Sunday

April 26, 2015

April 26, 2015

Total # Posts: 10,489

**math**

A = 0.5*AB*AC*sin A=0.5 *5 * 5 * sin 60 = 10.8 cm^2.
*December 26, 2014*

**physics**

X = 8.70 m. Y = -6.04 m. a. D^2 = X^2 + Y^2 = 8.7^2 + (-6.04)^2 = 112.2 D = 10.6 m. b. Tan A = Y/X = -6.04/8.7 = -0.69425 A = -34.8o CW. A = 360 - 34.8 = 325.2o CCW.
*December 25, 2014*

**Chemistry**

What are the different ways to split a mixture of benzene and methanol into its constituents using a 2 phase system separation ?
*December 25, 2014*

**Algebra 1**

y = 4x + 3 Eq1: -4x + y = 3 Eq2: -8x + 2y = 3 m1 = -A/B = 4/1 = 4 m2 = 8/2 = 4 Eq1: y-int. = C/B = 3/1 = 3. Eq2: Y-int. = 3/2 = 1.5 No Solutions.
*December 23, 2014*

**Algebra 1**

x = -4y + 4 Eq1: x + 4y = 4 Eq2: 2x+8y = 8 Divide Eq2 by 2: x+4y = 4 The Eqs are identical and represent the same line. Therefore, we have an infinite number of solutions.
*December 23, 2014*

**Algebra**

$X To start with. 1st. Store: Bal. = X - ((X/2)+1) = X - X/2-1 = X/2-1 2nd Store: Bal=X/2-1 - (X/2-1)/2+1 = X/2-1 - X/4+1/2-1 = X/4 - 1/2. 3rd. Store: Bal = X/4-1/2 - (X/4-1/2)/2+1 = X/4-1/2 - X/8+1/4)-1 = X/8 - 1 1/4 = X/8 - 5/4. 4th Store: X/8-5/4 - (X/8-5/4)/2+1 = X/8-5/4...
*December 22, 2014*

**Physics**

a. h=ho + (-Vo^2/2g)=100 + (-98^2)/-19.6 = 590 m. b. V = Vo + g*Tr = 0 98 - 9.8*Tr = 0 -9.8Tr = -98 Tr = 10 s. = Rise time. h = 0.5g*t^2 = 590 4.9*t^2 = 590 t^2 = 120.4 Tf = 11 s. = Fall time. Tr+Tf = 10 + 11 = 21 s. = Time to reach gnd. c. V^2 = Vo^2 + 2g*h Vo = 0 g = 9.8 m/s...
*December 22, 2014*

**Physics**

Note: The Fall time is not required.
*December 22, 2014*

**Physics**

h = 0.5g*t^2 = 4 m. 4.9*t^2 = 4 t^2 = 0.816 Tf = 0.904 s. = Fall time. V1^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4 V1 = 8.85 m/s., Downward. V^2 = V2^2 + 2g*h = 0 V2^2 = -2g*h = -2*(-9.8)*3 = 58.8 V2 = 7.67 m/s, Upward. a = (V2-V1)/t = (7.67-8.85)/0.01 = -118 m/s^2, Upward.
*December 22, 2014*

**physics**

F = 200N[30o] Fx = 200*Cos30 = 173.2 N. Fy = 200*sin30 = 100 N. M*g = 30kg * 9.8N/kg = 294 N. = Wt. of the object. a. Fk = u*Fn = u*(Mg-Fy) = 0.2(294-100) = 38.8 N. = Force of kinetic friction. b. a = (Fx-Fk)/M c. V = Vo + a*t Vo = 0 a = Value from part b. t = 5 s. Solve for V.
*December 21, 2014*

**physics**

Vs = 50mi/h * 1600m/mi * 1h/3600s = 22.22 m/s. = Velocity of the speedster. a. Ds = Dp 22.2*t = 0.5a*t^2 Divide both sides by t: 0.5a*t = 22.2 0.5*0.450t = 22.2 0.225t = 22.2 t = 98.8 s. b. Ds = Dp = 22.2*t = 22.2 * 98.8 = 2193 m. c. V^2 = Vo^2 + 2a*d Vo = 0 a = 0.450 m/s^2 d...
*December 21, 2014*

**Physics**

A 10.0 kg barrel is lifted by pulling up on a rope. The barrel accelerates at 1.50 m/s2. Find the force of tension on the rope.
*December 21, 2014*

**Calculas**

V = Vo + g*t Vo = 0 g = 9.8 t = 2.75 s. Solve for V.
*December 21, 2014*

**Science**

V = 60km/h = 60,000m/3600s = 16.7 m/s. Work = The change in KE = 0.5*M*V^2 = 0.5*1500*16.7^2 = 209,168 J.
*December 21, 2014*

**Science**

V^2 = Vo^2 + 2g*h = 0 Vo^2 = -2g*h = -2*(-9.8)*1000 = 19,600 Vo = 140 m/s.
*December 21, 2014*

**math**

a. I = Po*r*t = 5000*(0.09/360)*80 = $100 = Interest. b. V = Po + I = 5000 + 100 = $5100 = Maturity value. c. July 1.
*December 18, 2014*

**business math**

Ar = Ab - Ad Ar = 20,000 - 20,000*(0.085/360)*90 = 20,000 - 425 = $19,575 = Amount received (proceeds). Ab = Amt. borrowed. Ad = Amt. discounted.
*December 18, 2014*

**Physics**

h = 3.1*sin30 = 1.55 m. a. V^2 = Vo^2 + 2g*h = 0 + 19.6*1.55 = 30.4 V = 5.51 m/s. b. V^2 = Vo^2 + 2a*d a = (V^2-Vo^2)/2d = (0-(5.51^2))/10 = -3.04 m/s^2. M*g = 7.5 * 9.8 = 73.5 N. = Wt. of suitcase. Fp = 73.5*Sin30 = 36.75 N. = Force parallel to the incline. Fn = 73.5*Cos30 = ...
*December 18, 2014*

**physic**

250+160+420+F4 = 220+340+180+560 Solve for F4.
*December 18, 2014*

**math**

d = r*t = 12 km 3 * t = 12 t = 4 h 1. d = r*t = 12 r * 3 = 12 r = 4 km/h Change = 4 - 3 = 1 km/h = 1 km/h increase. 2. d = r*t = 12 km r * 5 = 12 r = 2.4 km/h. Change = 2.4 - 3 = -0.6 km/h = 0.6 km/h decrease.
*December 18, 2014*

**physics**

Vo[0o] Xo = Vo*Cos0 = Vo Yo = Vo*sin0 = 0 Vo[30o] Xo = Vo*Cos30 = 0.87Vo Yo = Vo*sin30 = 0.5Vo Vo[60o] Xo = Vo*Cos60 = 0.5Vo Yo = Vo*sin60 = 0.87Vo
*December 17, 2014*

**Phyiscs**

V^2 = Vo^2 + 2g*h Vo = 0 g = 9.8 h = 0.5 m. Solve for V.
*December 17, 2014*

**algebra**

(3,63),(5,57), (17,y). m = (57-63)/(5-3) = -6/2 = -3. m = (y-57)/(17-5) = (y-57)/12 = -3/1 y-57 = -36 Y = 57-36 = 21 Inches after 17 days.
*December 17, 2014*

**algebra**

What tank?
*December 17, 2014*

**physics**

W = Fx*d = 54*Cos10 * 3.4
*December 17, 2014*

**Algebra**

Correction: The Eqs are identical(divide the 1st Eq by 4). So we have an infinite # of solutions.
*December 17, 2014*

**Algebra**

m1 = -A/B = -16/-8 = 2 m2 = -4/-2 = 2 m1 = m2. So the lines do not intersect. Therefore, there is no solution.
*December 17, 2014*

**physic**

V^2 = Vo^2 + 2g*d = 0 + 19.6*50 = 980 V = 31.3 m/s.
*December 17, 2014*

**Algebra 1**

2x/3 + y = 7 Multiply both sides by 3: 2x + 3y = 21
*December 16, 2014*

**physics**

X = -5.4 m/s. Y = 3.6 m/s. Tan Ar = Y/X = 3.6/-5.4 = -0.66666 Ar = -33.7o = Reference angle. A = -33.7 + 180 = 146.3o = Direction. V = X/Cos A = (-5.4)/Cos146.3 = 6.49 m/s[146.3o].
*December 16, 2014*

**Algebra**

1. (-7)(3)-(5)(-3) = -21 - (-15) = -21+15 = -6 2. (x^2+6) - (3x^2-2x-5)=x^2+6-3x^2-2x-5 = =2x^2 - 2x + 1
*December 16, 2014*

**Algebra 2**

a. m = -A/B = -2/-5 = 2/5 b. m = -3/2
*December 16, 2014*

**Algebra**

Answer = A.
*December 16, 2014*

**Algebra 2**

K = y/x = 1/-3 = -1/3. y/x = -1/3 4/x = -1/3 -x = 12 X = -12.
*December 16, 2014*

**Algebra 2**

a. Y = 2x - 3 Y = mx - b m = 2 = Slope. b. 3y - 7x = 1? -7x + 3y = 1 m = -A/B = 7/3
*December 16, 2014*

**math**

X = Tony's future age. 4x = Mr. Jacob's future age. 4x - x = 55-7 3x = 48 X = 16 4x = 4 * 16 = 64 16-7 = 9 years. 64-55 = 9 years. So Mr. Jacobs will be 4 times as old as Tony in 9 years.
*December 16, 2014*

**Math**

D. All real numbers.
*December 16, 2014*

**geography**

2+2\4[o]
*December 16, 2014*

**Algebra 1**

Y = mx + b Y = 6x - 4
*December 15, 2014*

**Physics**

V = Vo + g*t = 16 - 9.8*1.5 = 1.3 m/s. Your answers are correct.
*December 15, 2014*

**physics**

a. X = 181*Cos66.2 b. Y = 181*sin66.2
*December 14, 2014*

**physics**

V = Vo + a*t = 0 + 231*0.0442 = 10.21 m/s. a. V*Cos A = 10.21*Cos55 = 5.86 m/s. b. 10.21*sin55 = 8.36 m/s.
*December 14, 2014*

**math**

X = Footballs sold @ $22 ea. Y = Footballs sold @ $6 ea. Eq1: x + y = 6000 Eq2: 22x + 6y = $76,000 Multiply Eq1 by -6 and add the Eqs: -6x - 6y = -36000 22x + 6y = 76,000 Sum: 16x = 40,000 X = 2500 In Eq1, replace X with 2500: 2500 + y = 6000 Y = 6000-2500 = 3500
*December 14, 2014*

**7th grade math**

3. Correct. 9. Correct. 12. X = 305/18 = 16 17/18 13. Correct. 14. C < -12.8 21. Correct. 26. Correct.
*December 14, 2014*

**physics**

Fp = Mg*sin50 Fn = Mg*Cos50 Fk = u*Fn = u*Mg*Cos50 Mg*sin50-0.1Mg*cos50 = M*a Divide both sides by M: g*sin50-0.1g*Cos50 = a a = 9.8*sin50-0.98*Cos50 = 6.88 m/s^2.
*December 14, 2014*

**physics**

Answer: The object with the greatest momentum. Object A: M*V = 1.9 * 8 = 15.2 kg-m/s Object B: M*V = 2.0 * 5 = 10.0 kg-m/s
*December 13, 2014*

**Physics**

Conservation of momentum: (M1+M2)V1/8 = M1*V1-M2*0 Multiply both sides by 8/V1: M1 + M2 = 8M1 3.32 + M2 = 8*3.32 M2 = 8*3.32 - 3.32 = 23.24 kg.
*December 12, 2014*

**Physics**

Conservation of momentum: (M1+M2)V = M1*V1-M2*V2 M1 = 4.66 kg M2 = 3.00 kg V1 = 4.77 m/s. V2 = 3.33 m/s. Solve for V.
*December 12, 2014*

**physics**

KE2 = KE1 * (V2/V1)^2 KE1 = 5*10^5 J. V1 = 36 mi/h V2 = 115 mi/h Solve for KE2.
*December 12, 2014*

**Science**

Vo = 27.4m/s[48o] Xo = 27.4*Cos48 = 18.3 m/s. Yo = 27.4*sin48 = 20.4 m/s. h = 0.5g*t^2 = 860 m. 4.9*t^2 = 860 t^2 = 175.5 Tf = 13.2 s. = Fall time. Dx = Xo*Tf = 18.3m/s * 13.2s = 242.4 m. b. Y = Yo + g*t = 0 + 9.8*13.2 = 129.4 m/s. Tan A = Y/Xo = 129.4/18.3 = 7.07104 A = 82o.
*December 12, 2014*

**Math**

P = Po*e^(r*t) e^(r*t) = P/Po r*t = 0.03*10 = 0.30 e^0.3 = P/5 Solve for P. Answer in millions.
*December 12, 2014*

**physics**

V = Vo + a*t V = 6.4 m/s. Vo = 0 t = 0.27 s. Solve for a. F = m*a
*December 12, 2014*

**Chemistry**

But these are the numbers which have been provided in the question... If u could show me just the way to tackle it, it wud be very helpful,plz.
*December 12, 2014*

**Chemistry**

0.250 moles of HCL and 0.100 moles of acetic acid are mixed in 0.365 moles of water. Determine the following: [CH3COOH], [CH3COO^-], pH and pOH. Ka=1.8E-5. Calculate the pH of 0.05M CH3COO-Na+ . Plz show me step by step, really confused right now.
*December 12, 2014*

**physics**

V = Vo + a*t V = 20 m/s. Vo = 25.1 m/s. t = 5.7 s. Solve for a. (It should be negative.) F = M*a
*December 9, 2014*

**physics**

V^2 = Vo^2 + 2a*d Vo = 0 a = F/m = 196.3/0.39 = d = 1.28 m. Solve for V.
*December 9, 2014*

**physics**

Incomplete.
*December 9, 2014*

**math**

P = Po + Po*r*t P = 39,995 + 39,995*(0.0249/12)*60 =
*December 8, 2014*

**Physics**

a = (F1-F2)/m F1 = 1141 N. F2 = 947 N. M = 2300 kg Solve for a. V^2 = Vo^2 + 2a*d V = 3 m/s. Vo = 0 a: Calculated in step #1. Solve for d in meters. .
*December 8, 2014*

**AP physics**

KE = M*V^2/2 KE = 5400 J. M = 2840 kg Solve for V. Units: m/s.
*December 8, 2014*

**physics**

a. 30o/h *2h = 60o. b. 360o/h * 2h = 720o. c. 4320o/h * 2h = 8640o
*December 8, 2014*

**Physics**

a. d = 0.5a*t^2 = 1.92 m. 0.5a*1.9^2 = 1.92 m 1.81a = 1.92 a = 1.06 m/s^2 b. M*g = 2.84 * 9.8 = 27.8 N. = Wt. of block. Fp = 27.8*sin30 = 13.9 N. = Force parallel to the incline. Fn = 27.8*Cos30 = 24.1 N. = Normal force Fp-Fk = m*a 13.9-Fk = 2.84*1.06 = 3.01 -Fk = 3.01-13.9...
*December 8, 2014*

**Physics**

d = 0.5a*t^2 d = 0.79 m. t = 0.64 s. Solve for a. F = m*a
*December 8, 2014*

**physics**

See previous post: Sat, 12-6-14, 4:06 PM
*December 8, 2014*

**physics**

V^2 = Vo^2 + 2a*d V = 0 Vo = 30 m/s. d = 75 m. Solve for a.(It should be negative). F = m*a. The force will be negative, because it acts against the motion.
*December 8, 2014*

**physics**

X m.=Distance the heavier person moves 2x m.= Lighter person distance. x + 2x = 12 3x = 12 X = 4 m.
*December 8, 2014*

**Physics**

M*g = 63 * 9.8 = 617.4 N. = Wt. of block Fp = Fs @ critical angle. Mg*sinA = us*Mg*CosA Divide both sides by Mg: sin A = us*Cos A Divide both sides by Cos A: sin A/Cos A = us = 0.47 = Tan A A = 25.2o = The critical angle. Fp = 262.6
*December 7, 2014*

**physics**

M*g = 16 N. Solve for g.
*December 7, 2014*

**Physics**

PE = M*g*h
*December 7, 2014*

**Math**

22n^2 + n - 5 Use AC Method: A*C = 22*(-5) = -110 = -10*11. Select the pair of factors whose sum = B =1: -10, and 11. Form 2 factorable binomials: 22n^2 + (-10n+11n) - 5 (22n^2+11n) - (10n+5) 11n(2n+1) - 5(2n+1) (2n+1)(11n-5)
*December 7, 2014*

**math**

See previous post: Sat, 12-6-14, 2:02 PM
*December 7, 2014*

**Physics**

V = Vo + a*t V = 0 Vo = 24 m/s t = 0.16 s. Solve for a.(It will be negative) F = M*a
*December 7, 2014*

**ALG 1**

12a^4b^3 + 8a^3b^2 = 4a^3b^2(3ab+2). GCF = 4a^3b^2.
*December 7, 2014*

**physics**

Incomplete.
*December 7, 2014*

**Physics**

X = M1*V1 = 24*3.3 = 79.2 Y = M2*V2 = 33*2.4 = 79.2 1a. Tan A = Y/X = 79.2/79.2 = 1.00 A = 45o = Direction. 1b. Momentum = X/Cos A = 79.2/Cos45 = 112 (M1+M2)*V = 112[45o] (33+24)V = 112[45] V = 1.96 m/s.[45o]
*December 7, 2014*

**problem solving**

See previous post: Sun,12-7-14, 1:41 AM.
*December 7, 2014*

**problem solving**

Difference = 35 - 0.75*40 = 35 - 30 = 5.
*December 7, 2014*

**problem solving**

Cost = Po + 0.05Po = $315 1.05Po = 315 Po = $300 = Initial price. (315-300)/6blks = $2.50/blk 300/(2.50/blk) = 120 Blocks bought.
*December 7, 2014*

**problem solving**

400rev/min * 60min/h = 24,000 rev/h. Circumference(C) = 47,520m/24,000rev = 1.98 m/rev C = pi*2r = 1.98 m 6.28r = 1.98 r = 0.315 m = 31.5 cm = Radius of the wheels.
*December 7, 2014*

**problem solving**

see 1:52 AM post.
*December 7, 2014*

**Physics**

a. d = Xo*Tf = 150 m. 450 * Tf = 150 Tf = 0.333 s. d = 0.5g*Tf^2 = 4.9*0.333^2 = 0.544 m. Below the center.
*December 6, 2014*

**Physics**

F = 27in/39in * 29Lbs = 20.1 Lbs.
*December 6, 2014*

**Physics**

F1 = 400N[32o] F2 = 536N.[10o] a. X = 400*Cos32 + 536*Cos10 = Y = 400*sin32 + 536*sin10 = Tan A = Y/X A = Fr = X/Cos A = Resultant force @ Ao. b. a = Fr/m = Acceleration @ A Degrees.
*December 6, 2014*

**math**

Eq1: Vp + Vw = 570/3 = 190 Eq2: Vp - Vw = 570/5 = 114 Sum: 2Vp = 304 Vp = 152 mi/h In Eq1, replace Vp with 152: 152 + Vw = 190 Vw = 38 mi/h.
*December 6, 2014*

**Algebra 1**

(-10,3), m = -1/5. Y = mx + b Y = 3 m = -1/5 x = -10 Solve for b(y-intercept).
*December 6, 2014*

**physics**

Momentum = M*V
*December 6, 2014*

**science**

V^2 = Vo^2 + 2g*h V = 0 Vo = 10 m/s. g = -9.8 m/s^2 Solve for h.
*December 6, 2014*

**Science**

h = 4.6 m. V1 = 16 m/s. V2 = 26 m/s. g = 9.81 m/s^2 h = 0.5g*t^2 Solve for t = Fall time. d = (V2-V1)*t = Distance apart.
*December 6, 2014*

**physics**

Wc = M*g = 92kg * 9.8N/kg = 902 N. = wt. of the clock. Fap-Fs = M*a 656 - Fs = M*0 = 0 Fs = 656 N. = Force of static friction. us = Fap/Wc = 656/902 = 0.727 uk = 560/902 = 0.621
*December 5, 2014*

**physics**

V^2 = Vo^2 + 2a*d V = 0 Vo = 8.6 m/s. a = u*g = 0.050*(-9.8) = -0.49 m/s^2 Solve for d.
*December 5, 2014*

**Math**

(24x^18/6x^3xX^8)/30x^6 = 4x^6/30x^6 = 2/15.
*December 5, 2014*

**Physics**

V = Vo + a*t V = 0 Vo = 8.5 m/s t = 3 s. Solve for a.(It will be negative) F = m*a
*December 5, 2014*

**Physics**

M1*V1 = (M1+M2)*V M1 = 1600 kg V1 = 13.7 m/s. M2 = 4600 kg Solve for V.
*December 5, 2014*

**physics**

Dp = Ds Vp*T = Vs*(T+95.7) 8*T = 4.5*(T+95.7) 8T = 4.5T+430.65 8T-4.5T = 430.65 3.5T = 430.65 T = 123 s. D = Vp*T = 8 * 123 = 984 km.
*December 5, 2014*

**physics**

L = V/F V = L*F = 5.8 * 1.15 = 6.67 m/s. Note: The distance between crests is the wavelength(L).
*December 5, 2014*

**physics**

T = d/V = 1340/457 = 2.93 s.
*December 5, 2014*

**Quantitative Business Analysis**

See previous post.
*December 5, 2014*

**Quantitative Business Analysis**

I = Po*r*t I = $9800 r = 0.08 t = 1 yr. Solve Po, the amount borrowed.
*December 5, 2014*

**physics**

Eq1: F1 + F2 = 376 N. Eq2: F1 - F2 = 162 N. Sum: 2F1 = 538 F1 = 269 N. In Eq1, replace F1 with 269: 269 + F2 = 376 F2 = 376 - 269 = 107 N.
*December 4, 2014*

**physics help PLZ!!**

Fx = -23.3 N. Fy = 409.5 N. 2nd Quadrant. Tan Ar = Fy/Fx = 409.5/-23.3 = -17.57511 Ar = -86.7o = Reference angle(Q4). A = -86.7 + 180 = 93.3o, CCW(Q2) = Direction.
*December 4, 2014*