Monday

November 30, 2015
Total # Posts: 11,452

**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*

**physics**

A quantity that shows magnitude AND direction.
*October 29, 2015*

**geometry**

x^2 + 3x = 180o. x^2 + 3x-180 = 0. Factor it. -180 = -12*15. -12 + 15 = 3 = B. (x-12)(x+15) = 0. X = 12. X = -15. Use positive value of X: X = 12. X^2 = 144o. 3x = 36o.
*October 29, 2015*

**physics**

See Related Questions: Fri, 3-7-14, 4:52 PM.
*October 29, 2015*

**Chemistry**

Methanol, a major industrial feedstock, is made by several catalyzed reactions, such as CO(g) + 2H2(g) ----> CH3OH (l) One concern about using CH3OH as an auto fuel is its oxidation in air to yield formaldehyde, CH2OH(g) which poses a health hazard. Calculate ÄG o at ...
*October 29, 2015*

**physics math science**

Vo = 200m/s[37o]. Xo = 200*Cos37 = 160 m/s. Yo = 200*sin37 = 120 m/s. A. Y = Yo + g*t = 120 + (-9.8)*2 = 100 m/s. = Ver. component of velocity @ 2 s. Tan A = Y/Xo = 100/160 = 0.625. A = 32o. V = Xo/Cos A = 160/Cos32 = 189 m/s[32o]. B. Yf = Yo + g*Tr Yf = 0, g = -9.8m/s^2, Tr...
*October 29, 2015*

**Calculus**

i did. Dv=128pi. But the answer is 8pi/25
*October 28, 2015*

**Calculus**

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter 40 meters. (Recall that the volume of a sphere of radius r is V=(4/3)πr^3. Notice that you are given that dr=0.02.)
*October 28, 2015*

**math**

2 Years ago: Pam: X yrs. old. Sam: 2x/3 yrs. old. 3 yrs. later: Pam: x+3 yrs. old. Sam: 2x/3 + 3 yrs. old 2x/3 + 3 = 3/4(x+3). 2x/3 + 3 = 3x/4+9/4 8x + 36 = 9x + 27. X = 9. So Pam is 3 yrs. older than Sam. In 20 yrs.: x + (x+3) = 69. 2x = 66. X = 33. X+3 = 36. Today: Sam: 33...
*October 28, 2015*

**Physics**

F = 0.119m * 110N./m =
*October 28, 2015*

**Physics**

Wc = M*g = 1400 * 9.8 = 13,720 N. = Normal force, Fn. Fp = mg*sin 0 = 0. = Force parallel to the road. Fk = u*Fn = 0.0889 * 13,720 = 1220 N. a = (Fp-Fk)/M = (0-1220)/1400 = -0.871 m/s^2. Vf^2 = Vo^2 + 2a*d. Vf = 0. a = -0.871 m/s^2. Vo = 21.4 m/s. d = ?.
*October 28, 2015*

**physics**

Wt. = 450cm^3 * 0.998g/cm^3 = 449.1 g. = 0.449 kg of water displaced. M*g = 0.449kg * 9.8N/kg = 4.40 N.
*October 28, 2015*

**SCIENCE\PHYSICS**

Disp. = 0. d = 5 + 5 = 10 m.
*October 28, 2015*

**PHYSICS**

d = 5 + 5 = 10 m.
*October 28, 2015*

**PHYSICS**

Displacement = 0, because it ended at the starting point.
*October 28, 2015*

**PHYSICS**

a. Yf^2 = Yo^2 + 2g*h = 0. Yo^2 = -2g*h = -2*(-9.8)27.5 = 539. Yo = 23.2 m/s. = Vertical component of initial velocity. sin A = Yo/Vo = 23.2/47.5 = 0.48877. A = 29.3o. Xo = Vo*Cos A = 47.5*Cos29.3 = 41.4 m/s = Horizontal component of initial velocity = Total velocity at max ht...
*October 28, 2015*

**physics**

The velocity of the truck must be equal to the hor. component of the airplane's velocity: V = 180*Cos47.
*October 28, 2015*

**Calculus**

Let f(x)=x^4. If a=1 and dx=Δx=1/2, what are Δy and dy?
*October 28, 2015*

**Physics**

Wb = M*g = 200 N., M = 200/g = 200/9.8 = 20.4 kg. sin A = 3/10 = 0.30, A = 17.5o. Fp = 200*sin17.5 = 60 N. Fn = 200*Cos17.5 = 191 N. a. Work = F*d = 120 * 10 = 1200 J. b. Mg*h-0 = Mg*h = 200 * 3 = 600 J. c. Conservation Energy : Change in KE = Change in PE = 600 J. d. Fap-Fp-...
*October 27, 2015*

**maths**

P = Po(1+r)^n. r = (6%/2)/100% = 0.03 = Semi-annual % rate expressed as a decimal. n = 2comp./yr. * 1yr. = 2 Compounding periods. P = Po(1+0.03)^2 = Po*1.03^2 = 1.0609Po. I = P-Po = 1.0609Po-Po = 0.0609Po. I/Po * 100% = 0.0609Po/Po = 6.09% = Effective APR.
*October 27, 2015*

**Physic**

V = Vo^2 + 2g*h. Vo = 0, h = 20 m, V = ?
*October 27, 2015*

**physicsn**

Notes: 1. Fp = Force parallel with the hill. 2. Fn = Normal force. 3. Fk = Force of kinetic friction. 4. Ws = Wt. of skier.
*October 27, 2015*

**physicsn**

Ws = M*g = 9.8M. Fp = 9.8M*sin30 = 4.9M. Fn = 9.8M*Cos30 = 8.49M. Fk = u*Fn = u*8.49M. Fp-Fk = M*a. 4.9M-u8.49M = M*(-1.4) Divide by M: 4.9-8.49u = -1.4 -8.49u = -1.4-4.9 = -6.3 u = 0.74.
*October 27, 2015*

**Physics**

a. a=(Vf-Vo)/t = (5-0)/20 = 0.25 m/s^2. d1 = Vo*t + 0.5a*t^2 = 0 + 0.5*0.25*20^2 = 50 m.
*October 27, 2015*

**math**

a scooter is priced between $1000 and $2000. its price is a multiple of 10. All the digits in the price, except for the thousands digit, are even numbers. The value of the hundreds digit is 30 times the value of the tens digit. what is the price of the scooter?
*October 26, 2015*

**science**

Wb = M*g = 10 * 9.8 = 98 N. Fn = 98-40*sin30 = Normal force. Fk = u*Fn = Force of kinetic friction. a = (Fap*Cos30-Fk)/M.
*October 26, 2015*

**Physics**

a = F/m = 5/0.3 = 16.7 m/s^2. V = Vo + a*t. Vo = 20 m/s. t = 0.03 s. V = ?
*October 26, 2015*

**Physics**

Correction: Yf = 36.6 m/s.
*October 26, 2015*

**Physics**

h = 0.5g*t^2 = 68.2 m. 4.9t^2 = 68.2 t^2 = 13.92. t = 3.73 s. Dx = Xo*t = 4.2m/s * 3.73s. = 15.7 m. Yf^2 = Yo^2 + 2g*h. Yf^2 = 0 + 19.6*68.2 = 1336.72. Yf = 36.6 m. Vf = sqrt(Xo^2 + Yf^2) = Final velocity.
*October 26, 2015*

**math 1350**

Bob: X baseball cards. Ted: 2x " " Reggie: 2x-5 " " 2x-5 + x = 82. X = ?. 2x = ?
*October 26, 2015*

**science**

h = 0.5g*t^2.
*October 26, 2015*

**science**

d = 0.5*g*t^2.
*October 26, 2015*

**science/physics**

F = 9C/5 + 32.
*October 26, 2015*

**Physics**

Wc = M*g = 1300 * 9.8 = 12,740 N. = Normal force, Fn. Fp = Mg*sin A = Mg*sin 0 = 0. = Force parallel to the plane. Fk = u*Fn = 0.8 * 12,740 = 10,192 N. a = (Fp-Fk)/m = (0-10,192)/1300 = -7.84 m/s^2. d = (Vf^2-Vo^2)/2a = (0-(28^2))/-15.68 = 50 m.
*October 25, 2015*

**Physics - Check Answers ASAP**

1. M = 70 kg. Slope = 35o? L = 60 m. h = 60*sin35 = 34.4 m. Work = F*h = Mg*h = 70*9.8 * 34.4 = 23,608 J. 3. 19kW = 19kJ/s. = 19,000 J/s 6.8*10^7J * 1s./ 1.9*10^4J = 3580 s. = 0.994 h. or 1 h. 4. d = 0.5g*t^2 5. 24h/day * 365days/yr. * 3600s./h = 6. 200km[60o] N. of E. Dx = ...
*October 25, 2015*

**Urgent physics help**

A. D = 230mi/h * 2.5h = 575 mi. B. Dx = 575*Cos60 = C. Dy = 575*sin60 = D. Vr = 575mi[60o]/2.5h = Resultant velocity. Vx = Vr*Cos60 =
*October 25, 2015*

**Quick Physics check**

How did you arrive at your answer? See previous post: Wed, 10-21-15, 5:38 PM.
*October 25, 2015*

**physics**

Effective Intensity = 0.12 * 250W/m^2 = 30W/m^2. Area=(1200W * 1m^2/30W) = 40 m^2.
*October 25, 2015*

**physics**

Ws = M*g = 18 * 9.8 = 176.4 N. Fn = 176.4 - 45*sin30 = Normal force. Fk = u*Fn = Force of kinetic friction. F(net) = 45*Cos30-Fk
*October 24, 2015*

**physics**

See previous post: Fri, 10-23-15, 10:39 PM.
*October 24, 2015*

**physics**

Ws = M*g = 18 * 9.8 = 176.4 N. a. Fn = 176.4 - 45*sin30 = 153.9 N. = Normal force = Force perpendicular to the plane.
*October 23, 2015*

**Physics**

a = Fap/M = -250/50,000 = -0.005 m/s^2. Vf = Vo + a*t = 0. t = -Vo/a = 0.30/-0.005 = 60 s.
*October 23, 2015*

**Physics**

a. Wo = M*g = 4 * 9.8 = 39.2 N. d = 39.2N. * 1m/150N. = 0.261 m.
*October 23, 2015*

**MAth**

2^4 = 16. 2^-4 = 1/2^4 = 1/2^2*2^2 = 1/4*4 = 1/16.
*October 23, 2015*

**Physics**

h = 9.4*sin8.0 = 1.31 m. h = 0.5g*t^2 = 1.31 m. g = 9.8 m/s^2. t = ?
*October 22, 2015*

**Physics**

Vf = Vo + a*t = 20 + 2*4 = 28 m/s. Dc = 20*4 + 0.5*2*4^2 = 96 m. in 4 sec. Dc = 28m/s * 6s = 168 m. in 6 sec. Dc = 96 + 168 = 264 m. in 10 sec. Dt = 18m/s * 10s. = 180 m. in 10 sec. (264-8) - 180 = Distance in front of the truck.
*October 22, 2015*

**physics**

d = 2.06*sin28.8 = 0.992km, Due North.
*October 22, 2015*

**Science**

Vf = 100,000m/3600s. = 27.8 m/s. a = (Vf-Vo)/t = (27.8-0)/10 = 2.78 m/s^2
*October 22, 2015*

**Algebra**

Sun: X tickets were sold. Sat: 3x " " " Fri: 2x " " " $2(x + 3x + 2x) = $2160. X = ?
*October 22, 2015*

**math**

Assume 5 cm is measured on the drawing. d = 5cm * 3m/1cm = 15 m. = Actual distance.
*October 22, 2015*

**math**

(2/3)/6 = 2/3 * 1/6 = 2/18 = 1/9 cup.
*October 22, 2015*

**physics**

M*g = 1000 * 9.8 = 9800 N. Fp = 9800*sin33 = 5337 N. = Force parallel to the incline. a = Fp/M = 5337/1000 = 5.34 m/s^2.
*October 22, 2015*

**sri krishna**

3-4-5 triangle with each side multiplied by 4: hyp. = 20 = 5*4. 3*4 = 12. 4*4 = 16. So X = 4. so
*October 22, 2015*

**Math**

3/4 * M = 48. M = 4/3 * 48 = $64.
*October 22, 2015*

**Physics**

a. Vf = Vo + g*Tr = 0. Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise time. b. h1 = Vo*Tr + 0.5g*Tr^2 = Ht. above the roof. g = -9.8 m/s^2. h = h1 + 20. c. h = 0.5g*Tf^2. h = h1 + 20. g = 9.8 m/s^2. Tf = Fall time. Tf = ? Tr + Tf = Total flight time. d. Vf^2 = Vo^2 + 2g*h. Vo = 0. g...
*October 22, 2015*

**Physics**

Need more INFO.
*October 22, 2015*

**math**

d = 5days * 2(13/2)mi/day = 65 Miles. 65mi/4gal. = 16.25 mi/gal.
*October 21, 2015*

**math**

144/32 = 4 R16 = 4 Thirty-two inch pieces with 16 inches left.
*October 21, 2015*

**math**

(3 2/3)/2/3 = (11/3)/2/3 = 11/3 * 3/2 = 33/6 = 5 3/6 = 5 1/2 or 5 pieces.
*October 21, 2015*

**physics**

Yf^2 = Yo + 2g*h = 0. Yo = -2g*h = -2*(-9.8)*0.449 = 8.8 m/s. Yo = Vo*sin3.40 = 8.8 m/s. Vo = 8.8/sin3.4 = 148 m/s.
*October 21, 2015*

**math**

Sole X necklaces. Sold 3x watches. 4*x + 7*3x = $300. X = ?.
*October 21, 2015*

**Physics- PLEASE HELP!!**

Hannah and Tammy please show how you arrive at your answer.
*October 21, 2015*

**maths**

1 - (2/5 + 1/3) = 15/15 -(6/15 + 5/15)= 15/15 - 11/15 = 4/15 Left.
*October 21, 2015*

**physics**

D = 26km[90o] + 62km[60o]. X = 62*Cos60 = 31km. Y = 26 + 62*sin60 = 79.7 km. Q1. Tan A = Y/X = 79.7/31 = 2.57076. A = 68.7o = Direction. D = Y/sin A = 79.7/sin68.7 = 85.5 km[68.7o].
*October 21, 2015*

**Quick physics check**

Fr = 60[135o]CCW + 5[330o]CCW. X = 60*Cos135 + 5*Cos330 = -38.1 N. Y = 60*sin135 + 5*sin330 = 39.9 N. Q2. Tan A = Y/X = 39.9/-38.1 = -1.04794. A = -46.3o = 46.3o N. of W. = Direction. Fr=Sqrt(X^2 + Y^2)=Sqrt(38.1^2 + 39.9^2) = 55.2 N. = Resultant force.
*October 21, 2015*

**Physics**

M*g = 9990 N., M = 9990/9.8 = 1019 kg. V = 64.87 mi/h * 1600m/mi * 1h/3600s. = 28.83 m/s. KE = 0.5M*V^2 = 0.5*1019*(28.83)^2 = 423,481 J. PE = Mg*h = KE. 9990*h = 423,481. h = ?
*October 21, 2015*

**physics 11**

X = -2.7 m/s. Y = 1.3 m/s. Q2. a. Tan A = Y/X = 1.3/-2.7 = -0.48148. A = -25.7 m/s = 25.7o N. of W. = 64.3o W. of N. = Direction. Vr = Sqrt(X^2 + Y^2) = Sqrt(2.7^2+1.3^2)= 3.0 m/s = Resultant velocity. Dr = Sqrt(20^2 + 41.5^2) = 46.1 m. = Resultant distance. Dr = Vr*T = 46.1 m...
*October 20, 2015*

**Physics**

1. Vf^2 = Vo^2 + 2a*d Vo = 0. Vf = ?. 2. d = 0.5a*t^2.
*October 20, 2015*

**College Physics**

Wc = M*g = 1670 * 9.8 = 16,366 N. N/m = 16,366N./6.5m.
*October 20, 2015*

**physcial science**

D.
*October 20, 2015*

**Physics**

See previous post: Tue, 10-20-15, 11:26 AM.
*October 20, 2015*

**Physics**

See previous post: Tue,10-20-15, 11.26 AM.
*October 20, 2015*