Friday

April 18, 2014

April 18, 2014

Total # Posts: 8,519

**Physics**

Incomplete.

**Math**

Tax=$125000*($6.35/$1000) + $125000($0.00587/$1.00) = 793.75 + 733.75=$ = $1527.50

**P&P**

d = 40m/min*6min + 200m = 440 m.

**physics**

Correction: V = 170 km/h[270o].

**physics**

V = 150[270o] + 20[270o] = 170km[270o].

**Intrest?**

See Related Questions: Fri,4-12-13,10:57AM.

**physics**

No. Magnitude AND direction must be equal: V1 = 70 km/h V2 = -70 km/h.

**Algebra**

3/(x+1) + 4/x = 2 Multiply both sides by X(x+1): 3x + 4(x+1) = 2x(x+1) 3x + 4x+4 = 2x^2 + 2x 7x - 2x -2x^2 = -4 -2x^2 +5x + 4 = 0 Use Quad. Formula and get: X = -0.64,and 3.14.

**Please Help! Rate of change...**

See Related Questions: Wed,4-17-13,12:03pm.

**Math**

See previous post.

**Algebra**

P = -25x^2 + 300x. h = Xv = -B/2A = -300/-50 = 6 Clerks. P = k = Yv = -25*6^2 + 300*6 = $900

**Math**

Jenny = X Pages. Ken = 2x Alicia = x+18 = 38+11 = 49. Mark = 38. x+18 = 49 X = 49-18 = 31. 2x = 62.

**algebra**

Please clarify your problem.

**Physics**

r = 1.1/sin65 = 1.21 m.

**geometry**

Incomplete

**Trig**

sec(A/2) = 1.4275 cos(A/2) = 1/1.4275 = 0.70053 A/2 = 45.53 A = 91.06o

**math**

I = Po*r*t r = (8.25%/365)/100% = 0.000226 = Daily % rate expressed as a decimal. I = 25,000*0.000226*60 = $339.00

**calculus**

See previous post.

**calculus**

F(1/3) = -2*(1/3)^5 + 2*1/3 - 6 F(1/3) = -2/243 + 2/3 - 6 F(1/3)=-2/43 + 162/243 - 1458/243= -1298/243 = -5.34.

**calculus**

Same procedure as a previous post.

**calculus (gr 12)**

F(-2) = 2*(-2)^3 + 4*(-2)^2 -5*(-2) + 8 F(-2) = -16 + 16 + 10 + 8 = 18.

**calculus (gr 12)**

F(x) = x^4 - 3x F(3) = 3^4 - 3*3 = 81 - 9 = 72.

**Math Radical functions...**

X = Smaller side. X+6 = Larger side. X(X+6) = A X = A/((x+6).

**Physical Science**

An object will float if its density is =< the density of the fluid.

**Physics**

V^2 = Vo^2 + 2g*h h = (V^2-Vo^2)/2g h = (0-10^2)/-19.6 = 5.1 m.

**Geomtery please help??**

10/6 = 4/X 10x = 24 X = 2.4

**PHYSICS**

b. V^2 = Vo^2 + 2g*h h = (V^2-Vo^2)/2g h = (4^2-0)/19.6 = 0.82 m.

**Find the rate of change**

Rate of Change=(3060-2490)/(2011-2009) = $570/2yrs. = $285/yr.

**algebra**

Log 3*sqrt(X) + Log X^4 - Log X^3 Log 3sqrt(3)*X^4 - Log X^3 Log 3sqrt(3)*X^4/X^3.

**Algebra**

Log4 X^4Y^(7/4) Log4 X^4 + Log4 Y^^(7/4) 4*Log4 X + (7/4)*Log4 Y.

**Math**

a = V-Vo)/t = (1.8-1)/0.03 = 26.67 m/s^2

**math algebra**

See previous post.

**math algebra**

X^3-2197 x^3-2197 = 0 x^3 = 2197Take cube root of both sides: X = 13. x-13 = 0 (x^3-2197)/(x-13) Use synthetic Division and get: x^2+13x+169 Factors: (x-13)(x^2+13x+169) x^2+13x+169 Use Quad. Formula to solve for X and get: X = -6.5+11.25i, and -6.5-11.25i.

**maths**

d1 = 500*t d2 = 750*(t-2.5). d1 = d2 After fighter catches bomber. 500t = 750(t-2.5) 500t = 750t-1875 750t-500t = 1875 250t = 1875 t = 7.5 h. a. t-2.5 = 5 h. to catch the bomber. b. d = 500*t = 500*7.5 = 3750 km

**math**

The Law of Cosines: CosB = (-b^2 + c^2 + d^2)/2cd. cosB=(-1681 + 4225 + 2704)/6760=0.77633 B== 39.1o

**math**

Law of Sines: sinI/i = sin(O)/o sinI/52 = sin73/32 Multiply both sides by 52: sinI = 52*sin73/32 = 1.55. The sine cannot be > 1: NO SOLUTION.

**Algebra**

See Related Questions:Fri,8-20-10,6:17pm Your 1st term should be 3x^2.

**Science**

True. I = V/R.

**physics**

a. V^2 = Vo^2 + 2g*h h = (V^2-Vo^2)/2g h = (0-(25.5))/-19.6 = 33.2 m.

**Physics**

A = -7[30o] = -7*cos30 + i(-7)*sin30 A = -6.06 - 3.5i. Q3. tanB = Y/X = -3.5/-6.06 = 0.57756 Br = 30o = Reference angle. B = 30 + 180 = 210o.A = A = -6.06/cos210 = 7[210o].

**physics**

Wt. = m*g = 8kg * 9.8N/kg = 78.4 N. Fw = 78.4N.[0o] = Force of wagon. Fp = 78.4*sin(0) = 0 = Force parallel to sidewalk. Fv = 78.4*cos(0) = 78.4 N. = Force perpendicular to sidewalk. Fk = u*Fv = 0.3*78.4 = 23.52 N = Force of kinetic friction. Fap-Fp-Fk = m*a. Fap-0-23.52 = 8*0...

**Physics**

R(Lp) = 270 Ohms each? R(Htr) = 18 Ohms? a. I = 4 * (120/270) = 1.78A. b. I = 6 * (120/270) = 2.67A. c. I = 2.67 + (120/18) = 9.34A .

**Physics**

a. 9*0.04*300/1000 = 0.108kWh. C = 0.50/0.108 = $4.63/kWh. b. C = 0.07/kWh * 0.108kWh = $0.00756

**Algebra**

See Related problems: Sun.4-14-13,1:35pm

**precalc**

2. V = 300mi/h[20o] + 42mi/h[170o]. X = 300*cos20+42*cos170 = 240.5 mi/h. Y = 300*sin20+42*sin170 = 109.9 mi/h. tanA = Y/X = 109.9/240.5 = 0.45696 A 24.6o = Direction. V=X/cosA = 240.5/cos24.6=264.4 mi/h [24.6o].

**physics grade 12.**

See previous post.

**physics**

See previous post.

**physics**

a. K = F/d = 2N/0.004m = 500 N/m. b. F = m*g = 0.15kg * 9.8N/kg = 1.47 N. (1.47N/500N) * 1m = 2.94*10^-3 m=2.94 mm

**physics**

a. Ek = 0.5m*V^2 = 0.5*0.0467*0^2 = 0. Note: The velocity is 0 at the max. ht. b. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*(25.9-5.85) = 391 V = 19.8 m/s.

**math**

C = pi*2r = 3.14 * 16in = 50.3 In. L = (0.6rad/6.28rad) * 50.3in = 4.81 In. NOTE: The central angle measures 0.6 radians or approximately 0.1 of a circle. Therefore, the length of the arc is approximately 0.1 of the circumference.

**Find the equation....**

P(-4,-2), Q(-2,1). m = (1-(-2))/(-2-(-4)) = 3/2. Y = mx + b Y = (3/2)*(-4) + b = -2 -6 + b = -2 b = 4. Eq: Y = 3x/2 + 4

**Find the distance**

P1(-1,5) P2(2,4). D^2 = (2-(-1))^2 + (4-5)^2 = 10 D = 3.16.

**physics**

R1 = 8 Ohms. R2 = 4 )hms. V2 = 9 Volts. I = V2/R2 = 9/4 = 2.25A E = V1 + V2. E = I*R1 + 9 E = 2.25*8 + 9 = 27 Volts.

**Physics**

Since you did not tell how the 3 resistors and inductor are connected, I assumed your circuit to be the same configuration as a selected Related problem. If the assumption is not correct, you cannot get your book's answer. So please explain how your circuit is connected.

**Physics**

See Related Question: Tue,4-9-13,3:30pm.

**physics**

t = 85.4rad/(18-14.7) = 25.9 s.

**math**

Gal. = 15000mi/30mi/gal = 500. C = (15000/M)P*T C = Cost. 15000 miles = Distance. M = Mileage in miles per gallon. P = Price per gallon. T = Time in years. C = (15000/30)3.61*1 = $1805. C = (15000/20)3.61*1 = $2707.50. Savings = 2707.50 - 1805 = $902.50/yr. 902.50/yr * 5yrs. =...

**math**

As = L*W + 2W*h + 2*L*h As = 9*7 + 2*7*6 + 2*9*6 = 255 U^2.

**Science 8**

Ek = 0.5m*V^2. Answer = C.

**maths- urgent**

sin18o - sin66o + (sqrt3)/2 = 0.26,150

**Science**

m*g = 24,500 N. m = 24500/g = 24500/9.8 = 2500 kg Ek=0.5m*V^2 = 1250*(100/20)^2=31,250 J.

**conceptional physics**

C = pi * 2r = 3.14 * 70cm = 219.9 cm V = 72Rad/s * 219.9cm/6.28Rad=2521cm/s = 25.21 m/s.

**math**

1. P = Po + Po*r*t P = 500 + 500*0.05*2 = $550. 2. Same procedure as #1. 3. P = Po(1+r)^n r = 5%/100% = 0.05 = Annual % rate expressed as a decimal. n = 1comp/yr * 2yrs = 2 compounding periods. Plug the above values into the given Eq and get $661.50. 4. Same procedure as #3.

**physics**

a. I = E/R = 12/0.05 = 240A. 0.95I = 0.95*240 = 228A. Vr + Vc = 12 228*0.05 + 12/e^(t/T) = 12 11.4 + 12/e^(t/T) = 12 12/e^(t/T) = 12-11.4 = 0.6 e^(t/T) = 12/0.6 = 20 t/T = 3. T = L/R = 0.09/0.05 = 1.8=Time constant. t/1.8 = 3 t = 5.4 s. b. Energy = 0.5L*I^2=.5*0.09*(228)^2 = 2...

**physics**

Please ignore my 1st response. My computer malfunctioned. I will have to repeat the entire process.

**physSteady-state current.ics**

b. I = E/R = 12/0.05 = 240A, =

**algebra**

V = Vo + g*Tr Tr = (V-Vo)/g = (0-45)/-32 = 1.41 s. = Rise time or time to reach max. ht. hmax = -16t^2 + 45t + 1 hmax=-16(1.41)^2 + 45*1.41 + 1=32.64 Ft. = 9.89 m. Will not hit roof. Note: Your Eq is not correct.

**phics**

Incomplete.

**maths**

d1 = 90*t. d2 = 100*(t-1). 1. d1 = d2. 90t = 100(t-1) 90t = 100t-100 90t-100t = -100 -10t = -100 t = 10 h. d = 90*10 = 900 km 2. V = d/t=(2*900)km/(10+9)h=94.74 km/h.

**bonds**

P = Po + P*r*t = 20,000 Po + Po*(0.0461/360)*30 = 20000 Po + 0.0038417Po = 20000 1.0038417Po = 20000 Po = $19,923.46 = Amt. paid.

**Physics**

Work = Fx*d = 500*cos20*5.3 = 2490 J.

**physical science**

Density = 25g/13cm^3 = 1.92 g/cm^3.

**math**

Can you clarify your problem by writing it in a conventional manner. Do you mean sqrt(1.2) + sqrt(2.8). Sqrt = Square Root.

**finance**

P = Po + Po*r*t. P = 10000 + 10000*0.09*5 = $14,500. = Value after 5 years.

**physics**

Mass = 8kg Vo = 3.8 m/s. d = 4.7 m. F = 10 N. F = m*a a = f/m = 10/8 = 1.25 m/s^2. V^2 = Vo^2 + 2a*d V^2 = (3.8)^2 + 2.5*4.7 = 26.19 V = 5.1 m/s = Final velocity.

**Math**

1/cosA = =4*sqrt(5)/5 cosA = -5/(4sqrt5) = X/r X^2 + Y^2 = r^2 (-5)^2 + Y^2 = (4sqrt5)^2 25 + Y^2 = 16*5 = 80 Y^2 = 80-25 = 55 Y = sqrt55. sin A = Y/r = sqrt55/(4sqrt5). tan A = Y/X = sqrt55/-5. cot A = 1/tan A = -5/sqrt55. csc A = 1/sin A = 4sqrt5/sqrt55.

**science**

Work = F*d = mg*d = 60*9.8*12 7056 J. Power = 7056J/15s = 470.4 J/s.

**math**

The 2nd 89 REPEATS continuously.

**math**

P = (6/10) * 30 = 18 Inches.

**math**

X Boys (X+3) Girls. x + (x+3) = 27. 2x = 27-3 = 24 X = 12. X+3 = 15.

**Algerbra**

(Vb+Vc)1.5 = 10.5 Eq1: Vb + Vc = 7 (Vb-Vc)3.5 = 10.5 Eq2: Vb - Vc = 3 Add Eq1 and Eq2: Vb + Vc = 7 Vb - Vc = 3 2Vb = 10 Vb = 5 mi/h 5 + Vc = 7 Vc = 2 mi/h.

**Calculus Follow Up**

A farmer wishes to build a fence for 6 adjacent rectangular pens. If there is 600 feet of fencing available, what are the dimensions of each pen that maximizes total pen area? The image looks like this: box box box box box box also: If the interior fencing is $3.00 per foot an...

**Calculus Word Problem**

assume each pen encloses 200 square feet of area in the 2nd part sorry!

**Calculus Word Problem**

A farmer wishes to build a fence for 6 adjacent rectangular pens. If there is 600 feet of fencing available, what are the dimensions of each pen that maximizes total pen area? The image looks like this: box box box box box box also: If the interior fencing is $3.00 per foot an...

**Calculus 2**

That was the only thing the question said :/. It never gave a description of the pens. There is a picture with 6 boxes connected to each other 3 boxes on top and 3 on bottom: box box box box box box other than that, that was all the info I was given:(

**Calculus 2**

A farmer wishes to build a fence for 6 adjacent rectangular pens. If there is 600 feet of fencing available, what are the dimensions of each pen that maximizes total pen area? I keep getting x=150 but I have been told that is not enough fencing. Can anyone help?

**Calculus**

find g''(x) if g(x) = xe^x I have g' as e^x(x+1) then g'' as e^x(1) or just e^x am I missing the mark?

**physics**

Y^2 = Vo^2 + 2g*h h = (Y^2-Yo^2)/2g h = (0-Yo^2)/-19.6 = 30 m. -Yo^2 = -588 Yo^2 = 588 Yo = 24.2 m/s.

**math**

I need the slope or a 2nd point.

**PHYSICS - Waves**

L = V*T = 5m/s * (4/3)s = 6.67 m. NOTE: 4/3 is the time for 1 wave.

**PHYSICS - Waves**

D = = V*t = 343m/s * (2.3/2) = 394.5 m. NOTE: If your answer is correct, the lake is 0ver 2 miles deep.

**PHYSICS - Waves**

Correction: L = V*T = V/F.

**PHYSICS - Waves**

L = V*T = V/F = 30 F = V/L = 7.5/2.5 = 3 Hz = 3 cycles/s. 3cycles/s * 10s = 30 Cycles or waves.

**Phys ics**

1. a = (V-Vo)/t = (210-0)/10 =21rad/s^2. d = 0.5a*t^2 = 0.5*21*10^2 = 1050 Rad. A = 1050rad * 360o/6.28 = 50,134 Deg.. 2. C = pi*2r = 3.14 * 64cm=201 cm=2.01m = Circumference. V = 60km/h = 60000m/3600s = 16.67 m/s. Va = 16.67m/s * 6.28rad/2.01m=52 rad/s. = Angular velocity.

**Science**

Work = F*d = mg*d = 8*9.8*.75 = 58.8 J.

**Physics**

a. I1=I3 = E/(R1+R3)=14/(19+23)=0.333A. I2 = 0 b. The inductor is fully charged and acts like a short circuit(Rc = 0). Rt=R1 + (R2*R3)/(R2+R3)=Tot. Resistance. Rt = 19 + (23*23)/(23+23) = 30.5 Ohms. I1 = E/Rt = 14/30.5 = 0.46A I2 = I3 = I1/2 = 0.46/2 = 0.23A. c. I1 = 0 I2 = I3...

**physics**

I = E/R = 12/0.05 = 24q Amps = Steady-state current. 0.95I = 0.95 * 240 = 228A. Vr + Vc = 12 Volts 228*00.05 + Vc = 12 Vc = 12-11.4 = 0.6 Volts = Voltage across coil when i = 0.95I steady-state a. Vc = 12/e^(t/T) = 0.6 e^(t/T) = 12/0.6 = 20 t/T = Ln20 = 3.0 T = L/R = 0.09/0.05...

**physics**

Vo = 28m/s[30o]. Yo = Vo*sin30 = 28*sin30 = 14 m/s. Y^2 = Yo^2 + 2g*h h = (Y^2-Yo^2)/2g h = (0-(14)^2)/-19.6 = 10 m.

**physics**

Vo = 28m/s[30o]. Yo = Vo*sin30 = 28*sin30 = 14 m/s. Y^2 = Yo^2 + 2g*h

**UOG**

P = Po(1+r)^n. Po = $12,000 r = (4%/4)/100% = 0.01 = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 2yrs. = 8 Compounding periods. Plug the above values into the given Eq. and get $12,994.28.

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