Saturday

February 13, 2016
Total # Posts: 11,511

**algebra**

Oops! Posted pre-maturely. 2. Mark: 3y Linda: 3y - 6 In the end: Mark: 3y + 2y = 5y Linda: (3y-6) + 2y = 5y - 6
*January 7, 2015*

**algebra**

1. Sheila: 7y Craft sticks. Mark: 7y-4y = 3y Linda: 3y-y = 2y In the end: Sheila: 7y-2y-2y = 3y Mark: 3y + 2y = 5y Linda: 2y + 2y = 4y 2. Mark: 3y Linda: 3y - 6
*January 7, 2015*

**Math**

X = 8 Mi Y = -10 Mi a. Tan A = Y/X = -10/8 = -1.25 A = -51.3o = 51.3o S. of E. b. d^2 = X^2 + Y^2 = 8^2 + (-10)^2 = 164 d = 12.81 Miles. c. d = V*t Solve for t. d. A = -51.3 + 180 = 128.7o, CCW = 51.3o N. of W.
*January 6, 2015*

**physics please help**

See previous post: Tue, 1-6-15, 3:44 PM.
*January 6, 2015*

**Physics**

Ww = Vo*Dw = 4m^3 * 1000kg/m^3 = 4,000 kg. = Wt. of water displaced. = Wt. of the object in water.
*January 6, 2015*

**Physics**

Vb = L*W*h = 5 * 1 * 0.5 = 2.5 m^3 = Vol. of the box=Vol. of water displaced. Mass = Vb*Dw = 2.5m^3 * 1000kg/m^3 = 2500 kg.
*January 6, 2015*

**physics**

Vo = 157km/h = 157000m/3600s = 43.61 m/s V = 75.0km/h = 75000m/3600s = 20.83 m/s. a = -7.0 m/s^2 A. V = Vo + a*t Solve for t. B. V^2 = Vo^2 + 2a*d Solve for d.
*January 6, 2015*

**physics please help**

Tr = Tf = 4s./2 = 2 s. = Rise & fall time. h = Vo*Tf + 0.5g*Tf^2 = 0 + 4-9*2^2 = 19.6 m.
*January 6, 2015*

**Algebra**

2nd number = X 1st number = 3x/20 x + 3x/20 = 46 Multiply both sides by 20 and solve for x.
*January 5, 2015*

**Math**

I = Po*r*t = 10,000*(0.0225/52)*13 = 56.25 = Int. for 13 weeks. (56.25/13) * 52 = $225 = Int./yr. %I = (225/10,000)*100% = 2.25%
*January 5, 2015*

**Physics**

At 100 m. above gnd. PE max = Mg*h max = 80*100 = 8000 J. KE + PE = 8000 0 + 8000 = 8000 At 60 m. above gnd. PE = Mg*h = 80*60 = 4800 J. KE + PE = 8000 KE + 4800 = 8000 KE = 8000-4800 = 3200 J. The total energy, KE+PE, is constant at 8000 J.
*January 5, 2015*

**physics**

h = 4*sin37 = 2.41 m. V^2 = Vo^2 + 2g*h Vo = 0 g = 9.8 m/s^2 h = 2.41 m Solve for V.
*January 4, 2015*

**physics**

V^2 = Vo^2 + 2g*h V = 0 Vo = Initial velocity. g = -9.8 m/s^2 h = 1.28 m. Solve for Vo.
*January 4, 2015*

**Physics**

Tx = 20*Cos53 = 12 N. = Hor. component of tension. Tx-Fk = M*a 12-Fk = M*0 = 0 Fk = 12 N.
*January 4, 2015*

**physics**

h = 0.5g*t^2 h = 2 m. g = 9.8 m/s^2 t = Fall time. Solve for t. Dx = Xo * t Dx = Hor. distance from table. Xo = 10m/s = Initial hor. velocity. Solve for Dx.
*January 4, 2015*

**Math**

9(a^2+a+4) Answer: Neither. Difference of 2 squares = (a+b)(a-b) = a^2 - b^2
*January 4, 2015*

**Pre-calc/trig**

Log3(x^2+18) = 5 x^2 + 18 = 3^5 x^2 + 18 = 243 x^2 = 243 - 18 = 225 X = 15
*January 4, 2015*

**Algebra 1**

Use Long-hand division. (6x^3 - 3x^2 + x - 40)/(x-2)=6x^2+9x+19, Remainder 2.
*January 4, 2015*

**mathe**

Use your calculator.
*January 4, 2015*

**physics**

V^2 = Vo^2 + 2g*h Vo = 0 g = 9.8 m/s^2 h = 2 m. Solve for V. Note: The 10 m/s velocity is in the hor. direction.
*January 4, 2015*

**Physics**

V^2 = Vo^2 + 2a*d V = 17 m/s Vo = 0 d = 45 m Solve for a. F = m*a
*January 3, 2015*

**Physics**

a = (Fap-Ff)/M = (56-12)/25 = 1.76 m/s^2
*January 3, 2015*

**physics**

A = 5m[45o], CCW. B = 7m[330o], CCW C = 4m[240o], CCW. X=5*Cos45 + 7*Cos330 + 4*Cos240 = 7.60 m Y=5*sin45 + 7*sin330 + 4*sin240=-3.43 m Tan A = Y/X = -3.43/7.60 = -0.45113 A = -24.28o = 24.28o S of E = 335.7o, CCW. Resultant = X/CosA = 7.60/Cos335.7 = 8.34m[335.7]
*January 3, 2015*

**physics**

See previous post: Sat, 1-3-15, 4:01 AM.
*January 3, 2015*

**physics**

a. h = 50 + 20 = 70 m Above gnd. V^2 = Vo^2 + g*h = 0 Vo^2 - 9.8*20 = 0 Vo^2 = 196 Vo = 14 m/s. = Initial velocity. V = Vo + g*Tr = 0 14 - 9.8Tr = 0 9.8Tr = 14 Tr = 1.43 s. = Rise time. h = 0.5g*t^2 = 70 m. 4.9t^2 = 70 t^2 = 14.3 Tf = 3.78 s. = Fall time. Tr+Tf = 1.43 + 3.78...
*January 3, 2015*

**physics**

a. P = 5m[150o] X = 5*Cos150 Y = 5*sin150 b. 120-90 = 30o Clockwise from positive x axis = 330o counter clockwise from positive x axis. Q = 3.6m[330o], CCW. X = 3.6*Cos330 Y = 3.6*sin330
*January 3, 2015*

**physics**

D = 5m[37o] + 10m[150o] X = 5*Cos37 + 10*Cos150 = -4.67 m. Y = 5*sin37 + 10*sin150 = 8.00 m. Tan Ar = Y/X = 8/-4.67 = -1.71501 Ar = -59.75o = Reference angle. A = -59.75 + 180 = 120.25o = Direction. D=Y/sin A = 8.00/sin120.25=9.26m[120.25] = Magnitude and direction.
*January 3, 2015*

**physics**

a. h = 0.5g*t^2 h = 43.4 m. g = 9.8 m/s^2 t = Fall time. Solve for t. b. Dx = Xo * t Dx = 57.3 m. = Hor. dist. Xo = Initial hor velocity. t = Fall time calculated in part a. Solve for Xo. c. X = Xo=Initial hor velocity(part b). Xo does not change. d. Y = Yo + g*t Yo = 0 g = 9....
*January 2, 2015*

**physics**

X^2 + Y^2 = r^2 X = 25 cm Y = Ver. Displacement r = 50 cm = Hypotenuse. Solve for Y.
*January 2, 2015*

**physics**

V = Vo + a*t V = -40 m/s Vo = 30 m/s t = 0.04 s. Solve for a(It is negative).
*January 2, 2015*

**physics**

11. Ws = 5000 N. = Normal force(Fn). Fk = u*Fn = 0.05 * 5000 = 250 N. = Eorce of kinetic friction. Fap-Fk = M*a Fap - 250 = M*0 = 0 Fap = 250 N. = Force applied. Work = Fap * d = 250 * 1000 = 250,000 J. = 2.5*10^5 J. 12. Same procedure as #11.13. 13. Ws = 6000 N. = Normal ...
*December 31, 2014*

**Math- Max and Minimum**

Consider the function f(x) = 3/4x^4 –x^3 -3x^2 + 6x , find the relative extrema for f(x), be sure to label each as a maximum or minimum. How do I find the x-values of what it is asking?
*December 31, 2014*

**physics**

E = 6 Volts r = 0.4 Ohms R = 2.6 Ohms. I = E/(r+R) = 6/(0.4+2.6) = 2 Amps. = Current flowing in the circuit. Vt = I*R = 2 * 2.6 = 5.2 Volts. = Potential difference between terminals of the battery. Vr = I * r = 2 * 0.4 = 0.8 Volts. = Voltage lost across the internal resistance...
*December 31, 2014*

**physics**

1 Amp = 6.242*10^18 Electrons/s 2.4 Amps = 2.4 * 6.242*10^18 = 1.5*10^19 Electrons/s.
*December 31, 2014*

**physics**

R2 = R1 + 0.00392(T2-T1)R1 = 257.6 Ohms. 200 + 0.00392(T2-0)200 = 257.6 200 + 0.784(T2-0) = 257.6 200 + 0.784T2 = 257.6 0.784T2 = 57.6 T2 = 73.5o C.
*December 31, 2014*

**physics**

V1=180rev/min * 6.28rad/rev * 1min/60s = 18.84 Rad/s. V2 = 140/180 * 18.84 = 14.65 Rad/s. a = (V2-V1)/t = (14.65-18.84)/20 = -0.209 Rad/s^2. b. V3 = V1 + a*t = 0 t = -V1/a = -18.84/-0.209 = 90 s. To stop. a. V3^2 = V1^2 + 2a*d = 0 d = -V1^2/2a = -(18.84^2)/-0.418 = 849.2 ...
*December 31, 2014*

**Math**

A = (0.44/6.28) * 360o = 25.2o sin A = h/L L = h/sin A = 202/sin25.2 = 474 Ft.
*December 30, 2014*

**physics**

g + a = 6.4 -9.8 + a = 6.4 a = 6.4 + 9.8 = 16.2 m/s^2 Thrust = M*a = 5 * 16.2 = 81 N.
*December 30, 2014*

**Physics**

Yes, I noticed that the initial velocity was unusually low; but I assumed that a change in velocity would change the answer but not the procedure.
*December 29, 2014*

**Physics**

1. Correct 3. Correct 5. Correct 6. Correct. 7. Correct 8. Correct 9. Correct.
*December 29, 2014*

**physics**

See Related Questions: Thu, 12-20-12, 6:03 PM.
*December 29, 2014*

**physics**

a = (V-Vo)/t = (25-20)/4 = 1.25 m/s^2.
*December 28, 2014*

**physics**

Wt. = M*g = 60kg * 10N/kg = 600 N.
*December 28, 2014*

**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*