Wednesday

July 23, 2014

July 23, 2014

Total # Posts: 8,897

**physics**

d1 = 0.5a*t^2 = 0.6*5.3^2 = 16.85 m. V1 = a*t = 1.2m/s^2 * 5.3s = 6.36 m/s. a. V = Vo + a.t V = 6.36m/s - 2.3m/s^2*1.60s = 2.68 m/s. b. V^2 = V0^2 + 2a*d d2 = V^2-Vo^2)/2a d2 = (2.68^2-6.36^2)/-4.6 = 7.23 m. D = d1 + d2 = 16.85 + 7.23 = 24.1 m.

**Math**

(150,80), (50,30). m =(30-80)/(50-150) = -50/-100 = 0.5 Y = mx + b Y = 0.5*150 + b = 80 b = 80-75 = 5 Eq: Y = 0.5x + 5 Error: The student interchanged the slope and Y-intercept.

**jru-physics**

PE = mg*h = 50*9.8*2m = 980 Joules

**Math**

Another Approach: %Decrease = ((15%-10%)/15%)*100% = (5/15)*100% = 33.33%

**Math**

(10%/15%)*100%=(10/15)*100%= 66.67%. 66.67% - 100%=-33.33% or 33.33% Decrease

**Physics**

See your 5:59 pm post.

**Physics**

a. Fa = ((V-Va)/(V+Vb))*Fb Fa = ((343-0)/(343+68.6))*960 = 800 Hz b. Fc=((343+114.33)/(343-0))*480=640 Hz. c. Fb = ((343+68.6)/(343-114.33))*1550 Fb = 2790 Hz.

**Physics**

Fr = ((Vs+Vr)/(Vs-Vo))*Fo = 205 Hz ((336+0)/(336-Vo))*200 = 205 ((336)/(336-Vo))*200 = 205 67,200)/(336-Vo) = 205 205(336-Vo) = 67200 68,880-205Vo = 67200 -205Vo = 67200-68880 = -1680 Vo = 8.2 m/s. = Speed of the train.

**Physics**

CORRECTION: b. Fr = 638.33 Hz.

**Physics**

Vo=120km/h = 120,000m/3600s = 33.33 m/s. a. Fr = ((Vs+Vr)/(Vs-Vo))*Fo Fr=(345+0)/(345-33.33)*700Hz = 774.9 Hz b. Fr = ((345-0)/(345+33.33))*700=356 Hz = Freq. heard by the receiver(person).

**Physical Science**

Yes, the answer is C, because KE + PE = A constant.

**Physics**

a. d = 115-160 = -45 miles. The negative sign means the number decreased. b A decreasing marker number means you are traveling South or West. Therefore, John is 45 miles South or West of town.

**Math**

((-3)^2+(-3)*(-2))/4*(-2) = (9+6)/-8 = 15/-8 = -15/8 = -1 7/8.

**science**

d1 = 0.5*3.5*5.9^2 = 60.9 m V1=a*t = 3.5*5.9 = 20.65 m/s. = Velocity after 3.5 s. d2 = Vo*t + 0.5a*t^2 d2 = 20.65*5.98 + 2.98*5.98^2 = 176.8 m. V2 = a*t = 2.98*5.98 = 17.82 m/s. = Velocity @ 176.8 m. d3 = (V^2-Vo^2)/2g d3 = (0-17.82^2)/-19.6 = 16.2 m. = Free fall distance up. ...

**optical circuits**

Total loss = -5 + -5 + -1 = -11 db 10*Log(Po/Pi) = -11 db. 10*Log(Po/2) = -11 Log(Po/2) = -1.1 Po = 2*10^-1.1 = 0.159 mW.

**Physics**

d1 + d2 = 5.3+4.2 = 9.5 km = Distance between the runners. 8.8t + 7.9t = 9.5 16.7t = 9.5 t = 0.569 h. to cross paths. d1 = 5.3 - 8.8*0.569 = 0.293 km West of flagpole. d2 = 4.2 - 7.9*0.569 = -0.295 km West of Flagpole. So they meet at approximately 0.294 km west of flagpole.

**Algebra**

Steve, that was an excellent explanation.

**math**

a. V = Vo + g*t Tr = (V-Vo)/g = (0-15)/-32 = 0.469 s. To reach max. Ht. = Rise time. b. h = ho + Vo*t + 0.5g*t^2 h = 4 + 15*0.469 - 16*(0.469)^2=7.52Ft. Above Gnd. c. h = Vo*t + 0.5g*t^2 = 7.52 Ft. 0 + 16*T^2 = 7.52 T^2 = 0.469 Tf = 0.685 s. = Fall time. T = Tr + Tf = 0.469 + ...

**US History**

True or False Check. 1. During the Starving Time of 1609-10, some of the colonists turned to cannibalism as a means of survival. True 2. Early settlers of Jamestown spent many of their waking hours bowling. True 3. The location of Jamestown was selected partially out of fear o...

**phiycics**

t = 144km * 1h/48km = 3 h.

**Physic**

Correction: The speed of the car does not have to be reduced to zero, because the truck is moving forward. It must be reduced to 50 km/h(13.9m/s). a = (V^2-Vo^2)/2d a = (13.9^2-27.8^2)/400 = -1.45 m/s^2.

**Physic**

Car Speed=100km/h=100000m/3600s=27.8m/s Truck Speed =(50/100) * 27.8=13.9 m/s. V^2 = Vo^2 + 2a*d a = (V^2-Vo^2)/2d a = (0-(27.8-13.9)^2)/400 = -0.483 m/s^2 NOTE: The effective speed of the car is less than the actual speed, because the truck is moving forward.

**Science**

d=109km/day * 1mi/1.6km*7days/wk = 476.875 mi/wk 1.1euros/L*$1.26/euro*3.8L/gal=$5.27/gal Cost = 476.875m1*1gal/39mi*5.27/gal = $64.44

**Science , physics**

Change = 35.4 - 99.9 = -64.5oC The negative sign means the change was a decrease.

**m= ath**

4 Liters(L) = 4000 Milliliters(mL). 4000/150 = 26.7 or 26 Beakers can be filled.

**SAT Math**

2. a is greater because it has the smaller reciprocal. 3. P/4<q/12 Multiply both sides by 12: 3P<q or q>3P. (P-1)/4 Let P = 3 (3-1)/4 = 2/4 = 1/2. (q-9)/12 Let q = 10 which greater than 3P. (10-9)/12 = 1/12. Therefore, (P-1)/4>(q-9)/12. This is true for values of q...

**Math**

(KR)^2 = (-5-2)^2 + (3-9)^2 = 85 KR = 9.22 (RN)^2 = (-5-0)^2 + (3+7)^2 = 125 RN = 11.18 (KN)^2 = (0-2)^2 + (-7-9)^2 = 260 KN = 16.12 P = 9.22 + 11.18 + 16.12 = 36.5

**Algebra 2**

See previous post.

**Algebra 2**

5681mg = 5681*10^-3g 5681*10*-3g * 10^-3 = 5681*10^-6 kg OR 5681mg * 10^-6=5681*10^-6 kg =0.005681 kg.

**AP Physics**

D = 10m[90o] + 7m[120o]. X = 10*cos90 + 7*cos120 = -3.50 m. Y = 10*sin90 + 7*sin120 = 16.1 m. tanAr = Y/X = 16.1/-3.5 = -4.58857 Ar = -77.7o = Reference angle. A = -77.7 + 180 = 102.3o D = X/cosA = -3.50/cos102.3 = 16.43 m.[102.3o]

**algebra**

a. 50/10^9. 15/10^9. b. 50*10^-9 15*10^-9.

**Science Physics**

I'm assuming R and C are in series. R = 10k ohms RC = 34us = 34*10^-6 s. Vr = E/e^(t/RC) = 11 Volts. 27/e^(t/RC) = 11 e^(t/RC) = 27/11 = 2.45455 Take Ln 0f both sides: (t/RC)*Ln e = Ln 2.45455 (t/RC)*1 = 0.89794 Multiply both sides by RC: t = 0.89794RC t=0.89794*(34*10^-6)...

**Physics !!**

1a. d1 = 0.5g*t^2 = 4.9*2^2 = 19.6 m. = Distance fallen by the 1st stone after 2 s. V1^2 = Vo1^2 + 2g*d V1^2 = 0 + 19.6*19.6 = 384.16 V1 = 19.6 m/s = Velocity of 1st stone after 2 s. d1 = d2-19.6 m. Vo1*t + 0.5g*t^2 = Vo2*t+0.5t^2-19.6 Vo1*t-Vo2*t+4.9t^2-4.9t^2 = -19.6 19.6t-2...

**physics**

V=(4.15*10^-2m)^3 = 7.15*10^-5m^3 Density=1.2kg/7.15*10^-5m^3=16,783kg/m^3

**physics**

R1 = 2 Megohms? R2 = 3 megohms? R3 = 15 Megohms? I = 3 Microamps? Please clarify your problem.

**physics**

X = 30m/30s = 1 m/s. Y = 1 m/s . tanA = Y/X = 1/1 = 1. A = 45o = Direction. V = 1/cos45 = 1.414 m/s[45o].

**Physics**

X = ? Y = -2.9 km. R = 4.25 km = Resultant. X^2 + Y^2 = R^2 X^2 + (-2.9)^2 = (4.25)^2 X^2 = (4.25)^2 - (-2.9)^2 = 9.653 X = 3.1 km tanA = -2.9/3.1 = -0.93548 A = -43.1 o A = -43.1 + 360 = 316.9o CCW. A = 316.9 - 270 = 46.9o East of South.

**physics**

V^2 = Vo^2 + 2a*h V^2 = 79^2 + 7.8*980 = 13,885 V = 117.83 m/s. @ 980 m a. V = Vo + a*t = 117.83 79 + 3.9t = 117.83 3.9t = 117.83-79 = 38.83 Tr1 = 9.96 s. to rise to 980 m. V = Vo + g*t 117.83 - 9.8t = 0 9.8t = 117.83 Tr2 = 12.02 s to reach max. ht. hmax = ho + (V^2-Vo^2)/2g h...

**physics**

V^2 = Vo + 2g*d V^2 = 0 + 19.6*(12-6) = 117.6 V = 10.84 m/s. V^2 = 0 + 19.6*12 = 235.2 V = 15.34 m/s

**Math ASAP**

As = 2*22 + 308 = 352 Sq. In. Same as previous results.

**Math ASAP**

SORRY! The calculated surface area was for a rectangular prism NOT a hexagonal prism.

**Math ASAP**

As=2*Ab + Al = 2*22 + 308 = 352 Sq. In.

**Mathematics**

A. P = Po + Po*(r/360)*t = $10,000 Po + Po*(0.043/360)*96 = 10,000 Po + 0.01146667Po = 10000 1.01146667Po = 10000 Po = $9886.63. = Amt. paid. B. %Yield=((10000-9886.63)/9886.63 *365/96 = 4.36

**business math**

P = Po + Po(0.04/360)*45 20000 + 20000(0.04/360)*45 = $20,100 after 45 days. Bal. = 20100-8000 = $12,100

**MaTh**

Ag + As = At As = At - Ag As = 3.14*6^2 - 3.14*5^2 = Area of strip. At = Total area. Ag = Area of garden.

**science**

h = Vo*t + 0.5g*t^2 = 160 m. 0 + 4.9t^2 = 160 t^2 = 32.65 t = 5.7 s. = Time to strike ground.

**Math**

3x^5 - 12x^4 + 18x^3. 1. Locate the smallest constant which is 3. If 3 will divide into the other constants with no remainder, factor out 3. 2. Factor out the variable with the smallest exponent which is x^3. We have factored out 3x^3. 3. To determine what goes inside of the p...

**Physics**

L=V*T = V/F=300*10^6 / 5.35*10^5=561 m.

**algebra**

30-13 means 30 decreased by 13 or 30 less(minus) 13.

**algebra**

13<30.

**algebra**

Where is the "question above"?

**Math**

A. (11-1)games/team * 11teams = 110 games played. Team 1. 10 Games 2. 10 Games 3. 10 4. 10 5. 10 6. 10 7. 10 8. 10 9. 10 10 10 11.10 Games. Total: 110 Games. B. (12-1)games/team * 12teams = 132 Games played.

**MATH**

a. A = -100 + 360 = 260o.

**math**

2(x-3/4)+3(x+2/3)+4(x-1/5) = 9x+17 2x-6/4+3x+6/3+4x-4/5 = 9x+17 9x-3/10 = 9x+17 9x-9x = 17+3/10 0 = 17 3/10 = 173/10. No Solution! Check and make sure the prob. is copied correctly.

**Math (Finance)**

See previous post: Mon,6:41pm.

**Finance**

5.17 Ci=49,000=The # of copies sold 1st year. C3 = C1*(1+r)^3) C3 = 49000(1.21)^3 = 86,806 Copies sold after 3 years. C4 = 86,806*(1.13) = 98,091 Copies sold in 4th year. 5.21a P = Po(1+r)^n r = (8.9%/12)/100% = 0.0074167 = Monthly % rate expressed as a decimal. n = 12Comp/yr ...

**physics**

Circumference = pi*2r = 3.14*100 = 314 m V=1rev/4min*314m/1rev1*1min/60s=1.31m/s

**8th grade algebra**

5x = The # of bikes rented. 2x = The # of scooters renteb 15.50*1.12*5x + 160*1.12*2x = 1600. 86.8x + 358.4x = 1600 445.2x = 1600 X = 3.59 Bikes = 5 * 3.59 = 18 Scooters = 2 * 3.59 = 7.

**8th grade algebra**

5x = The # of bikes rented. 2x = The # of scooters renteb 15.50*1.12*5x + 160*1.12*2x = 1600. 86.8x + 358.4x = 1600 445.2x = 1600 X = 3.59 Bikes = 5 * 3.59 = 18 Scooters = 2 * 3.59 = 7.

**8th grade algebra**

5x = The # of bikes rented. 2x = The # of scooters renteb 15.50*1.12*5x + 160*1.12*2x = 1600. 86.8x + 358.4x = 1600 445.2x = 1600 X = 3.59 Bikes = 5 * 3.59 = 18 Scooters = 2 * 3.59 = 7.

**Physics**

The current entering a point is equal to the current leaving the point. Therefore, the current leaving the positive terminal of a battery is equal to the current entering the negative terminal. This is true for the light bulb or any load.

**math**

P = (Po*r*t)/(1-(1+r)^-t) Po = 1598 * 1.05 = $1677.90 r = (15%/12)/100% = 0.0125 = Monthly % rate expressed as a decimal. t = 3yrs * 12mo./yr = 36 Months. Plug the above values into the given Eq and get: P = 2093.94 Monthly Payment = P/t

**Physics**

The brightness increases.

**Science**

Wb = m*g = 17.25kg * 9.8N/kg = 169 N. = Wt of body. Fv = m*g = 169 N. = Force perpendicular to the table. Fs = u*Fv = 0.34*169 = 57.5 N. = Force of static friction.

**physics**

P = F * d/t = F * V = mg * V P=400*9.8 * 0.03m/s = 118 J/s.=118 Watts

**Math**

63miles/22gal. = 2.86 Miles/gal. D = 66gal * 2.86mi/gal = 189 Miles.

**physics**

Rmax = V/I = (500000*.97)/1000=485 Ohms=Max. conductor resistance. Dmax = 485 Ohms / 0.026Ohms/km=18,654 km.

**Physic**

Fp = mg*sin23 = 597.8*2in23=233.6 N.= Force parallel to the hill. Fv = mg*cos23 = 597.8*cos23=550.3 N.= Force perpendicular to the hill. Fk = 72 N. a = (Fp-Fk)/m = (233.6-72)/61=2.65 m/s^2 V^2 = Vo^2 + 2a*d V^2 = (3.5)^2 + 2*2.65*62 = 340.85 V = 18.5 m/s.

**Pre- Algebra**

1. Y = -2x - 4. (1,3). Parallel lines have equal slopes: m1 = m2 = -2. Y = mx + b = 3 -2*1 + b = 3 b = 3+2 = 5 Eq: Y = -2x + 5 2. Y = 3x+3. (1,1). The slope of the line is equal to the negat1ive reciprocal of the slope of the given line. m1 = 3 m2 = -1/3. Y = mx + b = 1 (-1/3)...

**hydromechanic**

Fap-Ff = m*a 124-Ff = m*0 = 0 Ff = 124 N. = Force of friction. u = Ff/Fl = 124/1026 = 0.121 = Coefficient of friction. Fap = Force applied.

**basic math**

The DIGIT in the units(ones) position is 2. TENS Position: 1. HUNDREDS position: 5. THOUSANDS position: 5. TENS of THOUSANDS position: 5.

**math**

The hypotenuse is greater than either of the other 2 sides but less than the sum. Therefore the answer is d. c^2 = 5^2 + 12^2 = 25 + 144 = 169 c = 13.

**Physics**

V^2 = Vo^2 + 2g*h Vo^2 = V^2 - 2g*h. Vo^2 = 0 + 19.6*23 = 450.8 Vo = 21.2 m. h = (0.5Vo)^2-Vo^2)/2g h = ((10.62)^2-(21.2^2))/-19.6=17.2 m.

**MaThS**

A. Y = -2x+1, and Y = -2x+3. m1 = m2 = -2. The lines are parallel, because their slopes are equal. B. x+y/3+5 = 0, and 2y+6x = 1 x+y/3=-5, and 6x+2y = 1. 3x+y = -15, and 6x+2y = 1 m1 = -A/B = -3/1 = -3 m2 = -6/2 = -3 The slopes are equal; therefore, the lines are parallel.

**algebra**

10. 16-9+25 = 32 11. 32/(-7+5)^3 = 32/-8 = -4. 12. 4^3-(2-5)^3 = 64 - (-27) = 91

**math**

Gregg: p + q Coins. Brother: 8p + 4q Coins. Sum = 8p+p + q+4q = 9p + 5q

**math**

1. 0x+6y-5x = 5x+6y. 2. 9x^2+10+4x^2+7 = 13x^2+17.

**algebra**

P1(1,-3),P2(2,4). D^2 = (2-1)^2 + (4+3)^2 = 50 D = 7.07

**Maths**

tan 60o = (Y1+Y2)/X1 = 3000/X1 X1*tan 60 = 3000 X1 = 3000/tan60 = 1732 m. tan 30o = (Y1+Y2)/(X1+X2)=3000/(1732+X2) (1732+X2)*tan30 = 3000 1732+X2 = 3000/tan30 = 5196 X2 = 5196-1732 = 3464 m tan30 = Y2/X2 = Y2/3464 Y2 = 3464*tan30 = 2,000 m. Y1 = 3000-2000 = 1,000 m. = Distance...

**introduction to finance-16**

1. P = Po*(1+r)^n Po = $4,000. r = (4%/100%) = 0.04 n = 1Comp./yr * 5yrs = 5 Compounding periods. Plug the above values into the given Eq and get: P = $4,866.61. 2. P = Po(1+(y-1))*(1+r)^n Y = Length of loan = 5 years. P = 4000(1+4)*(1.04)^5 = $24,333.06

**math**

Eq1: x*y = -84 Eq2: x+y = 5 x+y = 5 y = 5-x In Eq1, replace y with 5-x: x*(5-x) = -84 5x-x^2 = -84 -x^2+5x+84 = 0 Use Quadratic Formula and get: X = -7, and 12.

**Math**

The angles are assumed to be complimentary(sum=90o). DCE + ECF = 90o 4x+15 + 6x-5 = 90 10x + 10 = 90 10x = 80 X = 8 DCE = 4*8 + 15 = 47o

**Mechanics, Waves and Optics**

Wt. of crate=m*g=18kg * 9.8N/kg=176.4 N. Fv = 176.4*cos0 = 176.4 N. = Force perpendicular to the surface. Fk = u*Fv = 0.5 * 176.4 = 88.2 N. = Force of kinetic friction. Fap*cos20-Fk = m*a Fap*0.94-88.2 = 18*0 = 0 0.94Fap = 88.2 Fap = 93.9 N. = Force applied b. Tension = Fap = ...

**physical science**

V = Vo + a*t = 10 m/s. 6 + a*2 = 10 2a = 4 a = 2 m/s^2.

**geometry**

a. RS + ST = RT 8y+4 + 4y+8 = 15y-9. Solve for y. b. Replace y with the value calculated in prob. a.

**Algebra**

P = Po(1+r)^n P = $5000 Po = $3500 r = (4.5%/12)/100% = 0.00375 = Monthly % rate expressed as a decimal. n = 12Comp/yr * T yrs = 12T = The # of compounding periods. P = 3500*(1.00375)^12T = 5000 (1.00375)^12T = 5000/3500 = 1.42857 Take Log of both sides: 12T*Log (1.00375) = Lo...

**Mathematic**

1. 2/(2/3) + 3/4/(4/5). 2/(2/3) + (3/4)/(4/5) = 2*(3/2) + (3/4)*(5/4) = 6/2 + 15/16 = 48/16 + 15/16 = 63/16 2. (2/5)(2/3)/(5/6)/(3/10) = (2/5)(2/3)(6/5)(10/3) = 240/225 = 48/45 = 16/15. The answer cannot be less than one.

**All subjects**

You are welcome! We are glad we can help.

**MatHS**

The angle of inclination is always between 0o and 180o and is measured CCW from the positive x-axis. A. x+y+1 = 0 x+y = -1 Tan A = m = -A/B = -1/1 = -1 Ar = -45o = Reference angle. A = -45 + 180 = 135o. B. 3-2y = 9x 9x+2y = 3 Tan A = m = -A/B = -9/2 = -4.5 Ar = -77.47o A = -77...

**maths**

See previous post: Tue,8-6-13,12:09 PM

**maths**

See previous post: 12:09 PM.

**maths**

Y = 2x Tan A = m1 = 2 A = 63.43o (-1,7/3),(0,2). m2 = Tan B = (2-7/3)/(0+1) = (-1/3)/1 = -1/3. B = -18.43o = -18.43o C = A-B = 63.43 - (-18.43) = 81.9o = Angle between A and B.

**Math**

What is the question?

**maths**

CORRECTION: m1 = m2 = -A/B = 6/2 = 3 Y = mx + b = -1 3*3 + b = -1 b = -10 Eq: Y = 3x - 10

**maths**

(3,-1). -6x + 2y = 1 Parallel lines have equal slopes: m1 = m2 = -A/B = 6/-2 = -3 Y = mx + b = -1 -3*3 + b = -1 b = 8 Eq: Y = -3x + 8

**mathematics**

P(-6,-2). 3x + 2y = 6 m1 = -A/B = -3/2 m2 = -(-2/3) = 2/3 = Negative reciprocal Y = mx + b = -2 2x/3 + b = -2 (2*-6)/3 + b = -2 -4 + b = -2 b = 2. Eq: Y = 2x/3 + 2

**physics**

See previous post.

**physics**

Assuming no wind: Vo*t + 0.5g*t^2 = 1.4 m. 0 + 0.5*1.63*t^2 = 1.4 0.815t^2 = 1.4 t^2 = 1.72 t = 1.31 s.

**algebra**

Yes, post your problem.

**Physics**

h1 = Vo*t + 0.5g*t^2 h1 = -9*18 + 4.9*18^2 = 1426 m Above ground when bag was released. h2 = 9*18 = 162 m.. Above point of release. h = 1426 + 162 = 1588 m. Above ground.

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