# Posts by Henry

Total # Posts: 13,500

**Electronics**

Even if an amplifier circuit does not have band-pass filters, there is a limit on the band of frequencies that it can pass. The lowest frequency is determined by the value of the coupling capacitors. The highest frequency is limited by any capacitance in parallel with the ...

**maths**

VAT = 0.05*(52,500+84,000+262,000+672,000+157,000) =

**science**

L = V/F = 3*10^5/40*10^6 = 0.0075 m.

**algebra or algebra1**

(20)(x^3y^4)^3 = (20)(x^9y^12) = 20x^9y^12.

**algebra or algebra1**

F(a) = 2a + 9 = 7, 2a = -2, a = -1.

**algebra or algebra1**

L = Sqrt(21^2+16^2)

**physics**

C = pi*D = 3.14 * 67 = 211 cm. = 2.11 m. Speed = 11.2rev/s * 2.11 m/rev = 23.6 m/s.

**Calculus**

Show that f(x) = 2000x^4 and g(x) = 200x^4 grow at the same rate I know that they grow at the same rate because they are both raised to the same power, but i don't know how to show it.

**Math**

(5,6), (8,9). m = (9-6)/(8-5) = 1, Y = mx + b = 6, 1*5 + b = 6, b = 1, Y = x + 1. (0,5), (6,3). m = (3-5)/(6-0) = -1/3, Y = mx + b = 5, (-1/3)0 + b = 5, b = 5, Y = (-1/3)x + 5. x + 1 = (-1/3)x + 5, x+1 = -x/3 + 5, x + x/3 = 4, Multiply both sides by 3: 3x + x = 12, X = 3 Min.

**Math**

1. 1cents = 2^0 = 2^(d-1). 2. 2cents = 2^1 = 2^(d-1) 3. 4cents = 2^2 = 2^(d-1). 15. 2^(d-1) = 2^(15-1) = 2^14 = 16,384 Cents(pennies) = $163.84.

**math**

1. Take the chicken across. 2. Take the fox across and bring the chicken back. 3. Take the corn across. 4. Take the chicken across.

**Calculus**

Find the range of the function ∫sqrt(4-t^2) dt where a=0 =x [0, 4π] [0, 2π] [-4, 0] [0, 4]

**physics**

a. Impulse = 0.224 * 22 = b. V = Vo + a*t = -22, 18 + a*0.06 = -22, 0.06a = -40, a = 667 m/s^2. F = M*a = 0.224* (-667) =

**Physics**

h = Vo*t + 0.5g*t^2

**Physics**

Y = 5.07*sin60.2 = 4.40 m/s. d = V*T = 24.7, 4.4*T = 24.7, T = 5.61 s.

**physics**

a. V^2 = Vo^2 + 2g*h, Vo = 0, g = 9.8 m/s^2, h = 0.54 m., V = ?. b. k = Mg/d = (1.7*9.8)/0.088m =

**Algebra**

See previous post.

**Algebra**

See previous post.

**Algebra**

See previous post: Wed, 3-2-16, 10:20 PM.

**Algebra**

(-5,3), (4,-5). Y = mx + b, m = (-5-3)/(4-(-5)) = -8/9, Y = (-8/9)(-5) + b = 3, b = 3 - 40/9 = 27/9-40/9 = -13/9. Eq: Y = (-8/9)x - 13/9.

**Algebra**

Incomplete.

**Algebra**

a. Y = a(x-h)^2 + k. h = -b/2a = -18/-6 = 3, k = -3*3^2 +18*3 + 53 = 80, Y = -3(x-3)^2 + 80. b. T max = k = 80o F.

**Algebra**

a. h(t) = 0.5g*t^2. b. h(1) = 16*1^2 = 16 Ft. c. h = 0.5g*t^2. h = 35 Ft., g = 32 Ft/s^2, t = ?.

**math**

son's age : (x+5) in 5 years. Parent's age: 2(x+5) in 5 years. 50 < ((x+5) + 2(x+5))< 100, 50 < (x+5 + 2x+10) < 100, 50 < 3x+15 < 100, 50-15 < 3x < 100-15, 35 < 3x < 85, 35/3 < x < 85/3. 11 2/3 < X < 28 1/3, 16 2/3 < x+5 <...

**Calculus**

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = 5 - x^2 on the interval 0 to sqrt 5 . If so, find the x-coordinates of the point(s) guaranteed by the theorem.

**math 5th grade**

5.29 + (2/3)*3.59 + (4/3)*4.79 = $14.07, Yes. 28.50-(2*3.59)-1.34 = 19.98 = Cost of 4 pks. of glass beads. 19.98/4 = $5.00 = Cost of 1 pk. of glass beads.

**Maths**

1. W/100 = 8/5, 5W = 800, W = 160g of wheat. 160 + 100 = 260 Grams of muesli. 2. 84/R = 7/3, 7R = 252, R = 36g of raisins. 84 + 36 = 120 grams.

**Math**

Did you mean? 12a + 15p = 100.

**Math**

Suzanne: X min. Jim: (x+30)min. X = 1/3 * 60min = 20 min. x+30 = 20 + 30 = 50 min.

**Banking Finance**

Higher.

**Algebra**

P = Po(1+r)^n. Po = $3500, r = (6.75%/12)/100% = 0.005625 = Monthly % rate expressed as a decimal. n = 1Comp./mo. * 6mo = 6 Compounding periods. P = ?.

**Algebra**

P = Po(1+r)^n. Po = $500, r = (4%/4)/100% = 0.01 = Quarterly % rate expressed as a deci8mal. n = 4Comp./yr. * 6yrs = 24 Compounding periods. P = ?.

**Algebra**

P = Po(1+r)^n. n = 1Comp./yr. * 5yrs = 5 Compounding periods. P = 4000(1.06)^5 =

**Math**

P = Po(1+r)^n. Po = $850, r = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 10yrs. = 40 Compounding periods. P = 850(1+r)^40 = 1100, (1+r)^40 = 1100/850 = 1.294, Raise both sides to the 1/40 power: 1+r = 1.294^(1/40) = 1.00647, r = 1.00647 - 1 = 0.00647 = Quarterly...

**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...

**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 = ?.

**math**

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

**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%.

**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.

**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.

**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.

**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...

**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]

**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?

**MATH**

Both ordered pairs at D satisfy the Eq .

**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 = ...

**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].

**MAth**

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

**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

**Calculus**

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

**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

**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 ...

**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(...

**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 &#...

**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.

**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...

**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 ...

**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.

**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.

**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. ...

**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....

**Physics**

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

**physics**

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

**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 ...

**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...

**Math**

The errors in your problem makes it difficult to understand.

**Algebra**

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

**Physical Science**

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

**Math**

My answer is B.

**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% ...

**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.

**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.

**Math/Finance**

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

**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.

**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.

**physics**

Incomplete.

**SCIENCE**

Incomplete.

**Math**

I chose b also.

**Math**

1. Correct. 2. Correct.

**Basic electricity**

The given units are incorrect.

**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 ...

**MATH HELP**

My answer is b.

**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...

**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 =

**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

**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.

**Maths**

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

**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 = ?.

**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.

**physics**

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

**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.

**physics**

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

**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:

**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.

**geometry**

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

**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

**physics**

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

**physical science**

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

**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.

**Pysics**

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