# Posts by henry

Total # Posts: 13,537

**ALGEBRA PLZ HELP!1**

L = 2in * 12ft/1n = 24 ft. W = 1.75in * 12ft/1n =

**Math**

10/35 = 8/W. W = ?.

**Math**

Length = 12 cm. Width = 1/1.414 * 12cm = 8.49 cm. Area = L * W.

**math**

65%/35% * 20Litres. = 37.1 Litres.

**Physics**

Va = 65mi/h * 1600m/mi * 1h/3600s = 28.9 m/s. F1 = (Vs+Vp)/(Vs-Va) * Fa. F1 = (330+0)/(330-28.9) * 500 = 548.0 Hz. F2 = (330-0)/(330+28.9) * 500 = 459.7 Hz. F1-F2 = 548 - 459.7 = 88.3 Hz. = Total shift in freq.

**math**

P = Po(1+r)^n. Po = 200 r = (9%/12)/100% = 0075/mo. n = 1comp./mo. * 24mo = 24 Compounding periods. P = 200(1.0075)^24 = $239.28

**FINANICIAL**

Po = 46,000/(1.0755)^12 = $19,205.82

**FINANICIAL**

P = Po(1+r)^n = 46,000. r = 7.55%/100% = 0.0755. n = 1comp./yr. * 12yrs. = 12 Compounding periods. Po = ?.

**Physics**

h = 0.5g*t^2 = 2.4-0.91. 4.9t^2 = 1.49. t1 = 0.551 s. d1 = Xo*t1 = 22.3 * 0.551 = 12.3 m. = Distance from back line to net. 0.5g*t2^2 = 2.4. 4.9t2^2 = 2.4, t2 = 0.70 s. to fall to gnd. d2 = Xo*t2 = 22.3 * 0.7 = 15.6 m. = Maximum distance travelled. a. Yes. b. d = 12.3 + 6.4 = ...

**algerba**

X pennies, and Y nickels. Eq1: 1x + 5y = 920 cents. Eq2: 5x + 1y = 2680 cents. -5x - 25y = -4600 +5x + 1y = 2680 Sum: -24y = -1920 Y = 80 nickels. In Eq1, replace Y with 80: x + 5*80 = 920, X = 520 Pennies.

**MATH**

Po = 46,000/(1.0755^12) = $19,205.82 = Amt. invested now.

**math algebra**

A = P*e^rt. P = $2,000 rt = 0.0475/yr. * 30yrs. = 1.425. A = ?.

**Math**

W = -(40*Cos30+60*Cos(theta)). -40*sin30 + 60*sin(theta) = 0. 60*sin(theta) = 40*sln30, sin(theta) = 2/3 * sin30 = 0.3333, theta = 19.5o W = -(40*Cos30+60*Cos19.5) = -91.2 N.

**physics**

6.28rad/rev * 2rev = 12.56 rad. 12.56rad/24h * 1h/3600s = 1.45*10^-4 rad/s.

**physics**

M*g = 30, g = 30/M.

**physics**

d = 0.5g*t^2.

**Science**

d = 0.5a*t^2 = 1500 m. 2.5*t^2 = 1500, t = 24.5 s.

**physics**

Range = Vo^2*sin(2A)/g.

**Physics**

See previous post.

**Science Physics**

See previous post: Wed, 11-30-16, 11:53 AM.

**Science, physics**

You have 2 vectors: one pointing north and one pointing west. Vr = -6 + 15i = 16.2km/h[-68.2o] = 16.2km/h[68.2o] N. of W.

**Science**

KE = 0.5M*V^2. KE = 0.5M*(V/2)^2 = 0.5M*V^2/4. So KE is reduced by a factor of 4(divided by 4).

**physics**

a. Vm - 2i = 4.2. Vm = 4.2 + 2i = 4.65m/s[25.6o] N. of E. = Velocity and Heading. b. Vm = 4.65m/s[25.6o]. c. d = Vm*t = 800. t = 800/Vm = 800/4.65 = 172 s.

**Algebra**

When B^2-4AC < 0, No real solutions. When B^2-4AC = 0, One real solution. When B^2-4AC > 0, Two real solutions.

**Algebra**

1. X = (-B +- Sqrt(B^2-4AC))/2A. X = (-0.8 +- sqrt(0.64+2.96))/-0.04 = -27.43, and 67.43. Solution: X = 67.43. 2. 0 = 5x^2+2x-12. B^2 - 4AC = 4 - (-240) = 244. Two solutions. 3. -8x^2 - 8x - 2 = 0. B^2-4AC = 64 - 64 = 0. One solution.

**consumers math**

Ar = Ab-I = Ab - Ab*r*t. Ar = 290 - 290*(0.16/360)*90, Ar = 290 - 11.60 = $278.40 = Amt. received(proceeds). I = $11.60.

**precalculus**

See previous post: Tue, 11-29-16, 3:33 PM.

**Math**

1. When X = 5, Y = 3*5 = 15. Since Y varies directly with X, Y will ALWAYS be 3 times X: Y = 3x. 2. Slope-Intercept Form: Y = (-2/3)x - 1. Multiply both sides by 3: 3y = -2x - 3, Add 2x to both sides: 2x + 3y = -3,

**pre-calculus**

A. 1rev/5.4min * 6.28rad/rev = 1.16 rad/min. B. Circumference = pi*2r = 3.14 * 78 = 245.0 feet. 5.6ft/s * 6.28rad/245ft. = 0.144 rad/s. C. 0.9rad/3s = 0.3rad/s.

**math**

See previous post: Tue, 11-29-16, 5:38 PM.

**maths**

a. 4.8kg/(0.25kg/pk.) = 19.2 or 19 packets. b. 250g/pk. * 0.2pk. = 50 g.

**pre calculus**

Eq2 was obtained by multiplying both sides of Eq1 by 2. Therefore, the two Eqs are identical. So we have an infinite number of solutions.

**pre calculus**

Eq2 was obtained by multiplying both sides of E

**science**

Circumference = pi*2r = 3.14 * 0.5 = 1.57 m. Vo = 72,000m/3600s * 6.28rad/1.57m = 80 rad/s. d = 20rev * 6.28rad/rev = 125.6 rad. V^2 = Vo^2 + 2a*d. V = 0, Vo = 80 rad/s, d = 125.6 rad, a = ?. "a" will be negative.

**Science**

Are you sure the time is only o.5 s.?

**Math**

Eq1: Vm - Vc = 15. Eq2: 0.5Vm + Vc = 12. Sum: 1.5Vm = 27, Vm = 18 mi/h. In Eq1, replace Vm with 18 and solve for Vc.

**Math**

d1 = d2. (Vp+Vw)*t1 = (Vp-Vw)t2. (150+Vw)*2 = (150-Vw)*3. Vw = ?.

**physics**

See Related Questions: Thu, 7-21-16, 1:58 AM.

**History**

The space is between the words the and act

**History**

The purpose of the act was to provide land to railroad companies so that a transcontinental railroad could be built.

**History**

Fill in the blank The purpose of the act was to provide land to railroad companies so that a transcontinental railroad could be built.

**History**

Yes

**History**

What are 2 things the railroad supplied from the west to other parts of the world ? What technology was used to show that the railway was complete ?

**precalculus**

Law of Sine: a/sin A = b/sin B. 9.4/sin150 = 4.8/sin B 9.4*sin B = 4.8*sin150, 9.4*sin B = 2.4, sin B = 0.25532, B = 14.8o. A+B+C = 180o. 150 + 14.8 + C = 180, C = 15.2o. a/sin A = c/sin C. c = ?.

**physics**

b. KE1 = 0.5M1*V1^2 + 0.5M2*V2^2, KE1 = 500*25^2 + (1000*15i^2), KE1 = 312,500 + 225,000 = 537,500 J. = KE before collision. KE2 = 0.5M1*V + 0.5M2*V. KE2 = 500*13^2 + 1000*13^2 = 253,500 J. = KE after collision. KE1-KE2 = KE lost after collision.

**physics**

a. M1V1 + M2V2 = M1V + M2V. 1000*25 + 2000*15i = 1000V + 2000V, 25,000 + 30,000i = 3000V, Divide by 1,000: 25 + 30i = 3V, V = 13m/s[50.2o] N. of E.

**Physics**

Correction: F = -1250 N.

**Physics**

V^2 = Vo^2 + 2a*d = 200^2. 300^2 + 2a*0.2 = 40,000, 0.4a = 40,000-90,000 = -50,000, a = -125,000 m/s^2. F = M*a = 0.01 * (-125000) = -11250 N. Resisting the bullet.

**Physics , static fluids**

Di/Dw = (920kg/m^3)/(1000kg/m^3) = 0.92 = 92% below the surface. 1.0-0.92 = 0.08 = 8% exposed.

**Physics**

F = 15 N.?. F = M*a, a = F/M = 15/14 = 1.07 m/s^2. V = Vo + a*t = 0 + 1,07*8 = 8.56 m/s.

**Physics**

Fr = 50N.[0o] + 30[30o]. X = 50 + 30*Cos30 = 76.0 N. Y = 30*sin50 = 15 N. Tan A = Y/X = 15/76 = 0.19737. A = 11.2o Fr = X/Cos A = 76/Cos11.2 = 77.5 N.]11.2o] N. of E.

**physics**

See previous post.

**physics**

M*g = 20 * 9.8 = 196 N. = Wt. of body. Fp = 196*sin30 = 98 N. = Force parallel to the incline. Fp-40 = M*a. 98-40 = M*a, a = 58/M = 58/20 = 2.9 m/s^2.

**Physics**

V = Vo + a*t = 2.0 rad/s. 0 + a*1.96 = 2.0, a = 1.02 rad/s^2. d = 0.5a*t^2 = 0.51*1.96*2 = 1.96 rad. (360deg./6.28rad) * 1.96rad = 112.4 deg. .

**Maths Ratio and Proportion**

Q1. (10/5)875 + (5/7)875 = 2375.

**Maths Ratio and Proportion**

Q2. 30,000L = 3cm. L = 1*10^-4 cm. Q3. 10/30 * 510 = 170 words.

**Physics**

Correction: A. PE = M*g*h Joules.

**Physics**

B. PE = M*g*h Joules. B. KE = PE = C. V^2 = Vo^2 + 2g*h = 0 + 19.6*10 = 196, V = 14 m/s.

**Physics**

V^2 = Vo^2 + 2a*d = 0. 400^2 + 2a*0.12 = 0, 160,000 + 0.24a = 0, a = -666,667 m/s^2. F = M*a = 0.03 * (-666,667) = -20,000 N.

**Math grade 5, two-part problem**

Adeline has X pens. Lisa has 2x/3 pens. A. (2x/3 + 16)/x = 2/1. 2x/3 + 16 = 2x 2x + 48 = 6x, X = 12. B. 2/3 * 12 = 8 pens. (8+16)/(12-6) = 24/6 = 4/1.

**Science Physics**

Disp. = X/Cos25.6 = 34.7km[25.6o].

**Science Physics**

Disp. = 20km[150o]+40km[0o]+10km[30o]. X = 20*Cos150+40+10*Cos30 = 31.3 km. Y = 20*sin150+10*sin30 = 15 km. Tan A = Y/X = 15/31.3 = 0.47923. A = 25.6o = Direction.

**Physics**

r/2 + r/2 + r = 2r = 2 in parallel + 2 in parallel + 1 in series.

**Physics**

r/2 + 3r = r/2 + 6r/2 = 7r/2 = 2 in parallel and 3 in series. r/3 + 2r = r/3 + 6r/3 = 7r/3 = 3 in parallel and 2 in series.

**physics**

See previous post.

**physics**

10 to 15 m/s: KE1 = 0.5M*V1^2 = 0.5M*10^2 = 50M. KE2 = 0.5M*V2^2 = 0.5M*15^2 = 113M. KE2-KE1 = 113M - 50M = 63M. = Change in KE. 15 to 20 m/s: KE1 = 0.5M*15^2 = 113M. KE2 = 0.5M*20^2 = 200M. 200M - 113M = 87M. = Change in KE.

**Physics**

V = Vo + g*t. Vo = 20 m/s, g = 9.8 m/s^2, t = 5 s, V = ?.

**Physics**

V^2 = Vo^2 + 2a*d. 0 = 300^2 + 2a*0.05, a = -900,000 m/s^2. V = Vo + a*t. 0 = 300-900,000t, t = ?.

**Math**

An angle ? satisfies the relation csc ? cos ? = -1. A) Use the definition of the reciprocal trigonometric ratios to express the left side in terms of sin ? and cos ?. B) What is the relation between sin ? and cos ? for this angle? C) Determine two possible values for ?. - ...

**physics**

Fr = F1+F2 = 10N[80o] + 3.5N[10o]. X = 10*Cos80 + 3.5*Cos10 = 5.18 N. Y = 10*sin80 + 3.5*sin10 = 10.5 N. Fr = = sqrt(x^2+y^2) = (5.18^2+10.5^2) = 11.7 N, = Resultant force. Fr = M*a = 11.7, a = 11.7/M = 11.7/0.5 = 22.14 m/s^2.

**Math**

Rate no. 1 = r mi/h. Rate no. 2 = r+2 mi/h. r*t = 2, t = 2/r. (r+2)*t = 3 (r+2)2/r = 3, 2 + 4/r = 3, 4/r = 1, r = 4 mi/h. (r+2) = 4+2 = 6 mi/h.

**science**

Exp. = 0.5cm/m * 100m = 50 cm.

**physic**

L = 120 cm, W = 80 cm, h = 40 cm. V = 120*80*40 = 384,000 cm^3. M = V*D = 384,000cm^3 * 19.3g/cm^3 = 7.41*10^6 g = 7410 kg.

**physics**

Note: 39.48 = (2pi)^2.

**physics**

T^2 = 39.48 * L/g = 1^2 = 1. L/g = 1/39.48, L = g/39.48 = 9.8/39.48 = 0.248 m. On The Moon: T^2 = 39.48*L/g = 39.48*0.248/1.63 = 6.0, T = 2.45 s.

**Maths**

1st number = X. 2nd number = X+1. (x+1) + x(x+1) = x+1 + x^2+x = x^2 + 2x + 1 = (x+1)(x+1) = (x+1)^2.

**Physics**

Fap = Force applied. Fap*d = 80 J. Fap*1 = 80, Fap = 80 N. Fap = M*a. a = Fap/M. = 80/2 = 40 m/s^2.

**Physics**

T = 136s/100cycles. = 1.36 s./cycle. T^2 = 39.48*L/g = (1.36)^2. L/g = 0.0468, g = 0.50/0.0468 = 10.7 m/s^2.

**Physics**

Ws = 1200 Lbs. Fp = 100*Cos30 = 86.6 Lbs. = Force parallel with the surface. Fn = 1200-100*sin30 = 1150 Lbs. = Normal force. Fk = u*Fn = u*1150 Lbs. Fp-u*Fn = Ws*a. u*Fn = Fp-Ws*0, u*Fn = Fp, u = Fp/Fn = 86.6/1150 = 0.075.

**Physics**

D = Vo*t + 0.5a*t^2. D = 1*10 + 0.05*10^2 = 15 rad. D = 15rad/(6.28rad/rev.) = 2.39 rev.

**Physics**

V^2 = Vo^2 + 2a*d. V = 0, Vo = 200 m/s, d = 0.045 m., a = ?.

**Physics**

C = pi*2r = 3.14 * 4 = 12.56 m. = circumference. Arc = (20/360) * 12.56m = 0.698 m.

**Physics**

F = M*a.

**Math**

P = Po(1+r)^n. First Year: Po = $100,000. r = 0.08/yr. * 0.5yrs. = 0.04. n = 2comp./yr. * 1yr. = 2 compounding periods. P = 100000(1.04)^2 = $108,160. Last Two Years: Po = 108,160. r = 0.12/yr. * 0.25yrs. = 0.03/qtr. n = 4comp./yr. * 2yrs. = 8 compounding periods. P = ?.

**physics**

M*g = 38.5, M = 38.5/g = 38.5/9.8 = 3.93 kg. = Wt. of fluid displaced. Density = M/V = 3.93kg/0.00735m^3 = 534.5 kg/m^3.

**Physics**

R = E/I = 40/5 = 8 Ohms. I = E/R = 120/8 = 15A.

**electronics and circuits**

I1 = Vs/(Rw1+RL1) = 4/(1+15) = 0.25A. = Current thru RL1. VL1 = I1*RL1 = 0.25 * 15 = I2 = (Vs/(Rw2+RL2) = VL2 = I2*RL2 =

**electronics and circuits**

I = V/r = 2/10 = 0.2A. Rbat = V/I = (4-2)/0.2 = 10 Ohms.

**physics**

See previous post: Thu, 11-17-16, 6:05 AM.

**math**

Salary = $X. X-x/4 = 3x/4. 3x/4-(1/4)3x/4 = 3x/4-3x/16 = 12x/16-3x/16 = 9x/16. 9x/16-(1/9)9x/16 9x/16-(1/9)9x/16 = 9x/16-x/16 = 8x/16 = x/2. x/2 = 3400, X = $6800.

**Physics**

See previous post: Fri, 11-18-16, 12:26 AM.

**Physics**

See previous post: Fri, 11-16-16, 12:26 AM.

**Physics**

See previous post.

**Physics**

V = (55-25)g * 1cm^3/g = 30 cm^3. Da = M/V = (48-25)g/30cm^3 = 0.767 g/cm^3 = Density of alcohol.

**Maths**

V = Vo + g*t. Vo = 0, g = 9.8 m/s^2, t is greater 0, but less than 1.5 s.

**Maths**

1. h = 0.5g*t^2 = 4.9*1.5^2 = 2. V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, V = ?.

**Physics**

M*g = 40 * 9.8 = 392 N. = Wt. of child. Fp = 392*sin35 = 224.8 N. = Force parallel to the incline. V^2 = Vo^2 + 2a*d = 2.5^2. 0 + 2a*4 = 6.25, a = 0.781 m/s^2. Fp-Fk = M*a. Fk = Fp-M*a = 224.8-40*0.781 = 193.6 N.

**physics**

c. F = 24.5N/m * 0.05m = 1.23 N.

**physics**

M*g = 0.5 * 9.8 = 4.9 N. = Wt. of object. a. k = F/d = 4.9N./(0.45-0.25)m = 24.5 N/m. b. 24.5N./m * (0.50-0.25)m = 6.125 N. 6.125 - 4.9 = 1.23 N. addition wt.

**Algebra**

Eq1: 5x - 4y = 10. Eq2: 2x + 8y = 4. Multiply Eq1 by 2 and add the Eqs: 10x - 8y = 20. +2x + 8y = 4. Sum: 12x = 24,m X = 2. In Eq2, replace x with 2: 2*2 + 8y = 4. Y = 0. M1 = -A/B = -5/-4 = 1.25. M2 = -2/8 = -0.25.