw = 2/3 = 0.667. Db=0.667Dw = 0.667 * 0.998=0.665 g/cm^3. = Density of the block. Notes: Vs = Volume submerged. Db = Density of the block. Dw = Density of water. Vb = Volume of the block.
*October 30, 2015*

**force**

Wb = M*g = 10*9.8 = 98 N. Fp = 98*sin 0 = 0 = Force parallel to the surface. Fn = 98*Cos 0 - 40*sin30 = 78 N. = Normal force. Fk = u*Fn = 0.3 * 78 = 23.4 N. = Force of kinetic friction. a=(Fap*Cos30-Fp-Fk)/M=(34.6-0-23.4)/10 = 1.12 m/s^2.
*October 30, 2015*

**Physics**

I had this same problem... θ > arctan(μs) is the answer, you're welcome fam.
*October 29, 2015*

**math**

d1 = 10km/h * (20/60)h = 3.33 km., N. d2 = 5km/h * (35/60)h=2.92 km, NE.(45o)= 2.06 + 2.06i. Disp.=2.06 + 2.06i + 3.33i=2.06 + 5.39i = sqrt(2.06^2 + 5.39^2)
*October 29, 2015*

**physics**

Disp. = 7m[90o] + 9m[E45oN] + 7m[W14oS] X = 9*Cos45 + (-7*Cos14) = -0.428 m. Y = 7 + 9*sin45 + (-7*sin14) = 11.7 m. Q2. Tan A = Y/X = 11.7/-0.428 = -27.26754. A = -87.9o = 87.9o N. of W. = Direction. Disp. = Y/sin A = 11.7/sin87.9 = 11.71 m
*October 29, 2015*

**physics**

Disp. = 7m[90o] + 9m[E45oN] + 7m[S76oW]. X = 9*Cos45 - 7*Cos76 = 1.84 m. Y = 7 + 9*sin45 - 7*sin76 = 6.57 m. Q1. Tan A = Y/X = 6.57/1.84 = 3.57168. A = 74.4o N. of E. = Direction. Disp. = Y/sin A = 6.57/sin74.4 = 6.82 m
*October 29, 2015*