Saturday

October 1, 2016
Total # Posts: 12,076

**math**

P = Po(1+r)^n. r = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 2yrs = 8 Compounding periods. P = 3800*(1+r)^8 = 4300, (1+r)^8 = 4300/3800 = 1.132, 8*Log(1+r) = Log1.132, Log(1+r) = Log1.132/8 = 0.00673, 1+r = 10^(0.00673), r = 10^(0.00673) - 1 = 0.0156, APR = 4...

*February 29, 2016*

**math**

P = Po(1+r)^n. r = (2%/2)/100% = 0.01 = Semiannual % rate expressed as a decimal. n = 2Comp./yr. * 2yrs = 4 Compounding peri0ds. P = Po*(1.01)^4 = 12,500. Po = ?.

*February 29, 2016*

**math**

See previous post: Mon, 2-29-16, 2:12 AM.

*February 29, 2016*

**math**

P = Po(1+r)^n. r = (10%/2)/100% = 0.05 = Semiannual % rate expressed as a decimal. n = 2Comp./yr. * 10yrs = 20 Compounding periods. P = 100,000(1.05)^20 = $265,329.77. Repeat the above procedure for 15% and 20%.

*February 29, 2016*

**math**

Cost = 60 + 11.05*11 = $181.55 P = Po + Po*r*t = 181.55, 170 + 170*r*(11/12) = 181.55, 170 + 155.83r = 181.55, 155.83r = 11.55, r = 0.047 = 4.7%, APR.

*February 29, 2016*

**Algebra 2**

49^3x = 343^(x+1), 3x*Log49 = (x+1)*Log343, Divide both sides by Log49: 3x = (x+1)*Log343/Log49, 3x = 1.5*(x+1), 3x = 1.5x + 1.5, 3x-1.5x = 1.5, X = 1. So the 2x in your Eq should be x.

*February 23, 2016*

**Algebra 2**

49^3x = 343^(2x+1). 3x*Log49 = (2x+1)*Log343, Divide both sides by the Log of 49: 3x = (2x+1)*Log343/Log49, 3x = 1.5(2x+1), 3x = 3x+1.5, 3x-3x = 1.5. No solution? Please make sure the problem is copied correctly.

*February 23, 2016*

**Calculus**

The rate at which water flows into a tank, in gallons per hour, is given by a differentiable function R of time t. The table below gives the rate as measured at various times in an 8-hour time period. t---------0-----2------3-------7----8 (hours) R(t)--1.95---2.5---2.8----4.00...

*February 23, 2016*

**Math**

All angles are measured CCW from +x-axis. Vr = 55mi/h[55o] + 550mi/h[25o] = 31.5+45.1i + 498+232i = 530 + 277i = 598mi/h[27.6o]

*February 23, 2016*

**Calculus**

Use the graph of f(t) = 2t + 3 on the interval [-3, 6] to write the function F(x), where F(x)= ∫f(t) dt where a=3 b=x. F(x) = 2x^2 + 6x F(x) = 2x + 3 F(x) = x^2 + 3x + 54 F(x) = x^2 + 3x - 18 Honestly have no idea where to start. Do i take the derivative of that or what?

*February 23, 2016*

**MATH**

Both ordered pairs at D satisfy the Eq .

*February 23, 2016*

**Math**

9 min. = 0.15h, 22.5 min = 0.375h. d = (Va+Vs)*t = 1.5 km. (Va+Vs) * 0.15 = 1.5. Eq1: Va + Vs = 10. (Va-Vs)*0.375 = 1.5. Eq2: Va - Vs = 4. Add the two Eqs: Va + Vs = 10. Va - Vs = 4. Sum: 2Va = 14, Va = 7 km/h. In Eq1, replace Va with 7: 7 + Vs = 10 km/h, Vs = 3 km/h. Va = ...

*February 22, 2016*

**Science**

All angles are measured CCW from +x-axis. D = 118km[345o] + 118[125o] = 114-30.5i + -67.7+96.7i = 46.3 + 66.2i = 80.8km[55o].

*February 22, 2016*

**MAth**

P = Po + Po*r*t = $5,040. 4500 + 4500*0.06*t = 5040. t = ?.

*February 21, 2016*

**Calculus**

Yes and b are the lower and upper and I plugged in my value of 1 and got 2.59 but that's not one of my answers. My options are 12 6 4 1/9

*February 21, 2016*

**Calculus**

I don't know because thats what the question says. Thanks

*February 21, 2016*

**Calculus**

How would I set this up in my calculator? Let F(x)=∫ ln(t^2) dt where a= 1 and b=3x . Use your calculator to find F"(1). I set it up and I got way the wrong answer. I got 2ln(1)=0

*February 21, 2016*

**Calculus**

Pumping stations deliver gasoline at the rate modeled by the function D, given by D(t)= 6t/(1+2t) with t measure in hours and and R(t) measured in gallons per hour. How much oil will the pumping stations deliver during the 3-hour period from t = 0 to t = 3? Give 3 decimal ...

*February 21, 2016*

**Calculus**

A particle moves along the x-axis with velocity v(t) = sin(2t), with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = π seconds. Do I have to take the integral of the equation like ∫ sin(...

*February 21, 2016*

**Calculus**

f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(2x)]. The table below gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h '(1). (4 points) x 1 2 3 4 5 6 f(x) 0 3 2 1 2 0 g(x) 1 3 2 6 5 0 f '(x) 3 2 1 4 0 2 g &#...

*February 21, 2016*

**Math Fractions**

Total = X pages. Monday: 1/6 * x = x/6 Pages read. x-x/6 = 6x/6 - x/6 = 5x/6 Remaining. Tuesday: 2/5 * 5x/6 = x/3 Pages read. 5x/6 - x/3 = 5x/6 - 2x/6 = 3x/6 Remaining. Wednesday: 1/3 * 3x/6 = x/6 Pages read. 3x/6 - x/6 = 2x/6 = x/3 Remaining. x/3 = 60, X = 180 Pages.

*February 21, 2016*

**physcis**

a. V = Vo + g*Tr = 0, Tr = -Vo/g = -20/-32 = 0.625 s. = Rise time. Tf = Tr = 0.625 s. = Fall time from max ht. to top of bldg. h = Vo*T + 0.5g*T^2. h = 60 Ft., Vo = 20 Ft/s, g = 32 Ft/s^2, T = ?. Tr+Tf+T = Time to reach gnd. b. V^2 = Vo^2 + 2g*h. Vo = 20 Ft/s, g = 32 Ft/s^2, h...

*February 21, 2016*

**Calculus**

Compare the rates of growth of f(x) = x + sinx and g(x) = x as x approaches infinity. f(x) grows faster than g(x) as x goes to infinity. g(x) grows faster than f(x) as x goes to infinity. f(x) and g(x) grow at the same rate as x goes to infinity. The rate of growth cannot be ...

*February 20, 2016*

**Calculus**

Which of the following functions grows the fastest as x goes to infinity? 3^x ln(x) e^4x x^10 I put e^4x but I thought that 3^x might have been it too. I know for sure that it is not ln(x) and x^10 because those grow much slower.

*February 20, 2016*

**Calculus**

Which of the following functions grows the slowest as x goes to infinity? x^e e^x ex they all grow the same rate. I think it is c because it does not have any exponents.

*February 20, 2016*

**Calculus**

The graph of f ′(x) is continuous and decreasing with an x-intercept at x = 0. Which of the following statements is false? (4 points) The graph of f has an inflection point at x = 0. The graph of f has a relative maximum at x = 0. The graph of f is always concave down. ...

*February 20, 2016*

**Physics**

Oops!! Please disregard my first post. First Student: Xo = Cos60.34 = 0.4949 Yo = sin60.34 = 0.8690 Y = Yo + g*Tr = 0, Tr = -Yo/g = -0.8690/-9.8 = 0.0887 s. = Rise time. Tf = Tr = 0.0887 s. T = Tr+Tf = 0.0887 + 0.0887 = 0.1774 s. = Time in air. Dx = Xo*T = 0.4949 * 0.1774 = 0....

*February 20, 2016*

**Physics**

Xo = Cos60.34 = 0.4949 Yo = sin60.34 = 0.8690

*February 20, 2016*

**physics**

M1*V1 + M2*V2 = M1V + M2*V. 3250*44 - 1500*17 = 3250V + 1500V. V = ? Direction: North.

*February 20, 2016*

**electronics**

Normally, the zener is connected in parallel with the load; it tends to maintain a constant voltage across the load. The zener is protected by a series resistor through which the zener and load current flows. If the load is removed, all of the current flows through the zener ...

*February 20, 2016*

**electronics**

1. The voltage dropped across the diodes might not be equal or the voltages from the center-tapped transformer might not be equal. Therefore, you could have a defective transformer or diode. 2.In cases where there are large variations in the input voltage and the load, a zener...

*February 20, 2016*

**Math**

The errors in your problem makes it difficult to understand.

*February 20, 2016*

**Algebra**

(T1+T2+T3)/3 = 19, T1+T2+T3 = 3*19 = 57. (57+T4)/4 = 21, T4 = 27 Degrees.

*February 20, 2016*

**Physical Science**

F = M*a = 100,000-82,000 = 18000, 30,000*a = 18000, a = 0.6 m/s^2.

*February 19, 2016*

**Math**

My answer is B.

*February 19, 2016*

**Physics**

Energy = 0.5C*V^2 = 0.5*14.1*45^2 = 14,276 uJ. = 0.01428 J. Energy = 0.5*14.1*V^2 = 0.001428, V^2 = 2.03*10^-4, Vc = 1.425 Volts when 90% of it's energy is lost. 45/e^x = 1.425, e^x = 31.6, X = 3.45 = t/RC = t/1.13, 1.13*3.45 = 3.90 mS. = 3.90*10^-3 s. = Time to loose 90% ...

*February 18, 2016*

**Physics**

T = R*C = 0.08k * 14.1uF = 1.13 Milliseconds. a. 45/e^x = 4.5, e^x = 45/4.5 = 10, X = 2.303 = t/T = t/1.13, t = 1.13 * 2.303 = 2.60 mS = 2.6*10^-3 s. b. i = v/R = 4.5/80 = 0.056 Amps.

*February 18, 2016*

**Physics**

T1 = 460.8N.[68.9o] N. of W. a. -T1*Cos68.9 = -T2*Cos33.8, -T2 = -T1*Cos68.9/Cos33.8, T2 = 0.433T1 = 0.433*460.8 = 199.6 N.[33.8o]. b. Y = 460*sin68.9 + 199.6*sin33.8 = 540.2 N. M*g = 540.2, M = 540.2/9.8 = 55.12 kg.

*February 18, 2016*

**Math/Finance**

Assuming simple interest. P = Po + Po*r*t = 4000. 2000 + 2000*r*2 = 4000. r = APR expressed as a decimal.

*February 18, 2016*

**Physics**

d = 27.8 * 20 = 556 m. Head start. 0.5a*t^2 = 556 + 27.8t. 1.5t^2 = 556 + 27.8t. 1.5t^2 - 27.8t - 556 = 0. t = Time required for B to catchup. 0.5a*t^2 = Distance at which B catches up.

*February 18, 2016*

**Physics**

a. db = 10*Log(I/Io) = 130. 10*Log(I/10^-12) = 130. Log(I/10^-12) = 13. I/10^-12 = 10^13. I = 10 W/m^2. b. db = 10*Log(20/10^-12) = 133. Doubling the sound intensity increases the sound level by only 3 db.

*February 17, 2016*

**physics**

Incomplete.

*February 17, 2016*

**SCIENCE**

Incomplete.

*February 16, 2016*

**Math**

I chose b also.

*February 16, 2016*

**Math**

1. Correct. 2. Correct.

*February 16, 2016*

**Basic electricity**

The given units are incorrect.

*February 16, 2016*

**science**

20 minutes per hour makes no sense. So let's assume: Vo = 20m/s[53o]. Xo = 20*Cos53 = 12.0 m/s. Yo = 20*sin53 = 16.0 m/s. a. Y = Yo + g*Tr = 0, 16 - 9.8*Tr = 0, -9.8Tr = -16, Tr = 1.63 s. = Rise time. Tf = Tr = 1.63 s. = Fall time. T = Tr+Tf = 1.63 + 1.63 = 3.27 s. = Time ...

*February 16, 2016*

**MATH HELP**

My answer is b.

*February 15, 2016*

**Math College Algebra (urgent before 2hours)**

The zero of a function is the value of X when Y or F(x) = 0. In other words it is the solution to the function. Example: Y = 2x + 4 = 0, 2x = -4, X = -2. So x = -2 when y = 0: (-2,0). 2. g(y) = y^2-3y+2 = 0, C = 2 = (-1)*(-2), Sum = B = -3. (y-1)(y-2) = 0, y-1 = 0, Y = 1, y-2...

*February 15, 2016*

**Science**

10o W. of N. = 100o CCW. 10o E. of N. = 80o CCW. Fr = 1.8*10^6[100o] + 1.8*10^6[80o]=-3.126*10^5+1.773*10^5i + 3.126^10^5+1.773*10^5i = 3.546*10^5i = 3.546*10^5[90o] = 3.546*10^5N., Due North. = Resultant force. Work = Fr*d = 3.546*10^5 * 500 =

*February 15, 2016*

**science**

F1 = 15N[0o], F2 = 20N[135o]. F1+F2 = 15 + -14.1+14.1i = 0.9 + 14.1i = 14.2 N.[86.4o] = 0.89 + 14.18i F3 must be equal to F1+F2 in magnitude and 180o out of phase: F3 = 14.2N[86.4+180] = 14.2[266.4o].= 14.2[86.4o] S. of W. = -0.89 - 14.18i

*February 14, 2016*

**Physics**

a. Work = Change in kinetic energy: Work = 0.5M*V^2-0.5M*Vo^2, Vo = 0, Work = 0.5M*V^2 = 0.5*845*25^2 = 264,063 J. Po = W/t = 264063/60 = 4401 J/s. = 4401 Watts = Power out. Pi = 4401/0.95 = 4633 Watts = Power in. Pi = 12*I = 4633, I = 4633/12 = 386 Amps.

*February 14, 2016*

**Maths**

r = 35/4 = 8 3/4% = 8.75% = 0.0875. I = Po*r*t = 3000*(0.0875/12)*14 = 90.0875*

*February 14, 2016*

**Physics**

1. V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 80 m, V = ?. 2. 0.5g*t^2 = 80. g = 9.8 m/s^2, t = ?.

*February 13, 2016*

**Physics**

Wc = M*g = 35*9.8 = 343 N. Fp = 343*sin26 = 150.4 N. = Force parallel to the incline. Fn = 343*Cos26 = 308.3 N. = Normal. Fp-Fk = M*a. 150.4-0.33*308.3 = 35*a. 35a = 48.66. a = 1.39 m/s^2. V = Vo + a*t = 0 + 1.39*2.5 = 3.48 m/s.

*February 13, 2016*

**physics**

V^2 = Vo^2 + 2g*h, V = 0, Vo = 5m/s, g = -10m/s^2, h = ?.

*February 13, 2016*

**physics**

177 + 0.00506(T-28.7)177 = 116. 177 + 0.8956(T-28.7) = 116. 177 + 0.8956T-25.7 = 116. 0.8956T + 151.3 = 116. 0.8956T = 116-151.3 = -35.3. T = -39.4oC.

*February 13, 2016*

**physics**

F = M*a, F = 37.5 N, a = 2.80 m/s^2, M = ?.

*February 13, 2016*

**Cal 2**

The region bounded by y=3/(1+x^2), y=0, x=0 and x=3 is rotated about the line x=3. Using cylindrical shells, set up an integral for the volume of the resulting solid. The limits of integration are:

*February 12, 2016*

**Biology**

How are the following terms related to the process of transcription and one another? Eukaryotes, Prokaryotes, membrane-bound nucleus, introns, exons, precursor RNA (pre-mRNA), mature RNA, heterochromatin, euchromatin, and structural gene.

*February 12, 2016*

**geometry**

Or 7.5/7 = x/18, X = ?.

*February 11, 2016*

**physics**

Vector V1: Vx = -6.5 units. Vy = 0. Vector V2: V = 8.3[60o] Vx = 8.3*Cos60 = Vy = 8.3in60 = V1 + V2 = -6.5 + 8.3[60] = -6.5 + 4.15+7.19i = -2.35 + 7.19i = 7.56[71.9o] N. of W. = 108o CCW

*February 11, 2016*

**physics**

Do you mean constant velocity? If so, d = V*t, t = d/V.

*February 11, 2016*

**physical science**

d = 0.5a*t^2. d = 10 m, t = 45 s, a = ?.

*February 10, 2016*

**mathematics**

David: X years old. John: x+4 years old. Joe: x-2 years old. x + x+4 + x-2 = 41. Solve for x and add 8 years to John's age.

*February 10, 2016*

**Pysics**

D = -400 + 200[135o] + 100i = -400 - 141.4+141.4i + 100i = -541.4 + 241.4i = 593 km[24o] N. of W.

*February 10, 2016*

**Science**

1. V = Vo + a*t. V = 3.2 m/s, Vo = 1.2 m/s, t = 2.5 s, a = ?. 2. V = Vo + a*t. V = 34 m/s, Vo = 0, t = 2.5 s, a = ?. d = 0.5a*t^2. 3. r = 30m/4.2s. = 4. d = r*t, t = ?.

*February 10, 2016*

**Math**

The simplification process starts inside the expression: ((3x)^4/(6x)^3)^2 = (81x^4/216x^3)^2 = (3x/8)^2 = 9x^2/64.

*February 9, 2016*

**Math**

Y = 3x. 3 = The number of wheels per tricycle. X = The number of tricycles. Y = The total number of wheels to be ordered.

*February 9, 2016*

**math**

15, 5, 3, 0.8, -5.

*February 9, 2016*

**Math**

Po*e^(rt) = 2Po. Divide both sides by Po: e^rt = 2. rt = 0.08t. e^0.08t = 2. 0.08t*Log e = Log 2. 0.08t = Log2/Log e = 0.693 t = 8.7 yrs.

*February 9, 2016*

**math**

a. Tan 24 = h/500, h = ?. b. Tan 27 = (h+L)/500. L = Length of flag = ? h = Ht. of bldg. calculated in part a.

*February 9, 2016*

**physics**

See previous post: Mon, 2-8-16, 7:33 PM.

*February 9, 2016*

**science**

30kW = 30 kJ/s. Work = 30kJ/s * 15s = 450 kJ.

*February 9, 2016*

**Calculus 2**

Sketch the region in the first quadrant enclosed by y=7/x, y=2x, and y=x/2. Decide whether to integrate with respect to x or y. Then find the area of the region.

*February 9, 2016*

**physics**

a. Vr = -100i + 70km[330o] = -100i + 60.6-35i = 60.6 - 135i = 148km[65.8o] S. of E = 24.2o E. of S. = Resultant velocity of the goose. Do you mean 20o W. of S.? If so, Direction = 24.2 + 20 = 44.2o W. of S. b. d = V*t = 1000, t = 1000/V = 1000/148 = 6.76 h.

*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 5000*0.04*2 = $400. (25/4)% = 6.25%. I = 5000*0.0625*2 = $625. Gain = (625-400)/2yrs. = $112.50 per year.

*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 0.125Po. Po*0.1*t = 0.125Po. t = 1.25 yrs. = 15 Months.

*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 1344-1200. Po = 1200, r = 0.06, t = ?.

*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 1770. r = 0.08 t = (15/2)yrs. = 7.5 yrs. Po = ?.

*February 8, 2016*

**SIMPLE INTEREST**

There are errors in your problem; please make corrections.

*February 8, 2016*

**science**

Sketch: From a point, draw 2 arrows pointing upward and one arrow pointing downward.

*February 8, 2016*

**science**

Fr = 5.3 + 2.2 - 10.7 = -3.2 N. = 3.2 N. downward.

*February 8, 2016*

**Engineering**

Mass = 0.00072kg/cm^3 * 37854cm^3/gal * 5gal. = 136.3 kg.

*February 7, 2016*

**physics**

h = V*t + 0.5g*t^2 = 2 m. V*0.1 + 4.9*0.1^2 = 2. 0.1V = 2-0.049 = 1.95. V = 19.5 m/s at top of the window. V^2 = Vo^2 + 2g*h. V = 19.5 m/s, Vo = 0, g = 9.8 m/s^2, h = ?.

*February 7, 2016*

**Finance**

P = Po(1+r)^n. P = $16,860. r = 5.75%/100 = 0.0575 n = 1Comp./yr. * 6yrs. = 6 Compounding periods. Po = ?.

*February 7, 2016*

**science**

a. h = 0.5g*t^2. h = 20 m., g = 9.8 m/s^2. t = ?. b. d = 10m/s * t. t = Value calculated in part "a".

*February 7, 2016*

**physics**

a. V^2 = Vo^2 + 2g*h = 0. Vo^2 = -2g*h = -(-19.6)*0.18 = 3.53, Vo = 1.88 m/s. b. KE = 0.5M*Vo^2 = 0.5*0.54*10^-6*1.88^2 = 9.54*10^-7 Joules.

*February 7, 2016*

**Physics**

Xo = 2.9 m/s. Tan 43.6 = Y/Xo = Y/2.9. Y = 2.9*Tan 43.6 = 2.76 m/s. V^2 = Vo^2 + 2g*d = 2.76^2. 0 + 19.6d = 7.63, d = 0.389 m. h = 1.14 - 0.389 = 0.751 m. Above the floor.

*February 6, 2016*

**physics**

See previous post: Thu,2-4-16, 11:20 AM.

*February 5, 2016*

**Physics**

Oops! The name is Henry.

*February 4, 2016*

**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.

*January 11, 2016*

**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 ...

*January 8, 2016*

**Calculus**

Okay I didn't know you could do that haha

*January 2, 2016*

**Calculus**

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

*January 2, 2016*

**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 ...

*January 2, 2016*

**Calculus**

Im think im going to go with B then

*January 2, 2016*

**Calculus**

Okay Thanks. That one confused me.

*January 2, 2016*

**Calculus**

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

*January 2, 2016*

**Calculus**

Wait so A is right or is it wrong?

*January 2, 2016*