# Posts by henry

Total # Posts: 13,498

**Math**

X dimes. 81-x nickels. 0.1x + 0.05(81-x) = 5.75 Multiply both sides by 100: 10x + 5(81-x) = 575. x = ?

**algebra**

-9m^-2*n^5*2m^-3*n^-6 = -18m^-5*n^-1 = -18/m^5*n. Answer: B.

**algebra**

Day 1. x miles. 2. 3x+52 miles. 3. x-83 miles. D = x + 3x+52 + x-83 = 5x - 31.

**how to set this up**

See previous post: Tue, 1-19-2016, 1:56 PM .

**8th grade Algebra**

49z^4-25y^4=(7z^2+5y^2)(7z^2-5y^2).

**i dont know how to solve this**

History: 0.93 = 93%.

**i dont know how to solve this**

English: (13/14)*100% = 92.9% Science: 93%. Math: (25/27)*100% = History: 9.93 = 93%.

**how to set up this problem**

Bal.=Ao - ((1/3)Ao+17.11+26.74) = Ao = $75 = Initial amount.

**math**

I = Po*r*t = $20. 325*(r/12)*2 = 20. 325*r/6 = 20. 325r = 120. r = 0.369 = 36.9% APR.

**Math**

P = Po(1+r)^n. r = (0.05/12)*1/2 = 0.0020833 = Semi-monthly % rate. n = 2Comp/mo. * 300mo = 600 Compounding periods. Solve for P. I = P-Po.

**Math**

P = Po + Po*r*t = 10,000. Po + Po*(0.04/52)*13 = 10,000. Po + 0.01Po = 10,000. 1.01Po = 10,000. Po = $9900.99

**Math**

P = Po(1+r)^n = 2M. Po(1.04)^35 = 2. 3.94609Po = 2. Po = 0.50683M. = $506,830.

**math**

P = Po(1+r)^n. r = (7%/12)/100% = 0.005833 = Monthly % rate expressed as a decimal. n = 12Comp/yr. * 1.5yrs. = 18 Compounding periods. P = 7000(1.005833)^18 = $7737.90. APY = ((P-Po)/Po)*100 =

**Math**

2. Given: P(0,5), m = 1/2. Y = mx + b = 5. (1/2)*0 + b = 5, b = 5. Eq: Y = (1/2)x + 5. 3. Given: (1,2), (2,3). m = (3-2)/(2-1) = 1/1 = 1. Y = mx + b = 2. 1*1 + b = 2, b = 1. Eq: Y = x + 1. 8. Same procedure as #3. 9. Given: (2,3), (x,y), m = -2. m = (y-3)/(x-2) = -2. Cross ...

**Math**

Y = (2/3)x + 2 x = 0 at the point where the graph crosses the y-axis. So in the Eq above, replace x with 0 and solve for y: Y = (2/3)*0 * 2 = 0 + 2 = 2. (x,y) = (0,2).

**Math**

P = 1,092 - (1/3+2/7)*1,092 = 1,092 - (7/21+6/21)*1092 = 1092 - (13/21)*1092 = 1092 - 676 = 416.

**physics**

Fap-Fk = M*a 550.27-Fk = M*0 = 0. Fk = 550.27 N.

**ECE**

Vb-a = 3 + 2 = 5 Volts. Va-b = -2 + -3 = -5 Volts. If the problem asked for Vb-a, I would say, +5 volts. If it asked for Va-b, I would say negative 5 volts. But it didn't ask for neither of the above; therefore, my answer would be "Five volts".

**Physics**

Vo = 1.67m/s[30o]. Yo = 1.67*sin30 = 0.835 m/s. Y = Yo + g*t = 0. 0.835 - 9.8*t = 0. 9.8t = 0.835. t = 0.0852 s. = 8.52*10^-2 s.

**Physics**

Fm = M*g=40kg * 9.8N/kg = 392 N. = Force of the mass. T1*Cos25 = -T2. T2 = -T1*Cos25 = -0.91T1. T1*sin25 = 392. T1 = 928 N. T2 = -0.91*T1 = -0.91 * 928 = -841 N.

**Maths**

a. 17/20 b. (17/20) * 100% =

**Maths**

4.39

**Maths**

5.078, 5.08, 5.287, 5.78, 5.8

**Math**

1. Friend's lunch = F. Your lunch = F+3. f + f+3 = 19. 2. Jason's wt. = 36 Lbs. His sister's wt. = w Lbs. w + 36 = 3w+4. 3. X packages sold last year. 35 = 2x-1. 4. 10 + x = 70. 5. 60 + x > 200. 6. 6.25h >= 143.75. your

**math help**

20 - 5 - 0.75x = 0. 20-5-0.75x = 0. 0.75x = 15. X = 20 Tickets.

**Algebra**

1. (3,8), (9,5). m = (5-8)/(9-3) = 2. (-7/2,-3), (-5,5/2). m = (5/2-(-3))/(-5-(-7/2)) = (5/2+6/2)/(-10/2+7/2) = (11/2)/(-3/2) = 11/2 * (-2/3) = -22/6. 3. (8,3), (x,y), m = 5. m = (y-3)/(x-8) = 5. Cross multiply: y-3 = 5(x-8).

**Algebra**

5k - 2k + 4k + 0.005k = 7.005k.

**math**

x^3 = 3.85*10^-17. Take the cube root of both sides by raising it to the 1/3 power: (X^3)^(1/3)=(3.85*10^-17)^(1/3). X = 3.38*10^-6.

**Trigonometry**

h/38 = 0.34/0.69

**Math**

42 < -6d. -7 > d. or d < -7.

**physics**

a = (V-Vo)/t. Initial Momentum = M*Vo. Final momentum = M*V. Change in momentum = M*V-M*Vo. F = M*a.

**Math**

See previous post.

**Math**

Multiply Eq1 by -2 and add: Eq1: -3x - 4y = -22. Eq2: +3x + 6y = 22. Sum: 2y = 0. Y = 0. In Eq2, replace y with 0 and solve for x: 3x + 6*0 = 22. X = 7 1/3. One solution.

**Math Check (thanks)**

1. 6/(4*5) = 6/20 = 3/10 = 30%. 3. 19/25 = 0.76 = 76%. 4. 0.09 * 100% = 9%. 5. 37.1%/100% = 0.371. 6. 0.8 = 80%; 7/8 = 87.5%. 19/25 = 0.76 = 76%. Order: 19/25, 0.8, 81%, 7/8. 7. (165/250) * 100% = 8. (4/15) * 100% = 9. (52/5) * 100% = 10. 365%/100% = 3.65 = 3 13/20. 11. 100...

**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.

**Algerbra**

y + 4x = 3. y + 4*(-1) = 3. Y = 7. (-1,7). y + 4*5 = 3. Y = -17. (5,-17).

**Physics**

Vo = 100m/s[30o]. Yo = 100*sin30 = 50 m/s. Y^2 = Yo^2 + 2g*h. h = (Y^2-Yo^2)/2g + ho. y = 0. g = -9.8 m/s^2. ho = 100 m. h = ?

**math**

(-6,1), (-3,2). m = (2-1)/(-3-(-6))/ = 1/3. Y = mx + b. (1/3)(-6) + b = 1. -2 + b = 1. b = 3. Y = (1/3)x + 3.

**maths**

T1*T2/(T1+T2) = T3. T1*(T1-10)/(T1+(T1-10)) = T1-18. (T1^2-10T1)/(2T1-10) = T1-18. Multiply by 2T1-10: T1^2-10T1 = 2T1^2-36T1-10T1+180 -T1^2 + 36T1 - 180 = 0 T1^2 - 36T1 + 180 = 0. -6*-30 = 180, -6 + -30 = -36 = B. (T1-6)(T1-30) = 0. T1-6 = 0, T1 = 6 h. T1-30 = 0, T1 = 30 ...

**maths**

T = Ta*Tb/(Ta+Tb) = 40*30/(40+30)= 17.14 Min. to fill with both pipes open. (17.14-5)/17.14 = 0.708 of a tank to be filled. 5 + 0.708*40 = 33.34 Min., total.

**Math**

400qts. = 1600 Cups. 400qts < 4000 Cups. or 4000Cups > 400qts.

**Math**

400qts * 32oz/qt * 1cup/8oz = 1600 Cups.

**Math**

X > c-b.

**physics**

Unless otherwise stated, all angles are measured CCW from the +x-axis. D = -120 + 210[315o] + 280[60o]. X = -120 + 210*Cos315 + 280*Cos60 = 168.5 m. Y=210*sin315 + 280*sin60 = 94 m. Tan A = Y/X = 94/168.5 = 0.55786 A = 29.2o D = X/Cos A = 168.5/Cos29.2 = 193 m[29o]. b. ...

**physics**

610MW = 610MJ/s. a. Energy = 610MJ/s * 3600s/h * 24h/day * 30days = 1.581*10*9 MJ.

**math**

I = Ab*r*t=749.25*(0.12/360)*120 = $29.97 Ar = Ab-I=749.25 - 29.97=$719.28 = Amount received(proceeds). I = Interest. Ab = Amt. borrowed. Ar = Amt. received.

**1 math question**

Correct.

**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 ...

**Math**

2. T-((T/2)-6 + (T/3)-2) = T-((3T/6)-6 + (2T/6)-2) = T - ((5T/6) - 8) = T - 5T/6 + 8 = 6T/6 - 5T/6 + 8 = T/6 + 8.

**Math**

1. #1. m. #2. 2m-1. #3. 4m-2-1 = 4m-3. #4. 8m-6-1 = 8m-7. #5. 16m-14-1 = 16m-15. #6. 32m-30-1 = 32m-31. m + 2m-1 + 4m-3 + 8m-7 16m-15 + 32m-31 = 63m - 57.

**Physics-Conservation of Momentum**

V1 = 90km/h = 90000m /3600s = 25 m/s. V = 80km/h = 80000m/3600s = 22.2 m/s. M1*V1 + M2*V2 = M1*V + M2*V. M1*25+6000*0=M1*22.2 + 6000*22.2 25M1-22.2M1 = 6000*22.2. 2.78M1 = 133,200. M1 = 47,952 kg. = 4.8*10^4 kg.

**Maths**

(2/3)*3.14*(30/2)^3=3.14*r^2*22.5 7069 = 70.69r^2. r^2 = 100. r = 10 cm.

**Math**

1. D = 200[115o] + 150[230o]. X = 200*Cos115 + 150*Cos230 = -181 km. Y = 200*sin115 + 150*sin230 = 66.4 km. D = Sqrt(X^2 + Y^2) = Distance from C to A. Tan A=Y/X = 66.4/-181 = -0.36685 A = 20.1o N. of W. = 159.9o CCW. = Bearing.

**Math**

Vc = 12*18*6 = 1296 in^3. Vb = 4*6*2 = 48 in^3. Vc/Vb = 1296/48 = 27 Boxes.

**ALGEBRA**

9m/2 - 6 = -7/2 + 4m. Multiply both sides by 2: 9m - 12 = -7 + 8m. 9m - 8m = 12 - 7. m = 5.

**area perimeter**

2L + 2W = 28. Divide both sides by 2: Eq1: L + W = 14. Eq2: L*W = 24. L = 24/W. In Eq1, replace L with 24/W: 24/W + W = 14. Multiply both sides by W: W^2 + 24 = 14W. W^2 - 14W + 24 = 0. 24 = -2*-12, -2 + -12 = -14 = B. (w-2)(w-12) = 0. w-2 = 0, W = 2. w-12 = 0, W = 12. (L,W) (...

**Typo (Test Marked)**

1. N = 0.73 * 215 = 156.95 2. 23/94 = 0.2447 = 24.47%. 3. 80 = 0.40*N. N = 200. 4. C = 0.05 * 1250 = $62.50 5. 100% + 6% = 106% = 1.06 Total Cost = 1.06 * 325 = $344.50

**pre-algebra**

1. Do you mean (y-5)/3 = 1 ? If so, y-5 = 3, Y = 8. If you mean y - 5/3 = 1, The answer is 8/3 or 2 2/3(Ms Sue's answer). Without the parenthesis, the tutor don't know what you mean ; because their are 2 possibilities.

**Algebra**

(4,-2), (0,-4). m=(-4-(-2))/(0-4) = -2/-4 = 1/2. Y = mx + b = -2. (1/2)4 + b = -2. 2 + b = -2. b = -4. Eq: Y = (1/2)x - 4.

**Math**

x-2y = 0. 2y = x. y = (1/2)x. Slope = 1/2 = Tan A. A = 26.57o. Cos 26.57 = 0.89443.

**Algebra**

m = (-5-r)/(3-1) = -4. (-5-r)/2 = -4. Multiply by 2: -5-r = -8. r = 3.

**Algebra**

Correct!

**Algebra**

Correct.

**Algebra.**

To break even: Income = Expense. 5x + 1300 = 3x + 1800 + 1500. 5x - 3x = 1800 + 1500 - 1300. 2x = 2000. X = 1000 Tickets.

**Algebra**

P = 2L + 2W. W = L. P = 2L + 2L = 4L. 20< P < 94. 20< 4L < 94. Divide by 4: 5< L < 23.5 in. 5< W < 23.5 in.

**Algebra 1**

1. (-3,2).

**Algebra 1**

Correct.

**Algebra 1**

Eq1: 4x + 2y = 10. Eq2: Y = -3x + 6. In Eq1, replace y with -3x+6 and solve for x: 4x + 2(-3x+6) = 10. 4x - 6x + 12 = 10. -2x = -2. X = 1. In Eq2, replace x with 1 and solve for y: Y = -3*1 + 6 = 3. Solution: (1,3).

**Algebra 1**

None of the give answers satisfy BOTH of the given equations. Remember that each answer must satisfy BOTH equations. My answer: (-2/3,-77/3) satisfies BOTH equations.

**Algebra 1**

Eq1: 4x - y = 23. Eq2: 7x - y = 21. Multiply Eq2 by -1 and add: +4x - y = 23. -7x + y = -21. Sum: -3x = 2. X = -2/3. In Eq1, replace x with -2/3 and solve for y. 4(-2/3) - y = 23. Y = ? You should get: Y = -77/2 = -25 2/3.

**Algebra 1**

Eq2: y + 4 = -3x. Y = -3x - 4. Eq1 and Eq2 are identical. Therefore, there are an infinite number of solutions. The graph is a single line.

**algebra 1**

A. Correct. B. In Eq1, replace x with 30+y and solve for y: 30+y + y = 60. 2y = 30. Y = 15 Minutes on free hand exercises. C. Yes, but the time spent in the gym would be 55 minutes(15+40) instead of 60 minutes.

**math**

7256chickens * 3290feathers/chicken = 23,872,240 Feathers to be plucked. 23,872,240fe * 2.6s/83.6fe * 1h/3600s = 206.2 h.

**Math**

Unless otherwise indicated, all angles are measured CCW from +x-axis. Ship #1: d1 = 11.5km/h[52o] * 2h = 22km[52o]. Ship #2: d2 = 13km/h[ 317o] * 2h = 26km[317o]. d2-d1 = 26[317o] - 22[52o]. X = 26*Cos317 - 22*Cos52) = 5.47 km. Y = 26*sin317 - 22*sin52 = -35.1 km. d2-d1 = Sqrt...

**physics**

V = Vo + a*t = 0. a = -Vo/t = -3/5 = -0.6 m/s^2.

**Calculus**

Okay I didn't know you could do that haha

**Calculus**

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

**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 ...

**Calculus**

Im think im going to go with B then

**Calculus**

Okay Thanks. That one confused me.

**Calculus**

Okay will do. So I'm presuming your saying the answer is B. I think D works.

**Calculus**

Wait so A is right or is it wrong?

**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) ∫&#...

**Calculus**

Oh shoot I misread my answer. Thanks

**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

**Calculus**

Yes Thank you

**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 ...

**Math**

Yes, it is I, your humble servant(LOL).

**Math**

T = 2 h. Ta = 3 h. Tb = ? T = Ta*Tb/(Ta+Tb) = 2. 3*Tb/(3+Tb) = 2. Tb = ?.

**Physics**

F = M*a, M = F/a = 26.5/6.5.

**English**

I chose C

**Maths**

g Girls in the class. 4g + 3 Boys in class. Total = g + (4g+3). The total should have been given.

**physics in kinematics**

Vo = 18m/s[37o]. Xo = 18*Cos37 = m/s. Yo = 18*sin37 = m/s.

**Physics**

A. Vr = -104 - 398i = 411.4km/h[75.4o] S. of W. = 14.6o W. of S. B. 411.4km/h[14.6o] E. of S.

**Science**

V^2 = Vo^2 + 2a*d = 0 + 2*0.5*100 = 100. V = 10 m/s. V=10m/s * 1km/1000m * 3600s/h = 36 km/h.

**physics**

Fr = 10N.[0o] + 30N.[120o] + 50N.[240o] X = 10 + 30*Cos120 + 50*Cos240 = -30 N. Y = 30*sin120 + 50*sin240 = -17.3 N. Tan A = Y/X = -17.3/-30 = 0.57735. A=30o S. of W. = 210o CCW from +x-axis. Fr = -30/Cos210 = 34.64 N. @ 210o CCW.

**PHYSICS**

Fr = Sqrt(F1^2 + F2^2) = Resultant force

**Calculus**

I tried to use partial fractions to do it, but it didn't work.

**Calculus**

Find the integral of 2/((2-x)(x+2)^2)dx.

**Physics**

I = (5^2/20^2)*X = (1/16)X W/m^2.