Thursday

October 2, 2014

October 2, 2014

Total # Posts: 9,185

**physics**

F = m*a = 9 * 3 = 27 N.
*December 6, 2013*

**physics**

To = 9C/5 + 32 = 9*35/5 + 32 = 95oF. T = 9*(-12)/5 + 32 = 10.4oF. a max = 7.9*10^-6/oF. Change = a*(T-To)ho 7.9*10^-6*(10.4-95)*170 = -11.36 cm. Change = 11.36 cm decrease
*December 6, 2013*

**physics**

Vo = 16m/s[20o] Yo = 16*sin20 = 5.47 m/s. Y = Yo + g*t = 0 At max. Ht. 5.47 - 9.8t = 0 9.8t = 5.47 Tr = 0.558 s. = Rise time Tf = Tr = 0.558 s. = Fall time. T = Tr+Tf = 0.558 + 0.558 = 1.12 s.
*December 6, 2013*

**Geometry**

As = 2W*h + 2L*W + 2L*h = 214 m^2. 70 + 14L + 10L = 214 24L = 144 L = 6 m. V = L*W*h = 6 * 7 * 5 = 210 m^3.
*December 6, 2013*

**physics**

A = 6m[30o[ Y = 6*sin30 = 3m = North Component. B = 7m[132o], CCW. Y = 7*sin132 = 5,20 m. = North component. The component along the north direction of person A is smaller.
*December 6, 2013*

**physics**

a = (V-V)/t = (44-0)/0.028 = 1571 m/s^2 F = m*a = 0.49 * 1571 = 770 N.
*December 6, 2013*

**Physics**

See previous post: Thu,12-5-13,3:00 PM.
*December 5, 2013*

**physics**

T= F = m*a = 454 * 3.51 = 1594 N.
*December 5, 2013*

**physics, buoyance**

A. Hb=(Do/Dw)*Ho=(730/1000)*0.2=0.146 m Below the the water. Ha = 0.2 - 0.146 = 0.054 m Above water. B.Mc=V*D = 0.2^3m^3 * 730kg/m^3=5.84 kg. = Mass of cube. (Mc+Mo)/Vc = Dw(Density of water). (5.84+Mo)/0.008 = 1000kg/m^3 5.84 + Mo = 8 Mo = 2.16 kg = Mass of object to be added.
*December 5, 2013*

**Physics**

See previous post: Tue,12-3-13,10:36 PM.
*December 4, 2013*

**physics**

Vc = 46.4mi/3600s * 1600m/mi = 20.62 m/s Circumference=pi*2r=3.14 * 0.568=1.784 m Period =(1.784m/20.62m)*1s. = 0.0865 s.
*December 4, 2013*

**Math**

a. x = 3 x-3 = 0 x = -1 x+1 = 0 x = -5 x+5 = 0 Y = (x-3)(x+1)(x+5) = 0 Multiply the 1st two parenthesis: Y = (x^2-2x-3)(x+5) = 0 Eq: Y = x^3 + 3x^2 -13x - 15 = 0
*December 4, 2013*

**Physics**

See previous post: Tue,12-3-13,10:36 PM.
*December 4, 2013*

**brief calc**

2q/3 = 8-p2q/3 = 8-6 = 2 2q/3 = 2 2q = 6 q = 3
*December 4, 2013*

**brief calc**

q = 130 2*11 = 108.
*December 4, 2013*

**physics**

I=V/(Ro+Ra) = 10/(1k+0.05k) = 9.5238 mA. = Measured current. I=10/1k = 10.00 mA=Theoretical current. Accuracy = ((10.00 - 9.5238)/10)*100%4.762 %.
*December 3, 2013*

**Physics**

At top of hill: KE + PE = mg*h 0 + PE = mg*h PE = mg*h At bottom of hill: KE + PE = mg*h KE + 0 = mg*h KE = mg*h = 0.5m*V^2 0.5m*V^2 = mg*h V^2 = 2g*h V = Sqrt(2g*h). V = Sqrt(19.6*h). Sqrt means Square root.
*December 3, 2013*

**physics**

See previous post.
*December 3, 2013*

**physics**

Fc = m*g = 29kg * 9.8N/kg = 284.2 N. = Force of crate. Fap-Fs = m*a 90-u*Fc = m*0 90 - u*284.2 = 0 u*284.2 = 90 us = 0.317
*December 3, 2013*

**math**

T = (2/5) * 2 = 4/5 or 0.8 h.
*December 3, 2013*

**Physics**

a. a=F/m = 3.34*10^7/2.77*10^6 = 12.06 m/s^2. b. t = 2.5min * 60s/min = 150 s. h = 0.5a(2t-1) h = 6.03*(300-1) = 1803 m.
*December 3, 2013*

**Maths**

100% - 40.25% - 30.76% = 28.99 % Greater than 5.5 Ft. People = 0.2899 * 120M=34.79 M Greater than 5.5 Ft.
*December 3, 2013*

**science**

P = 40.2hp * 746W/hp = 29,989 Watts. = 29,989 Joules/s. T = 5.9*10*5J/(2.9989*10^4J/s=19.67 s.
*December 3, 2013*

**Physics**

B. L = V/F = 343/239.7 1.43 When flying away.
*December 2, 2013*

**Physics**

A. L = V/F = 343/258 = 1.33 m. B. F = ((Vs+Vr)/(Vs-Vg))*Fo = 258 Hz ((343+0)/(343-12.6))*Fo = 258 (343/330.4) * Fo = 258 1.0381Fo = 258 Fo = 248.5 Hz=Initial Freq. of the goose F = ((343-0)/(343+12.6))*248.5 F = (343/355.6) * 248.5 = 239.7 Hz When flying away.
*December 2, 2013*

**Physics**

A. L = V*T =V*(1/F)=V/F L = 343/4.4*10^6 = 7.95*10^-5m. B. L = 1580/4.4*10^6 = 359*10^-6=3.59*10^-4 m.
*December 2, 2013*

**math **

what property is described in this equation: 24x5=(20x5)+(4x5)
*December 2, 2013*

**Physics**

Fs = m*g = 87.1kg * 9.8N/kg = 853.6 N. = Force of snowboarder. Fp = 853.6*sin36.7 = 510.1 N. = Force parallel to hill. Fn = 853.6*cos36.7 = 684.4 N = Normal = Force perpendicular to hill. Fk = u*Fn = 0.106 * 684.4 = 72.55 N. = Force of kinetic friction. a=(Fp-Fk)/m=(510.1-72....
*December 2, 2013*

**Algebra**

x^2 + y^2 + 8x + 2y + 8 = 0 x^2+8x+(8/2)^2+y^2+2y+(2/2)^2=-8+(8/2)^2 +(2/2)^2. x^2+8x+16 + y^2+2y+1 = -8+16+1 Eq: (x+4)^2 + (y+1)^2 = 9 C(-4,-1). r2 = 9 r = +-3
*December 2, 2013*

**Geometry**

See previous post: Mon,12-2-13,7:39 AM.
*December 2, 2013*

**Geometry**

See previous post:Mon,12-2-13,7:39 AM.
*December 2, 2013*

**Geometry**

The diagonals form 2 30o-60o rt. triangles. h = 24*sin30 = 12 In. W = 24*cos30 = 20.8 In. P = 2*(W+h) = 2*(20.8+12) = 65.6 In.
*December 2, 2013*

**physics**

v = Vmax*sinWt v = 100*sin(1000t) Vrms = 0.707 * 100 = 70.7 Volts. W = 2pi*F = 1000 6.28*F = 1000 F = 159 Hz. Xl = W*L = 1000 * 0.5 = 500 Ohms. Xc=1/W*C=1/(1000*4.6*10^-6)=-217.4 Ohms tan A = (Xl+Xc)/R tanA = (500+(-217.4))/400 = 0.7065 A = 35.2o Z = R/cos A = 400/cos35.2 = ...
*December 2, 2013*

**physics**

h = Vo*t + 0.5g*t^2 = 20 37.8t _ 4.9t^2 = 20 -4.9t^2 + 37.8t - 20 = 0 Use Quadratic Formula and get: t = 0.571, and 7.14 s. So, when the ball is rising; it reaches 20 m in 0.571 s. When it is falling, it reaches 20 m in 7.14 s.
*December 1, 2013*

**physics**

See your previous post: Sun,12-1-13,10:34 PM.
*December 1, 2013*

**physics**

See previous post: Fri,11-29-13,8:15 PM.
*December 1, 2013*

**physics. again.**

V== Vo + g*t = 10 m/s. 37.8 + (-9.8)t = 10 -9.8t = 10-37.8 =-27.8 t = 2.84 s. h = Vo*t + 0.5g*t^2 h = 37.8*2.84 - 4.9*2.84^2 = 67.83 m.
*December 1, 2013*

**physics**

Fs = m*g = 22kg * 9.8N/kg = 215.6 N. = Force of the sled. Fp = 215.6*sin6 = 22.54 N. = Force parallel to the incline. Fn = 215.6*cos6 = 214.4 N. = Normal = Force perpendicular to the incline. Fk = u*Fn = 0.2 * 214.4 = 42.88 N. = Force of kinetic friction. V = 64km/h = 64000m/...
*December 1, 2013*

**5th grade math**

I'm glad I could help, and thanks for your response.
*December 1, 2013*

**5th grade math**

4th Grade: 9 Left-handed students. Total = 3 * 9 = 27 Students. 9/27 = 1/3 Are left-handed. 5th Grade: 7 Left-handed students. Total = 30-2 = 28 Sudents. 7/28 = 1/4 Are left-handed. 6th Grade: 6 Left-handed students. Total = 27+3 = 30 Students. 6/30 = 1/5 Are left-handed. The ...
*December 1, 2013*

**Physics**

r2 = r1 + a*(T-To)r1 = 4.42 cm. 4.40 + 1.3*10^-4(T-20)4.40 = 4.42 4.40 + 5.72*10^-4(T-20) = 4.42 5.72*10^-4(T-20) = 4.42-4.40 = 0.02 5.72*10^-4T - 114.4*10^-4 = 0.02 5.72*10^-4T = 0.02 + 0.01144 = 0.03144 T = 314.4*10^-4/5.72*10^-4 = 54.97o
*December 1, 2013*

**Math**

Same procedure as previous post: Sun,12-1-13,7:39 PM.
*December 1, 2013*

**Math**

Given: A(-8,3), B(-6,1). m = -1 A(-8,3), P(x,y). m== -1. m = (y-3)/(x-(-8)) = -1 m = (y-3)/(x+8) = -1 Cross multiply: y-3 = -1(x+8) y-3 = -x-8 Eq: x + y = -5 Or x + y +5 = 0 Replace y with 0: x + 0 = -5 X = -5 = X-intercept. Replace X with 0: 0 + y = -5 Y = -5 = y-int.
*December 1, 2013*

**Physics**

Wr = m*g = 0.5kg * 9.8N/kg = 4.9 N = Wt. of the rat. Fp = 4.9*sin34 = 2.740 N. = Force to the plane. Fn = 4.9*cos34 = 4.062 N. Normal = Force perpendicular to the plane. Fk = u*Fn = 0.2 * 4.062 = 0.8125 N. Fc-Fp-Fk = m*a 0.4-2.74-0.8125 = m*a m*a = -3.15 a = -3.15/m = -3.15/0...
*November 30, 2013*

**Physics**

Fex = 15000 N. ? m = 500kg a = Fex/m = 15000/500 = 30 m/s^2.
*November 30, 2013*

**physics**

Change in L = Lo * a*(T-To) = 65.1 * 10^-5 * (-31.2-15.8) = -0.030597 m.
*November 30, 2013*

**physics**

Fap*cos45-u*mg = m*a Fap*cos45-0.02*588 = m*0 = 0 Fap * cos45 - 11.76 = 0 Fap*cos45 = 11.76 N. = Hor. component of Force applied. Work = 11.76 * 1000. = 11,760 Joules.
*November 29, 2013*

**physics**

a. Xl = 2pi*F*l = 6.28*50*0.405=127.2 Ohms Xc = 1/(6.28*50*4.43*10-6) = -718.9 Ohms tan A = (Xl-Xc)/R tan A = (127.2-718.9)/400 = -1.47925 A = -55.94o Z=R/cosA = 400/cos(-55.94) = 714.2 Ohms[-55.94o]= The Impedance. Vmax = Imax * Z=0.25 * 714.2 =178.6 Volts. b. The negative ...
*November 29, 2013*

**physics**

Fn = 75N[40o] + 85N[340o] X = 75*cos40 + 85*cos340 = 137.3 N. Y = 75*sin40 + 85*sin340 = 19.14 N. tan A = Y/X = 19.14/137.3 = 0.13940 A = 7.94o Fn=X/cosA = 137.3/cos7.94=138.6 N[7.94o]
*November 29, 2013*

**PHYSICS**

a. Xc = 1/2pi*F*c Xc = 1/(6.28*60*4*10^-6) = 663.5 Ohms. I = V/Xc = 120/663.5 = 0.181A rms. b. Same procedure as "a".
*November 29, 2013*

**Physics**

We need to know the length of the hill. Length = 100 m? A = 30o m = 75 kg h = 100*sin30 = 50 m. = Ht. of hill. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*50 = 980 V = 31.3 m/s At bottom of hill. V^2 = Vo^2 + 2a*d a=(V^2-Vo^2)/2d=(31.3^2-0)/200=4.9 m/s^2 V = Vo + a*t = 31.3 0 + 4.9*t = ...
*November 28, 2013*

**math**

CORRECTION: Cost = 2375*0.85*0.80*0.80 = $1292. Discount = 2375 - 1292 = $1083.
*November 27, 2013*

**math**

Discount = 2375*0.15*0.20*0.20 = $14.25
*November 27, 2013*

**Physics**

V1 = 6N[180o] X1 = -6N Y1 = 0 V2 = 4.5N[45o] X2 = 4.5*cos45 = 3.18 N. Y2 = 4.5*sin45 = 3.18 N. X = X1 + X2 = -6 + 3.18 = -2.82 N. Y = Y1 + Y2 = 0 + 3.18 = 3.18 N. tan Ar = Y/X = 3.18/-2.82 = -1.12766 Ar = -48.43o = Reference angle. A = -48.43 + 180 = 131.6o R=Y/sinA = 3.18/...
*November 26, 2013*

**Finance - Annuities**

P = Po*(1+r)^n P = $2250/mo. * 48mo. = $108,000 r = (3%/12)/100% = 0.0025 = Monthly % rate expressed as a decimal. n = 1comp./mo * 48mo = 48 Compounding periods. a. P = (Po-12000)*(1.0025)^48 = 108,000 (Po-12000)*1.12733 = 108000 Po-12000 = 95,801.75 Po = $107,801.75 = Initial...
*November 26, 2013*

**math**

A*(1-0.0653) = 1,225,000 A*0.9347 = 1225000 A = 1,310,580.94
*November 26, 2013*

**physics**

F1 = m*g = 1.6kg * 9.8N/kg = 15.68 N F2 = (0.05m/1m) * 400N = 20 N. F1-Fs = m*a 15.68-Fs = m*0 = 0 Fs = 15.68 = Force of static friction. Fs = u*F2 = 15.68 u*20 = 15.68 u = 0.784 = Coefficient of static friction.
*November 26, 2013*

**Physic: Projectile Motion**

Range = Vo^2*sin(2A)/g = 15 m. Vo^2*sin(40)/9.8 = 15 Vo^2*0.0656 = 15 Vo^2 = 228.7 Vo = 15.12 m/s.
*November 26, 2013*

**Math**

2. A cube has 6 edges. As = 6 * 3^2 = 6 * 9 = 54 Sq. In.
*November 25, 2013*

**finance**

1. P = (Po*r*t)/(1-(1+r)^-t) r = (9%/12)/100% = 0.0075 = Monthly % rate expressed as a decimal. t = 12mo/yr * 5yrs. = 60 Months. Plug th above values into the givenEq and get: P = $12,455.01 I = P-Po 2. Monthly = P/t 3. P = Po + Po*r*t Po = $10,000 r = 0.0075 t = 60 Months ...
*November 25, 2013*

**physics**

Fap-Fk = m*a 30-Fk = 2 * 10 = 20 Fk = 30-20 = 10 N. = Force of kinetic friction.
*November 25, 2013*

**geometry**

See Related Questions: Thu,5-25-11,11:28 AM.
*November 24, 2013*

**Science, Physics, math, calculus**

a. h = 0.5g*t^2 = 0.77 4.9t^2 = 0.77 t^2 = 0.1571 t = 0.396 s. = Fall time. Dx = Xo * t = 5.15 m. Xo * 0.396 = 5.15 Xo = 13 m/s = Break speed. b. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*0.77 = 15.09 V = 3.88 m/s
*November 24, 2013*

**Physics**

h = 0.5g*t^2 = 50 m. 4.9t^2 = 50 t^2 = 10.2 Tf = 3.19 s. = Fall time. d = V*Tf = 100m/s * 3.19s = 319 m.
*November 24, 2013*

**PHYSICS**

CORRECTION: T = 1.15 Milliseconds.
*November 24, 2013*

**PHYSICS**

CORRECTION: d. Vr = I*R = (0.8*0.857)*7 = 4.8 Volts Vi=6-4.8=1.2 Volts across the inductor. V/e^(T/0.714) = 1.2 6/e^(1.40T) = 1.2 e^1.40T = 6/1.2 = 5 1.40T*Ln e = Ln 5 1.40T = 1.609 T = 1.15 s. 7 =
*November 24, 2013*

**PHYSICS**

a. T.C.=L/R=5.0/7=0.714 Milliseconds b. T/TC = 0.25mS/0.714mS = 0.350 Vi = V/e^(T/TC) = 6/e^0.350 = 4.23 Volts after 250 uS. I=Vr/R = (V-Vi)/R = (6-4.23)/7= 0.253A. c. I=Vr/R = (V-Vi)/R = (6-0)/7 = 0.857A d. Vr=I*R = (0.8*0.857)*7 = 4.80 Volts. Vi = 6-4.8 = 1.2 Volts. = ...
*November 24, 2013*

**Math**

Incomplete.
*November 24, 2013*

**physics**

Fr = 975N[0o] + 745N[60o] + 1175N[90o] X=975 + 745*cos60 + 1175*cos90=1347.5 N Y = 745*sin60 + 1175*sin90 = 1820.2 N. tan A = Y/X = 1820.2/1347.5 = 1.31079 A = 53.5o Fr = Y/sin A = 1820.2/sin53.5=2264.3 N. [53.5o].
*November 24, 2013*

**physics**

Ht. = 50 m.? a. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*50 = 980 V = 31.3 m/s. b. V = Vo + g*t = 31.3 m/s. 0 + 9.8t = 31.3 t = 3.19 s. c. m = 2.0Lb * 0.454kg/Lb = 0.908 kg. KE = m*V^2/2 = 0.908*31.3^2/2 = 444.8 J.
*November 24, 2013*

**physics**

Ws = m*g = 77.7kg * 9.8N/kg = 761.5 N = Wt. of skier. Fp = 761.5*sin35.7 = 444.3 N. = Force parallel to slope. Fn = 761.5*cos35.7 = 618.4 = Normal = Force perpendicular to slope. Fk = u*Fn = 0.08 * 618.4 = 49.47 N. = Force of kinetic friction. Fp-Fk = m*a a=(Fp-Fk)/m=(444.3-49...
*November 22, 2013*

**science**

KE = 0.5m*V^2 = 2.5*4^2 = 40 J.
*November 22, 2013*

**Please help!- Physics**

d = 483[180o ] + 347[106o] + 347[106o] d = 483[180 + 694[106o] X = 483*cos180 + 694*cos106 = -674.3 km Y = 483*sin180 + 694*sin106 = 667.1 km tan Ar = Y/X = 667.1/-674.3 = -0.98935 Ar = -44.69o = Reference angle. A = 180-44.69 = 135.3o d = X/cosA = -674.3/cos135.3 = 948.7 km [...
*November 22, 2013*

**Math**

Incomplete.
*November 21, 2013*

**physics**

3.2 cm from equilibrium position: KE + PE = mg*d 0 + PE = mg*d/2 Halfway to the equilibrium position: KE + PE = mg*d mg*d/2 + mg*d/2 = mg*d KE = mg*d/2 = 0.5m*V^2 0.5m*V^2 = mg*d/2 = 3.92*0.032/2 0.2*V^2 = 0.06272 V^2 = 0.3136 V = 0.56 m/s.
*November 21, 2013*

**Physics**

CORRECTION: V = 81.6km/h[74.93o] D = 245 km[74.93o]
*November 20, 2013*

**Physics**

V = 100km/h[90o] + 30km/h[315o] X = 100*cos90 + 30*cos315 = 21.21 km/h Y = 100*sin90 + 30*sin315 = 78.79 km/h tan A = Y/X = 78.79/21.21 = 3.71461 A = 74.93o V=Y/sinA = 78.79/sin74.93=81.6 km/h [78.79] D=V*t = 81.6[78.79] * 3 = 245 km[78.8o]
*November 20, 2013*

**Math**

Parallel lines have equal slopes.
*November 20, 2013*

**Math**

Parallel lines have equal slopes.
*November 20, 2013*

**Math**

Parallel lines have equal slopes.
*November 20, 2013*

**Math**

2x - 3 = 4y Eq1: 2x - 4y = 3 m1 = -A/B = -2/-4 = 1/2 m2 = -(2/1) = -2 Y = mx + b Eq2: Y =-2x - 5/4.
*November 20, 2013*

**Math**

See previous post: Wed,11-20-13,6:18PM.
*November 20, 2013*

**Math**

Eq1: x + 2y = 10. m1 = -A/B = -1/2 The slope of the 2nd line = the negative reciprocal of m1: m2 = -(-2/1) = 2. Y = mx + b Eq2: Y = 2x - 10 or 2x - y = 10
*November 20, 2013*

**physics**

Energy = 120 * 5 *6h /1000=3.6 KWh Used Cost = 3.6kwh * 0.090/kwh = 0.324 or 32.4 Cents.
*November 20, 2013*

**Physics**

L = h/sin A = 2/sin 30 = 4 m. = Length of incline. KE = mg*h-Fk*L 0.5m*V^2 = mg*h-Fk*l 0.2*4.50^2 = 3.92*2 - Fk*4 4.05 = 7.84 - 4Fk 4Fk = 7.84-4.05 = 3.79 Fk = 0.948 N. = Force of kinetic friction.
*November 19, 2013*

**physics**

Fn = m * a =47 * 4.50
*November 19, 2013*

**Physics**

Yes. For a given initial velocity, the larger the angle the greater the vertical component of initial velocity. The increase in vertical velocity causes an increase in the max. ht.
*November 19, 2013*

**Physics - projectile**

V = Vo = 50 m/s.
*November 19, 2013*

**Physics**

b. h = Vo*t + 0.5g*t^2 = 3.05 8t + 4.9t^2 - 3.05 = 0 Use Quadratic Formula and get: t = 0.319 s. c. V^2 = Vo^2 + 2g*h V^2 = 8^2 + 19.6*3.05 = 123.78 V = 11.13 m/s.
*November 19, 2013*

**MATH**

A. Y = ho - 8X ho = The initial ht. B. When he got on the elevator, it was at the initial height(ho) and decreased at a rate of 8 feet per second for 12 seconds. C. Y = ho _ 8x = 240 Ft. ho - 8*12 = 240 ho = 240 + 96 = 336 Ft. above gnd. D. Y = 336 - 8x = 0 336 - 8x = 0 8x = ...
*November 19, 2013*

**math**

N = 0.57 * 11
*November 19, 2013*

**Physics**

d = 0.5g*t^2 = 75 4.9t^2 = 75 t^2 = 15.31 Tf = 3.912 s. = Fall time. d = V*Tf = 7-4 = 3 m. V * 3.912 = 3 V = 0.769 m/s.
*November 19, 2013*

**math**

Did your book give the total # of Robinsons? Something is missing.
*November 18, 2013*

**physical science**

P = F * d/t = 96 * 2.8/4.6 = 58.43 J/s = 58.43 Watts.
*November 18, 2013*

**college algebra**

4^(x-3) = 3^(2x) (x-3)*Log4 = 2x*Log 3 Divide both sides by Log 4: x-3 = 2x * 0.79248 = 1.5850x x - 1.5850x = 3 -0.5850x = 3 X = -5.1282
*November 18, 2013*

**science**

Magnitude = 210 N. Direction = 30o South of West = 210o CCW.
*November 18, 2013*

**science**

Magnitude = 200 N. Direction = 60o North of East = 60o CCW.
*November 18, 2013*

**science**

Magnitude = 125 N. Direction = North(Positive Y-axis).
*November 18, 2013*

**math**

Sally has X postcards. Marta has X postcards. X = 4(X-18) Solve for X.
*November 18, 2013*

**Math**

Y = 5 For all values of X. (0,5),(5,5),(2,5),(-3,5).
*November 18, 2013*

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