# Posts by Henry

Total # Posts: 14,538

Correct.
2. ### Math

P = Po(1+r)^n. r = 0.085/365 = 0.000233 = daily % rate. n = 13 Compounding periods. P = 5000(1+0.000233)^13 = \$5015.17 Int. = P - Po = 5015.17 - 5000 = \$15.17.
3. ### Algebra

1.X Lbs. of almonds. Y Lbs. of cashews. Eq1: x + y = 7 Eq2: 5x + 12y = 56. 2. Multiply Eq1 by -5 and add the Eqs.: -5x - 5y = -35 +5x + 12y = 56 Sum: 7y = 21 Y = 3 Lbs. In Eq1, replace Y with 3: x + 3 = 7 X = 4 Lbs.
4. ### Pre algebra

8^7 * 8^2 = 8^9 = 134,217,728.
5. ### math

5.99*10^12mi/Light-yr. * 4.2Light-yrs. = 2.516*10^13 Miles. If you had divided, your units would not have been miles. I hope this helps.
6. ### math

1. 5.2*10^5mi/2.5*10^4mi/rev = 2.08*10^1 = 20.8 Rev. or times around. Your answer is 20.8 revolutions; however, if you had multiplied, your answer would have been (mi)^2/rev. which does not make sense for your problem.
7. ### SAT math

(0,0), (6,m), (m,54). slope = m/6 = 54/m. m/6 = 54/m, m^2 = 324, m = 18.
8. ### Physics

V = 60mi/h * 1600m/mi * 1h/3600s = 26.7 m/s. Work = 0.5M*V^2 Joules.
9. ### Physics

V = Vo + a*t =-38. 19 + a*0.01 = -38. a = ? It will be negative. F = M*a.
10. ### Physics

Given: M1 = m, V1 = 485 m/s. M2 = 13m, V2 = 0. V3 = Velocity of proton(M1) after collision. Momentum before = Momentum after. M1*V1 + M2*V2 = M1*V3 + M2*V4. m*485 + 13m*0 = m*V3 + 13m*V4, 485m + 0 = m*V3 + 13m*V4, Divide both sides by m: 485 = V3 + 13V4, V3 = (V1(M1-M2) + 2M2*...
11. ### Math

75*2 + 75t = 125t, t = 3h. d = 125 * 3 = 375 Miles.
12. ### math

-4, -1, 1, 2, 2. They added 3, 2, 1, Then I added 0.
13. ### physics

D = 4m[0o] + 4m[60o]. X = 4*Cos0 + 4*Cos60 = 6 m. Y = 4*sin 0 + 4*sin = 60 = 3.46 m. D = sqrt(X^2 + Y^2) = 6.93 m. Displacement.
14. ### math

Vector Q makes an angle of 60o with the -X-axis. Vector R makes an angle of 60o with the +X-axis. Each vector forms a 60-30 right triangle. Using vector R, Cos60 = X/r = X/10, X = 10*Cos60 = 5 km. sin60 = Y/r = Y/10, Y = 8.66 km. = Ver. component of R and Q. Using Vector Q, ...
15. ### Precalculus 11

The Discriminant is greater than zero(100); therefore, we have 2 real roots.
16. ### math

8x = 40-10 = 30, X = 30/8 = 15/4 = 3 3/4.
17. ### Math

X buttons in bag A, x+18 buttons in bag B, 3(x+18) buttons in bag C. x + (x+18) + 3(x+18) = 4342,
18. ### Math

[(4^2)^2]^2 = [4^4]^2 = 4^8.
19. ### Trig

The angles are referenced to the Y-axis. so we'll assume CW rotation from +Y-axis. F1 = 100Lbs[79o] E. of S. = 180-79 = 101o CW, F2 = 60Lbs[68o] E. of N. = 68o CW. Fr = 100[101o] + 60[68o] = Resultant force, X = 100*sin101 + 60*sin68 = 153.8 Lbs. Y = 100*Cos101 + 60*Cos68...
20. ### Physics

V = Vo + a*t = 17 m/s, 18 + a*4*10^-4 = 17, a = ?. (It will be negative.) F = M*a.
21. ### Math

4x/15 = 500, X = \$1875/mo. 1875/mo. * 12mo./yr. = \$22,500/yr.
22. ### Trig

Correction: Replace 39.9 with 32.9: d^2 = (-32.9)^2 + (-42.33)^2 = 2874.24, d = 53.61 miles which agrees with Daniel and Reiny's answer.
23. ### Trig

Ship #1: 12mi/h[11o] W. of S. = 12mi/h[191o] CW, Ship #2: 9mi/h[75o] E. of N. = 9mi/h[75o] CW. d = 36mi[191o] - 27mi[75o], X = 36*sin191 - 27*sin75 = -32.9 mi, Y = 36*Cos191 - 27*Cos75 = -42.33 mi, d^2 =(-39.9)^2 + ( - 42.33)^2 = 3383.84, d = 58.2 miles apart.
24. ### physics

d = 8 + 6 = 14 km = Travel distance. D^2 = 8^2 + 6^2 = 100, D = 10 km = Displacement.
25. ### physics

d = 8 + 6 = 14 km = Travel distance. D^2 = 8^2 + 6^2 = 100, D = 10 km @ 37o. = Displacement = Straight line distance from starting point.
26. ### math

(4/9)^2 * (3/8)^2 = 16/81 * 9/64 = (9*16)/(81*64) = 9/81 * 16/64 = 1/9 * 1/4 = 1/36.
27. ### Maths

210m/lap * 5laps = 1050 m. = 1.05 km.
28. ### math

a. Speed = 27mi/h * 5280Ft/mi * 1h/3600s = 39.6 Ft/s. b. d = 39.6Ft/s * 2s =
29. ### Physics

M*g = 15 * 9.8 = 147 N. = Wt. of box, Fp = 147*sin A = Force parallel to the hill, Fn = 147*Cos A = Normal force, Fs = u * Fn = 0.45 *147*CosA = 66.2*CosA = Force of static friction, Fp-Fs = M*a, 147*sin A - 66.2*CosA = 15*0, 66.2*CosA = 147*sinA, 147*sinA/CosA = 66.2, sinA/...
30. ### science

Fp = 12*sin20 = 4.1 N. = Force parallel with incline, Fn = 12*Cos20 - 7.35*sin10 = 10 N. = Normal force, Fk = u * Fn = 0.2 * 10 = 2 N., F = 7.35*Cos10 - Fp - Fk = 7.24 - 4.1 - 2 = 1.14 N. = Net force, Work = F * d = 1.14 * 2 = 2.28 J. = Increase in KE.
31. ### Science

V^2 = Vo^2 + 2a*d = 0, 60^2 + 2a*0.5 = 0, a = -3600 m/s^2. V = Vo + a*t = 0,?. 60 + (-3600)t = 0, t = ?.
32. ### Math

y + 3 <= 5, y <= 5 - 3, y <= 2.
33. ### Math

D = 115km[48o] + 75km[120o], X = 115*sin48 + 75*sin120 = 150.4 km, Y = 115*Cos48 + 75*Cos120 = 39.45 km, D = 150.4 + 39.45i = 155.5km[75.3o] = Distance from X to Z. Tan A = X/Y For bearing notation. A = 75.3o.
34. ### Chemistry

Help me with this; For the second part of the analysis,you will be provided with hydrated crystal of oxalic acid (H2C204. 2H2O). weigh out accurately about 1.575g of oxalic acid in a weighing bottle. Dissolve the acid in a beaker, then transfer the solution completely to ...
35. ### Maths

V = Vo + a*t = 0, 30 + (-1.5)t = 0, t = 20 s.
36. ### Math

A triangle is formed: base = 2000 m., A = 180 - (37+49) = 94o, B = 37o, C = 49o, Use Law of Sine to solve triangle: sin94/2000 = sin37/b, b = 1207 m. = 1.207 km, sin94/2000 = sin49/c, c = 1513 m., d = V*t, t = d/V = 1.207/170 = 0.0071h = 25.6 s.
37. ### Physics

WL = wavelength, WL = V * T = V * (1/F) = 3*10^8 * (1/F) = 2.5*10^-9 m, 3*10^8/F = 2.5*10^-9, F = 1.2*10^17 Hz. = 1.2e17 Hz.
38. ### geometry

angle A = X degree, angle B = 2x-24 degrees, A + B = 90o, x + (2x-24) = 90, X = 38o, 2x-24 = 2*38 - 24 = 52o.
39. ### physics

A large bar magnet (mass of 0.4 kg) exerts a 5 N force on a small bar magnet (mass of 0.1 kg) located 20 cm away. Calculate the force exerted by the small bar magnet on the large one.
40. ### physics

One coulomb of charge passes through a point in an electric circuit every ten seconds. What is the current of the circuit?
41. ### physics

How close must two 1.0 C charged objects be for their mutual force to exceed 1.0 N? Compare that distance to something in your everyday life.
42. ### physics

Two insulating spheres with charges of -7.0 ?C and -12 ?C, are located 10.0 cm apart. What is the force between them? In what direction does it act?
43. ### physics

Two positively-charged objects are located 0.1 m apart. How does the value of their mutual electric force change if the distance between them is doubled? (A) quadruples (B) doubles (C) same (D) halved (E) one-fourth How does the value of their mutual electric force change if ...
44. ### math

Thanks to Millie for help with the question and Kiera for posting this up or I will have failed. Thank you very much guys.
45. ### math

At our school we can help each other. It is due Monday the 18th for me I need help with 3D for the math challenge. Please
46. ### math

(12/300) * 100% = 4% tagged, 0.04s = 150, S = 3,750 Squirrels, total.
47. ### Math

4 years ago: Joy was X years old, Paul was x + 20 years old, (x+20)/x = 7/3, 3x + 60 = 7x, X = 15. x + 20 = 15 + 20 = 35 years, Today: Paul's age = 35 + 4 = 39.
48. ### Math

Mean = (60+42+48+46)/4 = 49, (60+42+48+46+x)/5 = 49, (196+x)/5 = 49, 196+x = 245, X = 49. 9 (
49. ### Physics

a. V^2 = Vo^2 + 2a*d = 0, 40^2 + (-20)d = 0, d = 80 m. b. V = Vo + a*t = 0, 40 + (-10t = 0, t = 4 s.
50. ### Physics

2. d = 0.5a*t^2, t = 5min. = 300 s.
51. ### Pre-Calculus Help? Vectors?

B. The direction will change 180o, and the magnitude will decrease. I'm assuming you mean -pi/4.
52. ### Algebra, substitution method

Eq1: x + 5y = 10 Eq2: -2x - 6y = -8 Multiply both sides ofEq1 by 2 and add the Eqs: 2x + 10y = 20 -2x - 6y = -8 Sum: 4y = 12, Y = 3. In Eq1, replace Y with 3 and solve for X: x + 5*3 = 10, X = -5.
53. ### Math

(9m + 6) + (-5m-6) = 9m+6 + (-5m) -6 = 4m + 0 = 4m.
54. ### physics asap

1. V^2 = Vo^2 + 2g*h = 0, 4.7^2 + (-19.6)h = 0, h = 1.127 m. above the platform. 3.8 + 1.127 = 4.93 m. above gnd., V^2 = Vo^2 + 2g*h = 0 + 19.6*4.93 = 96.57, V = 9.83 m/s. 2. V^2 = Vo^2 + 2g*h = 0 + 19.6*11.6 = , V = ?.
55. ### Math

The vertical distance across the semi-circle = Dia./2 = 20/2 = 10 m., The width of the rectangle = 20 m., The length of the rectangle = 40-10 = 30 m., A = pi*r^2/2 + L*W = (3.14*10^2/2) + 30*20 = 757 m^2. P = Pi*Dia./2 + 2L + W = (3.14*20/2) + 2*30 + 20 = 111.4 m.
56. ### Physics

(800/900) * 60N = 53.33 N. = Wt. lost, Wt. in wax = Wt. in air - Wt. lost = 60 - 53.33 =
57. ### Physics

Tf = 0.5 * 4 = 2 s. = Fall time, h = 0.5g*Tf^2.
58. ### Quadratic Functions and Equations

A = 2 * 25 * 100 = 5000 Sq. Yds. = Area of the lawn, A = L * W = 5,000, 100W = 5,000, W = 50 Yds.
59. ### mathematics

Error = 10.60 - 10.06 = 0.54.
60. ### math

72p - (1/8)72p = 72p - 9p = 63p.
61. ### math

A. V = Vo + g*t = 0, 40 + (-32)t = 0, t = 1.25 s. B . h = -16*1.25^2 + 40*1.25 + 6 = 31 Ft.
62. ### maths

h1 = 2760*Tan26 = 1346 Ft. = Ht. of hill above gnd., h2 = Ht. of tower above top of hill, h1+h2 = 2760*Tan32, h2 = 2760*Tan32 - h1 = 1725 - 1346 = 379 Ft.
63. ### maths

h = d*Tan68 = (d+70)*Tan43, d*Tan68 = (d+70)*Tan43, d = (d+70)*0.38, d = 0.38d + 26.6, d = 42.9 Ft., A = 68o. h = d*Tan68.
64. ### physics

M*g = 400 * 9.8 = 3920 N. = Wt. of slab, Fp = 3920*sin35 = 2248.4 N. = Force parallel to incline, Fn = 3920*Cos35 = 3211 N. = Normal force, Fk = u*Fn = 0.6 * 3211 = 1927 N. = Force of kinetic friction, Fp-Fk = M*a, 2248.4 - 1927 = 400*a.
65. ### algebra

2*sqrt(5) + 3*sqrt(49) = 2*sqrt5 + 3*7 = (2*sqrt5) + 21.
66. ### math

2. P = Po + Po*r*t = \$230,000, Po + Po*0.075*2 = 230,000, 1.15Po = 230,000, Po =
67. ### math

36/24 * 2/3 = 3/2 * 2/3 = 6/6 = 1 Cup of butter.
68. ### Math

Triangle #1: a = 3, b = 4, c = 5. Similar Triangle: d = 6, e = 8, f = 10. Both triangles have the same shape as required but different size. The ratios of the corresponding sides are equal: a/d = 3/6 = 1/2, b/e = 4/8 = 1/2, c/f = 5/10 = 1/2.
69. ### Math

P = Po(1+r)^n, P = \$20M, Po = Initial investment, r = 0.06/4 = 0.015 = Quarterly % rate expressed as a decimal, n = 4 comp./yr. * 4yrs. = 16 Compounding periods.
70. ### Chemistry

a hydrocarbon is burnt completely in excess oxygen. it is found that 1.00g of the hydrocarbon gives 2.93g carbon dioxide and 1.80g water. find the empirical formula of the hydrocarbon (H=1,C=12,O=16)
71. ### physics

A. k = F/d = 30N./9cm = 3.33 N/cm. B. F = 3.33N/cm * 18cm = 60 N, Work = F * d = 60 * 18 Joules.
72. ### Math

28. P = Po(1+r)^n, Po = \$1500, r = 0.085/12 = 0.0071 = Monthly % rate. n = 1comp./mo. * 36mo. = 36 Compounding periods.
73. ### Math

23. 52,400 + 52,400*0.2*5 = 24. 36,725 + 36,725*0.11*6 = 25. 82,302 - 82302*05*4 =. I believe it should be 5% instead of 58%. Check textbook. 26. Same procedure #25, but change 1yr. to 12 mo. 27. P = Po(1+r)^n, Po = \$5,000, r = 0.04/4 = 0.01 = quarterly % rate. n = 4 Comp./yr...
74. ### Math

Ao = Ar + Ar*r*t, Ao = 12,345 + 12,345*(0.06/360)*120 = \$12,591.90 = Amt. owed or taken out.
75. ### physics

In the circuit at right, two resistors are connected in parallel. What is the voltage across the 4 ? resistor? What is the voltage across the 12 ? resistor?
76. ### physics

A circuit is powered by a 5 V battery and has 0.2 A of current through it. If the circuit contains two resistors connected in series and one of the resistors has a resistance of 5 ?, what is the resistance of the other resistor?
77. ### algebra 2

a. Tan (theta) = 0, Theta = 0. b. Theta = 90o.
78. ### physics

I = E/(R1+R2) = 10/(10+20) =
79. ### physics

R = 200 + 200 =
80. ### math

P = Po + Po*r*t = 28,500 + 28,500*(0.08/360)*118 =
81. ### physics

M*g = 9.8M = Wt. of vehicle = Normal(Fn). Fp = Mg*sin A = 9.8M*sin 0 = 0. = Force parallel with surface. Fk = u*Fn = 0.7 * 9.8M = 6.86M N. = Force of kinetic friction. Fp - Fk = M*a, 0 - 6.86M = M*a, a = -6.86 m/s^2. V^2 = Vo^2 + 2a*d = 0, Vo^2 - 13.72*24 = 0, Vo = 18.1 m/s.
82. ### physics

M*g = 100 * 9.8 = 980 N. = Wt. of crate = Normal (Fn), Fk = u*Fn = u * 100 = 100u = Force of kinetic friction, F - Fk = M*a, 300 - 100u = 100*0 = 0,, u = 3.
83. ### Math

You are welcome.
84. ### Math

4. (sqrt3+sqrt2)^2 - 3 + 2*sqrt(3*2) + 2 = 5 + 2*sqrt(6).
85. ### Math

1. sqrt(3c^2)/sqrt(27) = c*sqrt(3)/sqrt(9*3) = c* sqrt(3)/3*sqrt(3) = c/3. 2. sqrt(27)*sqrt(3) = sqrt(9*3)*sqrt(3) = 3*sqrt(3) * sqrt(3) = 3 * 3 = 9. 3. sqrt(75) - 4*sqrt(75) = -3*sqrt(75) = -3*sqrt(25*3) = -3*5*sqrt(3) = -15 * sqrt(3). The student should do #4.
86. ### Math & science?

F = 1/P = 1/6.8 = 0.147 cycles/s = 0.147 Hz.
87. ### Algebra 1

A = L * W, A = L*(x + 4) = 2x^3 - 29x + 12, L = (2x^3 - 29x + 12)/(x+4), Use long-hand division, L = 2x^2 - 8x + 3.
88. ### Math

(1200/2400) * 1h = 0.5h = 30 min. to print 1200 black-and -white posters, 50 - 30 = 20 min. to print 500 colored posters, (60min/20min) * 500 = 1500 colored posters in 1 hour.
89. ### math

A1 = 2(L*h) + 2(W*h) + 2(L*W) = Surface area when ht. = 1in , A2 = 2(L*h) + 2(W*h) + 2(L*W) = Surface area when ht. = 2 in, A2 - A1 = The increase in surface area.
90. ### math

Fraction = n. n*0.9/0.5 = 1.8n. 1.8n/n = 1.8 = 180%, Change = 180% - 100% = 80%.
91. ### math

By what percent will a fraction change if its numerator is decreased by 10% and its denominator is decreased by 50%?
92. ### Algebra

P = Po(1+r)^n, Po = \$2600, r = 0.04/yr. n = 1comp./yr. * 7yrs. = 7 Compounding periods. P = 2600(1.04^7) =
93. ### science

d = V * t = 10-2 = 8, V * 4 = 8, V = 2 m/s.
94. ### Math

P = Po + Po*r*t = 1000 + 1000*0.16*1 = \$1,160, 1160 - (350 + 200 + 400) = \$210 Due on 1 Jan. 2003.
95. ### Math

P = Po + Po*r*t = 1,000 + 1000*0.145*1 = \$1,145, 1145 - (200 + 400) = \$545 = Bal. due in 1 yr.

Incomplete.
97. ### finance

Use same Eq I used in your 12:38 AM post.
98. ### finance

750/mo. * 360mo. = \$270,000 = Total cost, P = (Po*r*t)/(1-(1+r)-t) = 270,000, r = 0.04/12 = 0.000328/mo. a. (Po*0.000328*360)/(1-1.000328^-360) = 270,000, (Po*0.1180)/(0.1114) = 270,000, 1.059Po = 270,000, Po = \$254,898.31 = Max. amount that can be borrowed. b. 254,898.31 + 20...
99. ### Math

You are right!
100. ### algebra 2

1. x^2 - (y-12)^2 = 144, and Y = -x^2. x^2 - (y^2 - 24y + 144) = 144, Replace x^2 with -y: -y - y^2 + 24y -144 = 144, -y^2 + 23y - 144 = 144, -y^2 + 23y - 288 = 0, Use Quadratic Formula: Y = (-23 +- sqrt(529 - 1152))/(-2) , Y = (-23 +- 24.96i)/(-2) = 11.5 - 12.5i, and 11.5 + ...
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