Wednesday

February 10, 2016
Total # Posts: 11,494

**Calculus 2**

Sketch the region in the first quadrant enclosed by y=7/x, y=2x, and y=x/2. Decide whether to integrate with respect to x or y. Then find the area of the region.
*February 9, 2016*

**physics**

a. Vr = -100i + 70km[330o] = -100i + 60.6-35i = 60.6 - 135i = 148km[65.8o] S. of E = 24.2o E. of S. = Resultant velocity of the goose. Do you mean 20o W. of S.? If so, Direction = 24.2 + 20 = 44.2o W. of S. b. d = V*t = 1000, t = 1000/V = 1000/148 = 6.76 h.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 5000*0.04*2 = $400. (25/4)% = 6.25%. I = 5000*0.0625*2 = $625. Gain = (625-400)/2yrs. = $112.50 per year.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 0.125Po. Po*0.1*t = 0.125Po. t = 1.25 yrs. = 15 Months.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 1344-1200. Po = 1200, r = 0.06, t = ?.
*February 8, 2016*

**SIMPLE INTEREST**

I = Po*r*t = 1770. r = 0.08 t = (15/2)yrs. = 7.5 yrs. Po = ?.
*February 8, 2016*

**SIMPLE INTEREST**

There are errors in your problem; please make corrections.
*February 8, 2016*

**science**

Sketch: From a point, draw 2 arrows pointing upward and one arrow pointing downward.
*February 8, 2016*

**science**

Fr = 5.3 + 2.2 - 10.7 = -3.2 N. = 3.2 N. downward.
*February 8, 2016*

**physics**

h = V*t + 0.5g*t^2 = 2 m. V*0.1 + 4.9*0.1^2 = 2. 0.1V = 2-0.049 = 1.95. V = 19.5 m/s at top of the window. V^2 = Vo^2 + 2g*h. V = 19.5 m/s, Vo = 0, g = 9.8 m/s^2, h = ?.
*February 7, 2016*

**Finance**

P = Po(1+r)^n. P = $16,860. r = 5.75%/100 = 0.0575 n = 1Comp./yr. * 6yrs. = 6 Compounding periods. Po = ?.
*February 7, 2016*

**science**

a. h = 0.5g*t^2. h = 20 m., g = 9.8 m/s^2. t = ?. b. d = 10m/s * t. t = Value calculated in part "a".
*February 7, 2016*

**physics**

a. V^2 = Vo^2 + 2g*h = 0. Vo^2 = -2g*h = -(-19.6)*0.18 = 3.53, Vo = 1.88 m/s. b. KE = 0.5M*Vo^2 = 0.5*0.54*10^-6*1.88^2 = 9.54*10^-7 Joules.
*February 7, 2016*

**Physics**

Xo = 2.9 m/s. Tan 43.6 = Y/Xo = Y/2.9. Y = 2.9*Tan 43.6 = 2.76 m/s. V^2 = Vo^2 + 2g*d = 2.76^2. 0 + 19.6d = 7.63, d = 0.389 m. h = 1.14 - 0.389 = 0.751 m. Above the floor.
*February 6, 2016*

**physics**

See previous post: Thu,2-4-16, 11:20 AM.
*February 5, 2016*

**Physics**

Oops! The name is Henry.
*February 4, 2016*

**Math**

A semicircular shape is in a (fills it) rectangle. If the perimeter of the rectangle is 30 cm, fine the perimeter of the semicircular shape.
*January 11, 2016*

**Calculus**

A water tank is made by rotating f(x)=2^x-1 between [0,2] about the y-axis. The water tank is initially full, when a hole is opened at the bottom tip so that the water drains at a rate of 4 unit^3/second. How fast is the height of the water decreasing when the height has ...
*January 8, 2016*

**Calculus**

Okay I didn't know you could do that haha
*January 2, 2016*

**Calculus**

Wait, but how can you let u=x^3 when there is not an x^3 in the equation
*January 2, 2016*

**Calculus**

Last one for me. Evaluate (3x^2)/sqrt(1-x^6) dx I changed sqrt (1-x^6) into 1^(1/2)- x^(3) and let u=1^(1/2)- x^(3) then I got du/dx=3x^2 and subbed dx=du/3x^2 into the equation and got x-(x^4/4) but its not right. Options 7x^3/(3sqrt(1-x^7) + C cos-1(3x) + C sin-1(x^3) + C ...
*January 2, 2016*

**Calculus**

Im think im going to go with B then
*January 2, 2016*

**Calculus**

Okay Thanks. That one confused me.
*January 2, 2016*

**Calculus**

Okay will do. So I'm presuming your saying the answer is B. I think D works.
*January 2, 2016*

**Calculus**

Wait so A is right or is it wrong?
*January 2, 2016*

**Calculus**

Which of the following integrals cannot be evaluated using a simple substitution? I think it is A because if you would substitute there would be nothing left in the equation? Is that right? Options ∫√(x-1) ∫1/√(1-x^2) ∫x/√(1-x^2) ∫&#...
*January 2, 2016*

**Calculus**

Oh shoot I misread my answer. Thanks
*January 2, 2016*

**Calculus**

∫(x^3-x^2)/x^2 I got (x^2)/(2)-x + C but that's not one of the answers Options x - 1 + C (x^2/2)-(x^3/3) + C (x^4-x^3)/4x^2 + C (x^2/2) -x + C
*January 2, 2016*

**Calculus**

Yes Thank you
*January 2, 2016*

**Calculus**

∫((cos^3(x)/(1-sin^(2)) What is the derivative of that integral? I have been trying to use trig identities but can't find one to simplify this equation. I can't find one for (cos^3(x) or (1-sin^(2)) My options -sin(x) + C sin(x) + C (1/4)cos^(4)(x) + C None of ...
*January 2, 2016*

**Calculus**

I tried to use partial fractions to do it, but it didn't work.
*December 27, 2015*

**Calculus**

Find the integral of 2/((2-x)(x+2)^2)dx.
*December 27, 2015*

**Calculus**

Solve the following problem by separation of variables. T^2y'-t=ty+y+1 Y(1)=0 Y is a function of t.
*December 25, 2015*

**Calculus**

Wait, why is it the integral from 3-5 instead of 0-2?
*December 23, 2015*

**Calculus**

Sorry I forgot the dx after the integral.
*December 23, 2015*

**Calculus**

Find the integral from 0 to 2 of xsqrt(5-sqrt4-x^2)). The hint said to use substitution of u=sqrt(4-x^2), and that I needed one more substitution, but I don't know how to do it.
*December 23, 2015*

**Math**

(Y+1)+4
*December 16, 2015*

**Mathematics**

Show that the root equation (x-a)(x-b)=k^2 are always real numbers
*December 16, 2015*

**Mathematics**

Factorice 16(a-2b)^2-(a+b)^2 leaving your answer in the simplest form
*December 16, 2015*

**math-unit rates**

Jeff hikes 1/4 mile every 2/5 hour. John hikes 1/2 mile every 1/7 hour. A) Who hikes farther in one hour. Show work. B) Who walks faster, Jeff or John. Explain your reasoning
*November 28, 2015*

**science**

I did a physics experiment and now i need help with the matlab fit. can someone tell me how to do this fit in matlab? a1*[sin(a2(x-a3))/(a2(x-a3)]^2*[cos(a4(x-a3))]^2 i have a program that fits a graph to a function i just ask how to write this function.
*November 27, 2015*

**Calculus**

Find the Absolute Maximum and Absolute Minimum of f on (0,3]. f(x)=(x^3-4x^2+7x)/x Multiple choice question I know the minimum is (2,3) but the maximum is either nothing or (0,7) but I can't tell which one
*November 7, 2015*

**math**

1. 6x4 in., P = 2*6 + 2*4 = 20 in. 2. 6*2 in., P = 2*6 + 2*2 = 16 in. 3. 4x2 in., P = 2*4 + 2*2 = 12 in. =
*November 5, 2015*

**Physics**

Units of time are seconds, minutes, and hours. Vf = Vo + a*t = 0. t = -Vo/a = -22/-6.1 = 3.61 s.
*November 5, 2015*

**Precalc**

a. P = Po(1+r)^n Po = $60,000. r = (4%/4)/100% = 0.01 = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 10yrs. = 40 Compounding periods. P = ? c. Same procedure as a.
*November 5, 2015*

**physics **

Vf = Vo + a*t = 3 + 1*5 = 8 m/s. d = Vo*t + 0.5a*t^2.
*November 5, 2015*

**universitu**

See Related Questions: Sun, 2-17-13, 5:09 AM.
*November 5, 2015*

**Calculus**

use tangent line approximation (linear approximation) to estimate The cube root of 1234 to 3 decimal places. Hint: the equation should be y=f'(x0)(x-x0)+f(x0) 11^3=1331 can be easily computed using binomial theorem. I used linear approximation and got 10.733, but it is not...
*November 4, 2015*

**Physics**

a. 0.5g*t^2 = 4780. 4.9t^2 = 4780. t^2 = 975.5 t = 31.2 s. Dx = 175m/s * 31.2s = 5465.8 m.
*November 3, 2015*

**math**

2. X = -30 km/h = velocity of the wind. Y = -295 km/h.=velocity of the aircraft. Q3. tan A = y/x = -295/-30 = 9.83333. A = 84.2o S. of W. = direction. Vr = y/sin A=-295/-sin84.2 = 296.5 km/h.
*November 3, 2015*

**physics**

Vf^2 = Vo^2 + 2g*h. g = 9.8 m/s^2.
*November 3, 2015*

**Science**

10cm3 * 10.5g/cm^3 = wt. in grams. mass = g/1000 = kilograms.
*November 2, 2015*

**physics**

140rev/60s * 20s =
*November 2, 2015*

**science**

T1*Cos26.5 - T2*Cos49.5 = 0 T1*Cos26.5 = T2*co49.5. T1 = 0.73T2. T1*sin26.5 + T2*sin49.5 = 155. Replace T1 with 0.73T2: 0.73T2*sin26.5 + T2*sin49.5 = 155. 0.324T2 + 0.760T2 = 155. 1.084T2 = 155. T2 = 143 N. T1 = 0.73*143 = 104 N.
*November 1, 2015*

**Physics**

Work = F*d = 23*Cos39 * 8 =
*November 1, 2015*

**physics**

Fs = M*g = 14 * 9.8 = 137 N. Fp = 137*sin 0 = 0 = Force parallel to the gnd. Fn = 137*Cos 0 - 25*sin30 = 124.5 N. = Normal Force. Fk = u*Fn = Force of kinetic friction. a = (25*Cos30-Fp-Fk)/M. Vf^2 = Vo^2 + 2a*d Vo = 1 m/s. d = 18 m. Vf = ?
*November 1, 2015*

**Physics**

See Relate Questions: Tue, 2-7-12, 1:11 AM.
*November 1, 2015*

**science**

Vf = Vo + g*Tr = 0. 24.5 + (-9.8)Tr = 0. -9.8Tr = 24.5 Tr = 2.5 s. = Rise time. Tf1 = Tr = 2.5 s = Time to fall back to roof. h = Vo*t + 4.9*t^2 = 24 m. 24.5*t + 4.9t*2 = 24. 4.9t^2 + 24.5t -24 = 0. Use Quad. Formula. t = 0.839 s. = Time to fall from top of roof to gnd. T = Tr...
*November 1, 2015*

**physics**

Fb = M*g = 13 * 9.8 = 127.4 N. Fp = 127.4*sin 0 = 0. = Force parallel to the floor. Fn = 127.4 - 86.4*sin64. = Normal force. Fk = u*Fn = Force of kinetic friction. a = (F*Cos64-Fp-Fk)/M.
*November 1, 2015*

**physics**

F1 = 2123 N. F2 = 1930 N. M = 1245 kg. a = (F1-F2)/M.
*November 1, 2015*

**Physics**

Vo = 90,000m/3600s = 25 m/s. d1 = Vo*t = 25m/s * 0.75s. = 18.75 m. d2 = 40 - 18.75 = 21.25 m. = Required stopping distance. Vf^2 = Vo^2 + 2a*d. Vf = 0. Vo = 25 m/s. a = -10 m/s^2. If d is => 21.25 m, the car will hit the barrier.
*October 31, 2015*

**physics**

4. Incomplete. Vo = 35m/s[33o]. Xo = 35*Cos33 = 29.4 m/s. Yo = 35*sin33 = 19.1 m/s. 5A. Yf^2 = Yo^2 + 2g*h. Yf = 0. g = -9.8 m/s^2. h = ? B. Dx = Vo^2*sin(2A)/g. A = 33o. g = +9.8 m/s^2. Dx = ?
*October 31, 2015*

**SPR,PHYSICS**

Vs = Db/Dw * Vb = 2/3 * Vb = 2Vb/3. Db/Dw * Vb = 2Vb/3. Db/Dw = 2/3 = 0.667. Db=0.667Dw = 0.667 * 0.998=0.665 g/cm^3. = Density of the block. Notes: Vs = Volume submerged. Db = Density of the block. Dw = Density of water. Vb = Volume of the block.
*October 30, 2015*

**force**

Wb = M*g = 10*9.8 = 98 N. Fp = 98*sin 0 = 0 = Force parallel to the surface. Fn = 98*Cos 0 - 40*sin30 = 78 N. = Normal force. Fk = u*Fn = 0.3 * 78 = 23.4 N. = Force of kinetic friction. a=(Fap*Cos30-Fp-Fk)/M=(34.6-0-23.4)/10 = 1.12 m/s^2.
*October 30, 2015*

**Physics**

I had this same problem... θ > arctan(μs) is the answer, you're welcome fam.
*October 29, 2015*

**math**

d1 = 10km/h * (20/60)h = 3.33 km., N. d2 = 5km/h * (35/60)h=2.92 km, NE.(45o)= 2.06 + 2.06i. Disp.=2.06 + 2.06i + 3.33i=2.06 + 5.39i = sqrt(2.06^2 + 5.39^2)
*October 29, 2015*

**physics**

Disp. = 7m[90o] + 9m[E45oN] + 7m[W14oS] X = 9*Cos45 + (-7*Cos14) = -0.428 m. Y = 7 + 9*sin45 + (-7*sin14) = 11.7 m. Q2. Tan A = Y/X = 11.7/-0.428 = -27.26754. A = -87.9o = 87.9o N. of W. = Direction. Disp. = Y/sin A = 11.7/sin87.9 = 11.71 m
*October 29, 2015*

**physics**

Disp. = 7m[90o] + 9m[E45oN] + 7m[S76oW]. X = 9*Cos45 - 7*Cos76 = 1.84 m. Y = 7 + 9*sin45 - 7*sin76 = 6.57 m. Q1. Tan A = Y/X = 6.57/1.84 = 3.57168. A = 74.4o N. of E. = Direction. Disp. = Y/sin A = 6.57/sin74.4 = 6.82 m
*October 29, 2015*

**physics**

A quantity that shows magnitude AND direction.
*October 29, 2015*

**geometry**

x^2 + 3x = 180o. x^2 + 3x-180 = 0. Factor it. -180 = -12*15. -12 + 15 = 3 = B. (x-12)(x+15) = 0. X = 12. X = -15. Use positive value of X: X = 12. X^2 = 144o. 3x = 36o.
*October 29, 2015*

**physics**

See Related Questions: Fri, 3-7-14, 4:52 PM.
*October 29, 2015*

**Chemistry**

Methanol, a major industrial feedstock, is made by several catalyzed reactions, such as CO(g) + 2H2(g) ----> CH3OH (l) One concern about using CH3OH as an auto fuel is its oxidation in air to yield formaldehyde, CH2OH(g) which poses a health hazard. Calculate ÄG o at ...
*October 29, 2015*

**physics math science**

Vo = 200m/s[37o]. Xo = 200*Cos37 = 160 m/s. Yo = 200*sin37 = 120 m/s. A. Y = Yo + g*t = 120 + (-9.8)*2 = 100 m/s. = Ver. component of velocity @ 2 s. Tan A = Y/Xo = 100/160 = 0.625. A = 32o. V = Xo/Cos A = 160/Cos32 = 189 m/s[32o]. B. Yf = Yo + g*Tr Yf = 0, g = -9.8m/s^2, Tr...
*October 29, 2015*

**Calculus**

i did. Dv=128pi. But the answer is 8pi/25
*October 28, 2015*

**Calculus**

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter 40 meters. (Recall that the volume of a sphere of radius r is V=(4/3)πr^3. Notice that you are given that dr=0.02.)
*October 28, 2015*

**math**

2 Years ago: Pam: X yrs. old. Sam: 2x/3 yrs. old. 3 yrs. later: Pam: x+3 yrs. old. Sam: 2x/3 + 3 yrs. old 2x/3 + 3 = 3/4(x+3). 2x/3 + 3 = 3x/4+9/4 8x + 36 = 9x + 27. X = 9. So Pam is 3 yrs. older than Sam. In 20 yrs.: x + (x+3) = 69. 2x = 66. X = 33. X+3 = 36. Today: Sam: 33...
*October 28, 2015*

**Physics**

F = 0.119m * 110N./m =
*October 28, 2015*

**Physics**

Wc = M*g = 1400 * 9.8 = 13,720 N. = Normal force, Fn. Fp = mg*sin 0 = 0. = Force parallel to the road. Fk = u*Fn = 0.0889 * 13,720 = 1220 N. a = (Fp-Fk)/M = (0-1220)/1400 = -0.871 m/s^2. Vf^2 = Vo^2 + 2a*d. Vf = 0. a = -0.871 m/s^2. Vo = 21.4 m/s. d = ?.
*October 28, 2015*

**physics**

Wt. = 450cm^3 * 0.998g/cm^3 = 449.1 g. = 0.449 kg of water displaced. M*g = 0.449kg * 9.8N/kg = 4.40 N.
*October 28, 2015*

**SCIENCE\PHYSICS**

Disp. = 0. d = 5 + 5 = 10 m.
*October 28, 2015*

**PHYSICS**

d = 5 + 5 = 10 m.
*October 28, 2015*

**PHYSICS**

Displacement = 0, because it ended at the starting point.
*October 28, 2015*

**PHYSICS**

a. Yf^2 = Yo^2 + 2g*h = 0. Yo^2 = -2g*h = -2*(-9.8)27.5 = 539. Yo = 23.2 m/s. = Vertical component of initial velocity. sin A = Yo/Vo = 23.2/47.5 = 0.48877. A = 29.3o. Xo = Vo*Cos A = 47.5*Cos29.3 = 41.4 m/s = Horizontal component of initial velocity = Total velocity at max ht...
*October 28, 2015*

**physics**

The velocity of the truck must be equal to the hor. component of the airplane's velocity: V = 180*Cos47.
*October 28, 2015*

**Calculus**

Let f(x)=x^4. If a=1 and dx=Δx=1/2, what are Δy and dy?
*October 28, 2015*

**Physics**

Wb = M*g = 200 N., M = 200/g = 200/9.8 = 20.4 kg. sin A = 3/10 = 0.30, A = 17.5o. Fp = 200*sin17.5 = 60 N. Fn = 200*Cos17.5 = 191 N. a. Work = F*d = 120 * 10 = 1200 J. b. Mg*h-0 = Mg*h = 200 * 3 = 600 J. c. Conservation Energy : Change in KE = Change in PE = 600 J. d. Fap-Fp-...
*October 27, 2015*

**maths**

P = Po(1+r)^n. r = (6%/2)/100% = 0.03 = Semi-annual % rate expressed as a decimal. n = 2comp./yr. * 1yr. = 2 Compounding periods. P = Po(1+0.03)^2 = Po*1.03^2 = 1.0609Po. I = P-Po = 1.0609Po-Po = 0.0609Po. I/Po * 100% = 0.0609Po/Po = 6.09% = Effective APR.
*October 27, 2015*

**Physic**

V = Vo^2 + 2g*h. Vo = 0, h = 20 m, V = ?
*October 27, 2015*

**physicsn**

Notes: 1. Fp = Force parallel with the hill. 2. Fn = Normal force. 3. Fk = Force of kinetic friction. 4. Ws = Wt. of skier.
*October 27, 2015*

**physicsn**

Ws = M*g = 9.8M. Fp = 9.8M*sin30 = 4.9M. Fn = 9.8M*Cos30 = 8.49M. Fk = u*Fn = u*8.49M. Fp-Fk = M*a. 4.9M-u8.49M = M*(-1.4) Divide by M: 4.9-8.49u = -1.4 -8.49u = -1.4-4.9 = -6.3 u = 0.74.
*October 27, 2015*

**Physics**

a. a=(Vf-Vo)/t = (5-0)/20 = 0.25 m/s^2. d1 = Vo*t + 0.5a*t^2 = 0 + 0.5*0.25*20^2 = 50 m.
*October 27, 2015*

**math**

a scooter is priced between $1000 and $2000. its price is a multiple of 10. All the digits in the price, except for the thousands digit, are even numbers. The value of the hundreds digit is 30 times the value of the tens digit. what is the price of the scooter?
*October 26, 2015*

**science**

Wb = M*g = 10 * 9.8 = 98 N. Fn = 98-40*sin30 = Normal force. Fk = u*Fn = Force of kinetic friction. a = (Fap*Cos30-Fk)/M.
*October 26, 2015*

**Physics**

a = F/m = 5/0.3 = 16.7 m/s^2. V = Vo + a*t. Vo = 20 m/s. t = 0.03 s. V = ?
*October 26, 2015*

**Physics**

Correction: Yf = 36.6 m/s.
*October 26, 2015*

**Physics**

h = 0.5g*t^2 = 68.2 m. 4.9t^2 = 68.2 t^2 = 13.92. t = 3.73 s. Dx = Xo*t = 4.2m/s * 3.73s. = 15.7 m. Yf^2 = Yo^2 + 2g*h. Yf^2 = 0 + 19.6*68.2 = 1336.72. Yf = 36.6 m. Vf = sqrt(Xo^2 + Yf^2) = Final velocity.
*October 26, 2015*

**math 1350**

Bob: X baseball cards. Ted: 2x " " Reggie: 2x-5 " " 2x-5 + x = 82. X = ?. 2x = ?
*October 26, 2015*

**science**

h = 0.5g*t^2.
*October 26, 2015*

**science**

d = 0.5*g*t^2.
*October 26, 2015*

**science/physics**

F = 9C/5 + 32.
*October 26, 2015*