Monday

May 4, 2015

May 4, 2015

Total # Posts: 10,530

**pre cal**

V(t) = Po - Po*r*t V(t) = 25,000 - 25000*0.05*t V(t) = 25000(1-0.05t)
*September 7, 2014*

**physics**

a = F/m = 390N[90o]/220 = 1.77m/s^2[90o] =1.77m/s^2, North.
*September 7, 2014*

**Math**

Change = -17 - (-10) = -17 + 10 = -7o C.
*September 6, 2014*

**Math**

Change = -17-(-10) = -17 + 10 = -7o C.
*September 6, 2014*

**physics**

a. V = a*t = 1.4 * 14 = 19.6 m/s. d = 0.5a*t^2 + V*t2 + (Vf-V)/2a b. d = 0.5*1.4*14^2 + 19.6*23 + (0-(19.6^2)/-2.6 = 137.2 + 450.8 + 147.8 = 736 m.
*September 6, 2014*

**physiscs**

Correction: Mass = 150kg/6 = 25kg on the moon.
*September 6, 2014*

**physiscs**

g = 9.8 on earth = 9.8/6 on the moon. Mass = 150g/6 = 25 Grams.
*September 6, 2014*

**physical science**

F2*d2 = F1*d1 F2*0.2 = 60 * 1.2 0.2F2 = 72 F2 = 360 N.
*September 6, 2014*

**(Please Help!) Analytic Geometry**

See previous post: Thu, 9-4-14, 6:20 AM
*September 6, 2014*

**physics**

Do you mean 2 meters away? If so, Dx = Vo^2*sin(2A)/g = 2 m. 6.8^2*sin(2A)/9.8 = 2. 46.24*sin(2A) = 19.6 sin(2A) = 0.42388 2A = 25.1o A = 12.54o
*September 6, 2014*

**Physics**

D = 745.6[298o] + 178[204o] + 490[55o] X=745.6*cos298 + 178*cos204 + 490*cos55= 468.5 Paces. Y=745.6*sin298 + 178*sin204 + 490*sin55= -329.3 Paces. Tan A = Y/X = -329.3/468.5 = -70291 A = -35.1o = 324.9o CCW = Direction. Displacement=X/Cos A = 468.5/Cos 324.9 = 573 Paces.
*September 5, 2014*

**physics**

D = 5.54km[47.9o] - 1.29km[24.60] X = 5.54*cos47.9 - 1.29*cos24.6=2.54 km. Y = 5.54*sin47.9 - 1.29*sin24.6=3.57 km. Tan A = Y/X = 3.57/2.54 = 1.40691 A = 54.6o Displacement = Y/sin A = 3.57/sin54.6 = 4.38km[54.6o]. a. Magnitude = 4.38 km. b. Direction = 54.6o N of E.
*September 5, 2014*

**AP Physics**

Since both vehicles are travelling at the same speed, they will meet at the midpoint. d = V*t = 7km/2 = 3.5 km. t = d/V = 3.5/85 = 0.04117647 h = 2.47 min.
*September 5, 2014*

**Business Math**

Correction: Cost = 40 * 3.20 = $128 A = 128 + 0.8*128 = $230.40 = Amt. to be received. (40-8)dinners * $D/dinner = 230.40 32D = 230.40 D = $7.20/dinner. (7.20-1.00)/7.20 = 0.861 = 86.1% % Markdown = 100 - 86.1 = 13.9
*September 5, 2014*

**Business Math**

Cost = 40*3.20 + 0.2*40*3.26 = $153.60 P=153.6 + 0.8*153.6 = $276.48 = Selling price for 40 dinners. 276.48/40 = $6.91 Each. (6.91-1.00)/6.91 = 85.5% % Markdown = 100%-85.5% = 14.5%
*September 5, 2014*

**Business Math**

P = Po - 0.1Po = 259.92 0.9Po = 259.92 Po = 288.80 Before 10% markdown. P = Po + 0.3Po = 288.80 1.3Po = 288.80 Po = $222.15 = Cost to Dylan.
*September 5, 2014*

**Business Math**

Correction: Cost = 600Lbs * 5.59/Lb. = $3354. A = 3354 + 0.6*3354 = $5366.40 = Amt. to be received for 60% markup. A = (600-60)Lbs. * $D/Lb. = 5366.40 540D = 5366.40 D = $9.94/Lb. to allow for 10% spoilage.
*September 5, 2014*

**Business Math**

Cost=600Lbs * 5.59/Lb. + 60Lbs*5.59/Lb. = $3689.40 P = 3689.40 + 0.6*3689.40 = $5903.04 = Selling Price for 600 Lbs. $5903.04/600Lbs. = $9.84/Lb.
*September 5, 2014*

**Math**

P = 5.12 + 0.55*5.12 = $7.94 /yd^2
*September 5, 2014*

**Physics**

a. d^2=X^2 + Y^2=1030^2 + (1950-1480)^2 = 1,281,800 d = 1132 m. b. Tan A = Y/X = (1950-1480)/1030 = 0.45631 A = 24.5o S. of E.
*September 5, 2014*

**Math**

In a close race for the batting title, the more decimal places you have, the easier it is to select the winner. For example, if two players batted 0.301 and 0.304 , respectively, 0ne or two decimal places would be a tie.
*September 4, 2014*

**Physics**

a. T = d1/V1 + d2/V2 = 45/20 + 10/15 = 2.92 Hours. b. T1 = d1/V1 = 45/20 = 2.25 Hours. T2 = d2/V2 = 25/15 = 1.67 Hours. Vavg=(d1+d2)/(T1+T2)=(45+25)/(2.25+1.67) = 17.86 km/h.
*September 4, 2014*

**Physics**

Incomplete.
*September 4, 2014*

**Electronis**

Explain how the resistors are connected in the circuit.
*September 4, 2014*

**physics**

Vo = 90km/h = 90000m/3600s. = 25 m/s. V^2 = Vo^2 + 2a*d. Solve for a. a = (V^2-Vo^2)/2d = (0-(25^2)/200 = -3.125 m/s^2. t = 0.3 + (-Vo/a) = 0.3 + (-25/-3.125) = 8.3 s.
*September 4, 2014*

**Please Help ! Analytic Geometry**

x+1 = 0 X = -1 = h = axis. F(-1,Y2) V(-1,k) D(-1,1) 1/a = 4 = Latus rectum a = 1/4 1/4a = 1/1 = 1 k = y+1/4a = 1 + 1 = 2. Y2 = k+1/4a = 2 + 1 = 3. Y = a(x-h)^2 + k Y = 1/4(x+1)^2 + 2 Use the following points for graphing: (x,y) (-3,3) (-2,9/4) V(-1,2) (0,9/4) (1,3)
*September 4, 2014*

**Physics**

The average velocity = The total distance traveled divided by the total time required to travel that distance: Vavg = (1.1-0.1)/0.76 = 1.32 m/s.
*September 3, 2014*

**Physics**

Vp + 28[0o] = 770[90-48] = 770[42o]CCW Vp + 28 = 770*cos42 + i770*sin42 Vp + 28 = 572.2 + 515.2i Vp = 544.2 + 515.2i = 749mi/h[43.6o] CCW = 749mi/h[46.4o] E of N.
*September 3, 2014*

**math**

200 * 1.2 = 240 mg.
*September 3, 2014*

**mathematics**

8/8-5/8 = 3/8 Boys. 3x/8 = 135 Multiply both sides by 8/3: X = 8/3 * 135 = 360 Children. 9
*September 3, 2014*

**Physics**

A.V=Vo + a*t = 0 + 2.3m/s^2*18s=41.4 m/s B. d = d1 + d2 d = 0.5a*t^2 + 301 = 0.5*2.3*18^2 + 301 = 673.6 m. C. V^2 = Vo^2 + 2a*d a = (V^2-Vo^2)/2d a = (15^2-41.4^2)/602 = -2.47 m/s^2 = deceleration. t2 = (V-Vo)/a = (15-41.4)/-2.47 = 10.7 s t1+t2 = 18 + 10.7 = 28.7 s. for the ...
*September 2, 2014*

**Algebra 2**

1. (2s-t^2)/(st^2) = (2*5-3^2)/(5*3^2)= 1/45. 2. -x-3y+4x-9y+2 = 3x - 12y + 2. 3. -4(-a+2b) - 3(a-5b) = 4a-8b-3a+15b = a + 7b. 4. 5-(3m+2n) = 5-3m-2n.
*September 2, 2014*

**Grade 11 Physics**

See previous post: Tue. 9-2-14, 6:17 PM.
*September 2, 2014*

**algebra 2**

8|x+3/4| < 2 8(x+3/4) < 2 8x+6 < 2 8x < 2-6 8x < -4 X < -1/2. 8|-x-3/4| < 2 8(-x-3/4) < 2 -8x-6 < 2 -8x < 2+6 -8x < 8 X > -1 Solution Set: -1< X < -1/2.
*September 2, 2014*

**gr 11 physics**

Amy's Speed = 14400m/2400s = 6 m/s. Bill's Speed = 14400m/3600s = 4 m/s. So Amy's speed is 2 m/s faster.
*September 2, 2014*

**Physics**

Vo = 29 m/s. V = -18 m/s.(opposite direction of Vo). t = 4 mS = 0.004 s. a = (V-Vo)/t = (-18-29)/0.004=-11,750 m/s^2.
*September 2, 2014*

**What is P?**

P = Po*(1+r)^n = $8600 r = 5%/100% = 0.05 n = 1comp./yr. * 3yrs. = 3 Compounding periods. Po*(1.05)^3 = 8600 Po = 8600/1.05^3 = $7429.00 To be set aside today.
*September 2, 2014*

**Math**

Incomplete.
*September 2, 2014*

**Physics**

h = 0.5g*t^2 = 7.8 m. 4.9*t^2 = 7.8 t^2 = 1.59 Tf = 1.26 s. = Fall time for each boat. d1 = V1*t = 22m/s * 1.26s. = 27.76 m. d2 = V2*t = 11m/s * 1.26s. = 13.88 m. d1-d2 = 27.76 - 13.88 = 13.72 m. Apart.
*September 2, 2014*

**finance**

P = (Po*r*t)/(1-(1+r)^-t) Po = $270,000 r = (12%/12)/100% = 0.01 = Monthly % rate expressed as a decimal. t = 25yrs. * 12mo/yr. = 300 Months. P = (270,000*0.01*300)/(1-1.01^-300) = $853,111.56 Monthly Payments = P/t
*September 2, 2014*

**math**

a. N = (17,568*100) + 24,574 = b. N = (15,615*100) - 17,615,638 = -
*September 2, 2014*

**physics**

Incomplete.
*August 31, 2014*

**phsycis**

F1 = 27N[20o] = 27*cos20 + i27*sin20 = 25.37 + 9.23i. F2 = -27N[20o] = -27*cos20 - 27*sin20 = -25.37 - -9.23i Tan Ar = Y/X = -9.23/-25.37 = 0.36382 Ar = 20o = Reference Angle. Since X and Y are both negative, we are in the 3rd Quad. A = 20 + 180 = 200o = The direction of F2.
*August 31, 2014*

**physics**

Correction: 1st Ball: d1 = Vo*t + 0.5g*t^2 d1 = 0 + 4.9*2^2 = 19.6 m. in 2 s. V = Vo + g*t = 0 + 9.8*2 = 19.6 m/s. 2nd Ball: d2 = d + 19.6 m. Vo*t + 0.5g*t^2 = V*t + 0.5g*t^2 + 19.6 Vo*t - V*t + 0.5g*t^2-0.5g*t^2 = 19.6 Vo*t - V*t = 19.6 30t- 19.6t = 19.6 10.4t = 19.6 t = 1.88...
*August 31, 2014*

**physics**

1st Ball: d1 = Vo*t + 0.5g*t^2 d1 = 0 + 4.9*2^2 = 19.6 m in 2 seconds 2nd Ball: d2 = d1 + 19.6 m Vo*t + 0.5g*t^2 = 4.9t^2 + 19.6 30t + 4.9t^2 - 4.9t^2 = 19.6 30t = 19.6 t = 0.6533 s. To overtake the 1st ball.
*August 31, 2014*

**Physics**

C = 8,420.10/2.9 =
*August 31, 2014*

**business**

P = Po*(1+r)n Po = $1,000 r = (16%/2)/100% = 0.08 = Semi-annual % rate. n = 10yrs. * 2comp./yr = 20 Compounding periods. P = 1000(1.08)^20 = $4660.96 I = P-Po
*August 30, 2014*

**Physical Science**

Wood will float in a liquid if it's density is less than the density of the liquid.
*August 30, 2014*

**physics**

1uA = 1*10^-6A = 1*10^-3 mA.
*August 29, 2014*

**Physical Science**

Wt. = m*g = 4.82kg * 9.8N/kg = 47.24 N. a = F/m = 10/4.82
*August 29, 2014*

**Physics**

a. Tr = -Vo/g = -40/9.8 = 4.08 s. = Rise time. b. h = ho + (V^2-Vo^2)/2g h = 80 + (0-(40^2))/-19.6 = 161.6 m. Above gnd. c. V = Vo + g*t = 40 - 9.8*2 = 20.4 m/s. d. V*2 = Vo^2 + 2g*(h-20) V^2 = 0 + 19.6*(161.6-20) = 2775.36 V = 52.7 m/s. e. h = 4.9t^2 = 161.6 m. 4.9*t^2 = 161....
*August 27, 2014*

**Physics**

a. Correct. b. Correct. c. Correct. d. h = ho-(V^2-Vo^2)/2g h = 80-(30^2-4^2)/19.6 = 34.9 m Or V = Vo + gt = 30 m/s. V = 4 + 9.8t = 30 9.8t = 26 t = 2.653 s. h = ho - (Vo*t + 0.5g*t^2) h = 80-4*2.653 + 4.9*2.653^2 = 34.9 m
*August 27, 2014*

**Matt**

X = The # of boys. X+250 = The # of girls x/(X+250) = 2/7 7x = 2x + 500 5x = 500 X = 100 Boys. X+250 = 100+250 = 350 Girls.
*August 27, 2014*

**math**

(3x-2)^2(3x-2)-(4x+5)^2 = (9x^2-12x+4)(3x-2)-(16x^2+40x+25) = 27x^3-36x^2+12x-18x^2+24x-8-16x^2-40x-25 = 27x^3-70x^2-4x-33.
*August 27, 2014*

**math**

The cable is the hyp. of a rt. triangle, and the ht. of the balloon is the ver. side. h = 842*sin65 = 763.1 Ft. = Ht. of the balloon. Tan4 = h/BC BC = h/Tan4 = 763.1/Tan4 = 10,913 Ft = Distance from the target to point C.
*August 27, 2014*

**algebra**

Investment @ 7% = Po Investment @ 6% = 32,000-Po Po*r*t + (32000-Po)*r*t = 2,050 Po*0.07*1 + (32000-Po)*0.06*1 = 2050 0.07Po + 1920 - 0.06Po = 2050 0.01Po = 2050-1920 = 130 Po = $13000 @ 7%. 32,000-13000 = $19,000 @ 6%.
*August 27, 2014*

**math**

Correct!
*August 27, 2014*

**Physics**

See previous post: Tue,8-26-14,10:29 PM.
*August 27, 2014*

**Math**

610EUR/800AUD * 1USD/0.753EUR = 1.0126 USD/1AUD Note: EUR in the Equation cancels.
*August 27, 2014*

**Physics**

V = 34[20o] + 28[53.1o] X = 34*cos20 + 28*cos53.1 = 48.75 Units. Y = 34*sin20 + 28*sin53.1 = 34.03 Units. V^2=X^2 + Y^2 = 48.75^2 + 34.03^2=3535. V = 59.5 Units.
*August 26, 2014*

**math**

20,000,000+484,000+100+60+3.
*August 26, 2014*

**Math**

2,5. LCM = 10. 5,10. LCM = 10.
*August 26, 2014*

**math**

(2-x)/8 = 3/4 Multiply both sides by 8: 2-x = 6 -x = 6-2 = 4 X = -4. Or 2 - x/8 = 3/4 Multiply both sides by 8: 16-x = 6 -x = 6-16 = -10 X = 10.
*August 26, 2014*

**Math**

Kendall: X in. tall. Rachel: 2x in. tall. Rich: (2x+3) in tall. Brian: (2x+3)-1 = 74 in. tall. (2x+3)-1 = 74 2x + 2 = 74 2x = 72 X = 36 in. = Kendall's ht.
*August 26, 2014*

**science**

100.0g of 4.0c water is heated until its temperature is 37c. if the specific heat of water is 4.18 j/g c, calculate the amount of heat energy needed to cause this rise in temperature.
*August 25, 2014*

**Physics**

Range = 15,000 Meters? Range = Vo^2*sin(2A)/g = 15,000 m. Vo^2*sin(60)/9.8 = 15000 0.08837Vo^2 = 15000 Vo^2 = 169,741 Vo = 412 m/s.
*August 25, 2014*

**physics**

Work = Fx*d = 180*cos20 * 5 = 845.7 J. I don't understand what you are saying in the last paragraph.
*August 25, 2014*

**Math**

0.06C = 0.30 C = $5.00
*August 25, 2014*

**math**

a. 120I = 100 W. I = 0.833A R = V/I = 120/0.833 = 144 Ohms. b. Same procedure as a. c. Same procedure as a. a. Imax = V/(72+0) = 115/(72+0) = 1.60A Imin = 115/(72+88) = 0.719A b. P = V^2/(Rc+R) = 90 W. 115^2/(72+R) = 90 90*(72+R) = 115^2 72+R = 146.94 R = 146.94-72 = 74.94 ...
*August 24, 2014*

**Algebra 2**

T1*T2/(T1+T2) = 3*2/(3+2) = 1.2 h/Lawn. T = 1.2h/Lawn * 12Lawns = 14.4 Hours Note: 1.2h = 6/5 h = Time to do 1 Lawn.
*August 24, 2014*

**physics**

X = 200 km. Y = -600 km. d^2=X^2 + Y^2=200^2 + (-600)^2=400,000 d = 632 km.
*August 24, 2014*

**math**

Undefined.
*August 24, 2014*

**Algebra**

(6-3w)(-W^2). We have an algebraic expression that's already factored. So what is the question? Did you mean (6-3W)(-W^2)=0 ? If so, 6-3W = 0 -3W = -6 W = 2 -W^2 = 0 W^2 = 0 W = 0 Solution Set: W = 0, and 2.
*August 24, 2014*

**algebra**

Eq1: 5r-9s = -76 Eq2: 9r+5s = 54 Multiply Eq1 by -9 and Eq2 by 5, then add: -45r + 81s = 684 +45r + 25s = 270 Sum: 106s = 954 s = 9 In Eq1, replace s with 9: 5r - 9*9 = -76 5r = 81-76 = 5 r = 1.
*August 24, 2014*

**Math**

Incomplete.
*August 24, 2014*

**Geometry**

L = 0.5W W = 2L A = L*W = 25 L * 2L = 25 3L = 25 L = 25/3 = 8.333 Sq. units. W = 2L = 2 * 8.333 = 16.666 Sq. units. P = 2W + 2L
*August 24, 2014*

**math**

Do you mean 18 is added? If so: 17,26,35. 35 + 18 = 53 So the number is 35.
*August 24, 2014*

**math**

1. X = +- Infinity. 2. X < 7.
*August 24, 2014*

**College Algebra 1**

4x^2 = 16 x^2 = 4 x^2-4 = 0. The difference of 2 squares: (x+2)(x-2) = 0 x+2 = 0 X = -2. x-2 = 0 X = 2.
*August 23, 2014*

**Science**

d = 0.5g*t^2 = ho-81 4.9*0.25^2 = ho-81 0.306 = ho-81 ho = 0.306 + 81 = 81.31 m. = Initial ht. of the balloon. V^2 = Vo^2 + 2g*ho V^2 = 0 + 19.6*81.31 = 1593.7 V = 39.9 m/s. = Final velocity.
*August 22, 2014*

**Physics**

m*g = 20kg * 9.8N/kg = 196 N = Wt. of the wagon. Fap*Cos30 - Fk = m*a Fap*cos30-30 = m*0 = 0 Fap*cos30 = 30 Fap = 30/cos30o = 34.6 N. = Force applied.
*August 22, 2014*

**science**

Density = 50g/8cm^3 = 6.25 g/cm^3
*August 22, 2014*

**math**

11,13,15,17,19,21,23. Select the number whose digits add up to 4: Answer = 13.
*August 21, 2014*

**physics**

Grade = 100%*sin A = 37.7% sin A = 37.7%/100% = 0.377 A = 22.1o To climb a hill, the angle of inclination must be less than 90o.
*August 21, 2014*

**physics**

X = 47 km. Y = 69 km. a. d^2 = X^2+Y^2 = 47^2 + 69^2 = 6970 d = 83.49 km. b. Tan A = Y/X = 69/47 = 1.46809 A = 55.74o W of S = 34.26o S of W.
*August 21, 2014*

**physics**

Tan30 = (49-14)/(X1+X2) Tan30 = 35/(19+X2) (19+X2)Tan30 = 35 (19+X2) = 35/Tan30 = 60.62 X2=60.62-19 = 41.62 Ft. from the rocks.
*August 21, 2014*

**physics**

Density = m/V = 0.70kg/(0.0642m)^3 = 2645 kg/m^3.
*August 21, 2014*

**math**

Specific Gravity
*August 21, 2014*

**Math**

N = The number. N + 0.2N = 28 1.2N = 28 N = 28/1.2 = 28/(1 1/5) = 28/(6/5) = 28 * 5/6 = 140/6 = 70/3.
*August 21, 2014*

**Algebra**

Velocity = 4.5 Ft/s. Downward. Speed = 4.5 Ft/s.
*August 21, 2014*

**physics**

distance = 4 + 2 + 4 + 2 = 12 m. Displacement = Df-Do = 0-0 = 0. Df = Final distance. Do = Starting point.
*August 21, 2014*

**physics**

Distance = 5 + 3 = 8 km. Displacement = 5-3 = 2 km.
*August 21, 2014*

**physics**

Vr = 186m[23o] + 327m[133o] X = 186*cos23 + 327*cos133 = -51.8 m/s. Y = 186*sin23 + 327*sin133 = 311.8 m/s. Tan Ar = Y/X = 311.8/-51.8 = -6.01986 Ar = -80.6o = Reference angle. A = -80.6 + 180 = 99.4o CCW = 9.4o W of N. Vr = Y/sin A = 311.8/sin99.4)=316 m/s [99.4o] = 316 m/s[9...
*August 21, 2014*

**physics**

Vs = 3.12 - 1.35 = 1.77 m/s.
*August 21, 2014*

**Physics**

Vo = 25m/s[30o] Vx = 25*cos30 = 21.65 m/s. Vy = 25*sin30 = 12.50 m/s. h=(V^2-Vy^2)/2g=(0-(12.5^2)/-19.6=7.97 m Range = Vo^2*sin(2A)/g Range = 25^2*sin(60)/9.8 = 55.23 m. V=Vy + g*t = 12.50 + (-9.8*2) = -7.1m/s 21.65 - 7.1i=22.78m/s[-18.16o]=Velocity 2 seconds after the launch...
*August 19, 2014*

**physics**

Cannot solve without graphs.
*August 19, 2014*

**Physics**

a. X = 30km/h Y = 40km/h Tan A = Y/X = 40/30 = 1.33333 A = 53.13o E. of N. if wind is coming from the West. = 53.13o W. of N. if wind is coming from the East. V = Y/sin A=40/sin53.13=50 km/h[53.13o]
*August 19, 2014*

**math**

Thanks! I thought it was 30o from the x axis, because the problem didn't say 30o East of North.
*August 18, 2014*

**math**

D = 90km[30o] + 150km[0o] X = 90*Cos30 + 150 = 228 km. Y = 90*sin30 = 45 km. Tan A = Y/X = 45/228 = 0.19742 A = 11.17o D=X/CosA = 228/Cos11.17=232 km[11.17o] Dx = 232*cos11.17 = 227.998 km East of starting point.
*August 18, 2014*

**math**

5*15 + 2 = 77 6*15 + 2 = 92
*August 18, 2014*