# Posts by Henry

Total # Posts: 13,538

**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.

**Precalc**

The "j" means there is a 90-degree phase difference between the two terms and they cannot be added directly. Another method of writing the polar form: 2.83[45o].

**physics**

Work = F*d = 1500*Cos60 * 30 =

**Calculus**

Solve the following problem by separation of variables. T^2y'-t=ty+y+1 Y(1)=0 Y is a function of t.

**Calculus**

Wait, why is it the integral from 3-5 instead of 0-2?

**Calculus**

Sorry I forgot the dx after the integral.

**Calculus**

Find the integral from 0 to 2 of xsqrt(5-sqrt4-x^2)). The hint said to use substitution of u=sqrt(4-x^2), and that I needed one more substitution, but I don't know how to do it.

**Algebra 1**

Since there are 2 variables and only 1 Eq, I have to solve for 1 variable in terms of the other.

**Algebra 1**

12AB + 7B = 5A. 12AB - 5A = -7B. A(12B - 5) = -7B. Divide by (12B-5): A = -(7B/(12B-5)).

**physics**

Glad I could help.

**physics**

By definition, acceleration means to increase speed. When speed is decreasing it is called Deceleration. So you can't do both at the same time. When an object is moving, it is either accelerating, decelerating or moving at a constant speed. In other words, there are 3 ...

**physics**

d = 75N * 1m/1500N = 0.05 m. PE=0.5F * d = 0.5*75 * 0.05 = 1.875 J.

**physics 1**

1a. V = Vo + a*T1 = 0. T1 = -Vo/a = -22.5/-3.95=5.70 s To stop. V = Vo + a*T2 = 22.5. T2 = (22.5-Vo)/a = (22.5-0)/4.91=4.58 s. To accelerate back up to speed. T1+45+T2 = 5.70 + 45 + 4.58 = 55.3 s. 2. Hare: r*t = 1000. 1.5t = 1000. t = 666.67 s., Non-stop. Tortoise: r*t = 1000...

**physics**

Assuming the 3.5 m/s is horizontal: h = 0.5g*t^2. g = 9.8 m/s^2. t = ? A3. h = Vo*t + 0.5g*t^2 h = 230 m. Vo = -7.9 m/s(plane moving upward). g = 9.8 m/s^2. t = ? K5. The problem doesn't say what the captain is trying to do.

**physics**

Ma*Va + Mb*Vb = Ma*V + Mb*V. 30.7*2.64 + 109*1.29i + 30.7V + 109V. 81.05 + 140.6i = 139.7V. 162.3m/s[60o] = 139.7V. V = 1.16m/s[60o] N. of E.

**physics**

M1*V1 + M2*V2 = M1*V + M2*V. 2864*V1 + 856*0 = 2864*7.5 + 856*7.5. V1 = ?

**physics**

M1*V1 + M2*V2 = M1*28.7 + M2*V. 0.192*44.4 + 0.459*0 = 0.192*28.7 + 0.459V. V = ?

**physics**

M1*V1 + M2*V2 = M1*(-0.43) + M2*V. 0.13*0.22+0.22*(-0.30)=-0.43*0.13 + 0.22V. V = ? Multiply your answer by 100 to convert to cm/s.

**physics**

M1*V1 + M2*V2 = M1*V + M2*25. 24.9*22 + 13.2*11.8 = 24.9V + 13.2*25. V = ?

**physics**

M1*V1 + M2*V2 = M1*V + M2*V. 13.2*6.8 + 3*(-3.5) = 13.2V + 3V. V = ?

**physics**

a. V = Vo + g*t = 0. Vo - 9.8*2 = 0. Vo = 19.6 m/s. b. h = Vo*t + 0.5g*t^2. g = -9.8 m/s^2. t = 2 s. h = ?. c. Tr+Tf = 2 + 2 = 4 s. To fall back to top of bldg. Tr = Rise time. Tf = Fall time. 8-4 = 4 s. To fall from top of bldg. to gnd. h = Vo*t + o.5g*t^2. Vo = 19.6 m/s. t...

**Language Arts Project/Essay HELP!**

Ugh, I am not really good at writing, sorry but I fell I'm useless here. I would better recommend you to contcat Supreme essay service. Those guys are real professionals in what they do. Best of luck!

**MATH HELP**

A = (3)^5/2 * (3)^5/2 = (3)^10/2 = 3^5 = 243 in^2.

**College Algebra**

1. P1 = 0.64 + 0.30*0.64 = P2 = 0.64 + 0.40*0.64 = P3 = 2. P = 0.64 + m*0.64 = 1.10 0.64m = 1.10 - 0.64 = 0.46. m = 0.719 = 71.9%.

**Math**

A = 2*(15*5) + 2(30*5) + 30*15=900 Ft^2. = Total area to be painted. 900Ft^2 * 1gal/250Ft^2 = 3.6 Gal. of paint needed. Buy 4-1gal. cans.

**PHYSICS! Really Confused! Test Tmrw**

V^2 = Vo^2 + 2g*h = 0. h = -Vo^2/2g + 10 = -(5^2)/-19.6 = 11.28 m. h = 0.5g*T^2 = 11.28 m. 4.9T^2 = 11.28. T^2 = 2.30. T = 1.52 s. d = Vt*T = 30 * 1.52 = 45.6 m.

**Math**

(Y+1)+4

**physics**

See previous post: Wed,8-12-15, 12:14 AM.

**physics**

Correction: r = (d1+d2)/(43.3+T) = 1.99. (170.2+1.13T)/(43.3+T) = 1.99. Solve for T.

**physics**

d1 = 3.93m/s * 43.3s = 170.2 m. d2 = 1.13*T. r = (d1+d2)/T = 1.99 m/s. (170.2+1.13T)/T = 1.99. Multiply by T: 170.2 + 1.13T = 1.99T. 1.99T - 1.13T = 170.2. 0.86T = 170.2. T = 198 s.

**=**

Diode Connected: Vd + (I2+Id)R1 = 6 Volts. 0.7 + (I2+1.0)R1 = 6. 0.7 + I2*R1 + R1 = 6. Eq1: I2R1 + R1 = 5.3. Diode disconnected: V2 + I2*R1 = 6 Volts. 2 + I2R1 = 6. Eq2: I2R1 = 4. In Eq1, replace I2H1 with 4: 4 + R1 = 5.3. R1 = 1.3k = 1300 Ohms. In Eq2, replace R1 with 1.3k: ...

**Algebra**

You show the Ht. of only 7 players and the Wt. of only 6 players.

**Mathematics**

Show that the root equation (x-a)(x-b)=k^2 are always real numbers

**Mathematics**

Factorice 16(a-2b)^2-(a+b)^2 leaving your answer in the simplest form

**pre algebra**

P = Po(1+r)^n. Po = $2000. r = 0.051/2 = 0.0255 = Semi-annual % rate expressed as a decimal. n = 2Comp./yr. * 7yrs. = 14 Compounding periods.

**Algebra**

1. x-1/5 = +-3. 5x - 1 = +-15. 5x = 16, and -14. X = 16/5, or -14/5. 2. D. decreasing nonlinear.

**Math**

0.64N. N = the number of boxes sold.

**physics**

Fr = 25N.[25o] + F2[-73o] Since the acceleration is parallel to the x-axis, the sum of Y components of velocity equals 0: 25*sin25 + F2*sin(-73) = 0. 10.57 -0.956F2 = 0. -0.956F2 = -10.57. F2 = 11.1N[-73o].

**physics**

Vo = 56m/s[35o]. Xo = 56*Cos35 = 45.9 m/s. Yo = 56*sin35 = 32.1 m/s. Y = Yo*t + 0.5g*t^2=32.1*2.9 -4.9*2.9^2 = 52 m. X = Xo*T = 45.9m/s * 2.9s = 133 m.

**Physics**

V^2 = Vo^2 + 2g*h. Vo = 0. V = ?

**Math**

1-3/4 = 4/4-3/4 = 1/4 Eaten.

**Science**

Eff. = 70/100 * 100%.

**Math**

a. d1 = 200/sin43 = 293 Ft. b. d2 = 200/tan43. c. d3 = d2/Cos16. d. h = 200 + d2*Tan16

**Business Math**

P = Po + Po*r*t P = 1700 + 1700*(0.12/12)*18 = $2,006.

**Physics**

V = 50,000m/3600s = 13.9 m/s. t = 20min ? = 1200 s. a = (V-Vo)/t = Vo = 0. a/g * 100% =

**Math**

The driver formed a rectangle with 10 km horizontal sides and 20 km vertical sides(northward).

**Physics Help**

Vo = 100m/s[60o]. Xo = 100*Cos60 = 50 m/s. Yo = 100*sin60 = 86.6 m/s. Y^2 = Yo^2 + 2g*h = 86.6^2 - 19.6*150 = 4560. Y = 67.5 m/s.

**Physics (angle Vector)**

Vb - 5.4i = 11.8. Vb = 11.8 + 5.4i = 13m/s[24.6o] N. of E. = Compass heading. 9-5.4i) =

**math**

Started with $X. x/3 + x/2 = x - 9.00. Multiply by 6 and solve for x.

**algebra**

Sold X none student tickets. Sold Y student tickets. Eq1: x + y = 937. Eq2: 5x + 3y = 3943. Multiply Eq1 by -3 and use Elimination Method.

**Algebra**

Right!

**physics**

Incomplete.

**physics**

Oops! I misread the numbers. Total=2*3.33 + 1*2.11 + 1*1 = 9.77 Lbs. 9.77Lbs * 5days/10Lbs. = 4.9 Days or 5 days.

**physics**

Total = 23*33 + 12*11 + 1*1 = 892 Lbs. 892Lbs * 5days/10Lbs = 446 Days.

**Mathematics**

(x-6)*Log4 = 3x*Log2. Divide by Log4: x-6 = 3x*Log2/Log4. x-6 = 3x*0.5 = 1.5x. x-1.5x = 6. -0.5x = 6. X = -12.

**8th grade algebra**

Correction: In your case, 48 is the area of the base: V = B*h = 48m^2 * 7m = 336 m^3 or 336 cubic meters. Again, take half of the base only when it is the diameter.

**8th grade algebra**

V = pi*r^2*h = 3.14*24^2*7 = 12,660.5 cm^3(Not m^3). V = B*h, B=Area of the base, V=pi*r^2*h 48 is not the area of the base; it is the diameter of the base. Diameter*height = Square units. Diameter/2 = Radius. So you take 1/2 of the diameter to get the radius. I hope this helps.

**Physics**

-2.8 + Vb = -10.4i. Vb = 2.8 - 10.4i. = 10.8m/s[-74.9o] = 74.9o S. of E. = 15.1o E. of S.

**algebra**

1/4 + 1/3 + 1/2 = 3/12 + 4/12 + 6/12 = 13/12 = 1 1/12. Impossible, because 13/12 is more than a whole pizza.

**Physics**

101kpa = 101*10^3pa = 1.01*10^5 pa.

**science**

Increasing the voltage increases the current and power(P = E*I); therefore, the voltage should be increased.

**Math**

15/0.6 =

**math**

V1 = 75i mi/h. V2 = 60 mi/h Tan A = V1/V2=75/60 = 1.2500, A = 51.3o. d1 = 250*sin51.3 = 195 mi. V1*T = 195, 75 * T = 195, T = 2.60 h. = 156 min.