Friday

November 28, 2014

November 28, 2014

Total # Posts: 27,046

**mathematics**

2OR = 2x+5 2OR = HE+5 HE = 2OR-5
*November 10, 2014*

**mathematics - incomplete**

can you not tell there is some missing information here?
*November 10, 2014*

**Science**

(8m/s)/(3s) = 8/3 m/s^2 (40m/s)/(8s) = 5 m/s^2 which is greater?
*November 10, 2014*

**Algebra**

the factors of -7 are -1 and 7 1 and -7 can you pick them so they add up to -6? (x+1)(x-7)
*November 10, 2014*

**math**

So, you must want the pair (cosθ,sinθ) which is the point on the unit circle associated with θ. 30: (0.866,0.5) 60: (0.5,0.866)
*November 10, 2014*

**math**

ok - I thought that might be it. So, now, what ordered pair did you have in mind? 30 degrees is just an angle. Is there supposed to be an ordered pair associated with an angle?
*November 10, 2014*

**math**

eh? Not sure what it is you are asking. ff?
*November 10, 2014*

**geometry - incomplete**

so, what is the question, again? You seem to have dropped some data, as well as what it is you are asking.
*November 10, 2014*

**math**

just write the data as given: a+b+c+d = 9000 a+c+d = 4b b+d = 4/5 (a+c) The first says that a+c+d = 9000-b using the second, that means that 9000-b = 4b b = 1800 So, now we have a+c+d = 7200 1800+d = 4/5 (a+c) Now we know that a+d = 7200-d, so 1800+d = 4/5 (7200-d) 1800+d = ...
*November 10, 2014*

**math(fractions)**

no idea. How long is the Blue Trail?
*November 10, 2014*

**calculus**

online, we generally use ^ for powers, and * for multiplication. Why bother changing the way it was posted? And the quotient rule says 5(3x+4)^4(3)(4x+1)^2 - (3x+4)^5(2(4x+1)(4)) ----------------------------------------- (4x+1)^4 (3x+4)^4 (15(4x+1)-8(3x+4...
*November 10, 2014*

**what is the rate of change of the angle of eleva**

see related questions below
*November 10, 2014*

**calculus**

when the elevator is x meters up, tanθ = (x-20)/20 So, sec^2θ dθ/dt = 1/10 dx/dt when x=10, tanθ = -1/2, so sec^2θ = 5/4 5/4 dθ/dt = 1/10 (5) dθ/dt = 2/5 m/s
*November 9, 2014*

**physics**

Look out below!! Was there a question in there somewhere?
*November 9, 2014*

**phyics**

If the tug pulls at speed v on a course of θ, then in x-y coordinates, 15 + 9cos60° + 35 sinθ = 0 19.5 + 35 sinθ = 0 sinθ = -0.5571 θ = N 33.9° W Now you can figure part B.
*November 9, 2014*

**Calculus**

revenue = price * quantity, so r = x(160-2x) = 160x - 2x^2 max revenue when r'=0, or x = 40 So, just plug that in as needed.
*November 9, 2014*

**Math 112**

not so. The value is a non-terminating decimal, so it cannot be written exactly. The calculator just shows the first few digits. The only way to write it exactly is: √3
*November 9, 2014*

**math**

14.1√2 = ?
*November 9, 2014*

**College Algebra**

27.95 + 0.15m = 45.95 find m
*November 9, 2014*

**Math**

how much in taxes? well, duh: 7% .07 * 43.99 = $3.08 so, add the price to the taxes to get the total
*November 9, 2014*

**Algebra**

If you mean 2/3y + 5 - 3 - 11/12y then collect like terms: 2/3 y - 11/12 y + 5-3 (2/3 - 11/12)y + 2 (8/12 - 11/12)y + 2 (-3/12)y + 2 -1/4 y + 2
*November 9, 2014*

**ap calc**

so, solve the equation for f'. You have 3sinx = -2cosx tanx = -2/3 So, that gives you x for maximum value of f(x). Just plug it in and evaluate f(x). Of course, you need to recall how to take sin(arctan(x)) and cos(arctan(x)) But that's just trig, right? Or, note that ...
*November 9, 2014*

**Can you check my answer - Physics**

clearly the net force is the sum of the downward and upward forces: -1.5N + 0.7N = -0.8N That is, 0.8N downward You just added the magnitudes, but did not take into account the directions
*November 9, 2014*

**math**

Not sure what your characters are, but (A∩B)∪C = {9,10}∪{11,13,15} = {9,10,11,13,15} (A∪B)∩C = {9,10,11,12,13,14}∩{11,13,15} = {11,13}
*November 9, 2014*

**ap calc**

f(x) = bx^2+6x if x <= 2 ax^3 if x>2 So, we have lim(x->2-) f(x) = 4b+12 lim(x->2+) f(x) = 8a That means we need 4b+12 = 8a Also, we need the derivatives to be equal: 2bx+6 -> 4b+6 3ax^2 -> 12a And they need to be equal: 4b+6 = 12a rearranging things a bit, ...
*November 9, 2014*

**Math**

Since y = kx, y/x = k That is, pick any two known values of x and y. Their ratio is the constant k.
*November 9, 2014*

**College algebra**

f(x) has 3 changes, so 3 or 1 + roots f(-x) has no sign changes, so no - roots check: http://www.wolframalpha.com/input/?i=6x^5-6x^4%2B7x^3-8
*November 9, 2014*

**consumer math**

799.99(1-0.40) = 479.99 or, $480
*November 9, 2014*

**calculus**

y = log |2-x| y' = 1/|2-x| (-1) = 1/(x-2)
*November 9, 2014*

**Pre-trig/Calc**

1 positive, 2 or 0 negative check: http://www.wolframalpha.com/input/?i=x^3%2B3x^2-18x-40
*November 9, 2014*

**Math**

start with x. The value after each step is listed below x 2x 2x+6 (2x+6)/2 = x+3 x+3-3 = x you always get your original number.
*November 9, 2014*

**calculus**

s = t^9 ln|t| s' = 9t^8 ln|t| + t^9/t = t^8(9ln|t| + 1)
*November 9, 2014*

**Math**

15.3g + 2q + 3.79 Note that you don't need an equation, just an expression
*November 9, 2014*

**calculus**

y' = (26x^2-26)e^-3x - 3(13x^2-26x+26)e^-3x = -13(3x^2-8x+8)e^-3x
*November 9, 2014*

**Math**

#1 False A triangle has three angles, which add up to 180 degrees. The two right angles use all that up. #2 True #3 True #4 False Just draw any obtuse triangle. If two of the legs happen to be equal, stretch one of them a little. #5 True #6 False. Draw a tall skinny isosceles ...
*November 9, 2014*

**Math -word problem**

Consider the room. The front wall is square ABCD, and the back wall is EFGH. The front outlet (P) is a foot from the floor (AB), and the back outlet (Q) is a foot from the ceiling (GH). A line directly up and across and down is 20 feet. But, you can do the following: Unfold ...
*November 9, 2014*

**calculus**

just use the product rule and chain rule: y = uv y' = u'v + uv' where u=(2x+1)^3 v=(4x+1)^-4 y' = 3(2x+1)^2(2)*(4x+1)^-4 + (2x+1)^3*(-4)(4x+1)^-5(4) = -2(2x+1)^2(4x+5) --------------------- (4x+1)^5
*November 9, 2014*

**math pre algebra**

how about 5x+7y
*November 9, 2014*

**Algebra 2**

plug in your coefficients at http://www.gregthatcher.com/Mathematics/GaussJordan.aspx to see all the details Or, you can try this: http://www.wolframalpha.com/input/?i=solve+{{3%2C4%2C1}%2C{1%2C3%2C-2}%2C{-2%2C1%2C-2}}*{{x}%2C{y}%2C{z}}%3D{{1}%2C{-2}%2C{-5}}
*November 9, 2014*

**Algebra 2**

(2x-1)^2 = x^2 + 33 4x^2-4x+1 = x^2+33 3x^2-4x-32 = 0 (3x+8)(x-4) = 0 ...
*November 9, 2014*

**math**

315 = 3^2 5 7 882 = 2 3^2 7^2 So, GCF = 3^2 7 = 63
*November 9, 2014*

**math count warm up**

just start writing down the facts. If the 4 digits, left to right are a,b,c,d, then: a+b=c b+c=d c+d=10a+b a+b=c d-b=c 10a+b-d=c a+b=d-b a+2b = d 10a+b-(a+2b) = a+b 8a = 2b b = 4a so, c=5a and, d=9a Since a,b,c,d are single digits, a must be 1. So, we get 1459
*November 9, 2014*

**algebra**

w(3w-2) = 16 w = 8/3 p = 2(8/3 + 6) = ?
*November 9, 2014*

**math**

6 - 4(2/3) = ?
*November 9, 2014*

**projectives**

time to hit: 4.9t^2 = 90 t = 4.29 s 4.29s * 50 m/s = 214.28 m
*November 9, 2014*

**calculus**

see related questions below
*November 9, 2014*

**Math**

Drop a perpendicular from G on CD to F. Now you have a rectangle GDEF with sides 3.5 and 4.5 Also, triangle CGF where CG = 3.5/√3 CF = 7.0/√3 Now the perimeter is easy - add up three sides of the rectangle, and the triangle.
*November 8, 2014*

**math**

I don't see any fractions If you meant (11+x)/(13+x) = 10/11 then x=9
*November 8, 2014*

**Calculus**

Do you mean ln√x or √lnx ?
*November 8, 2014*

**trig**

recall your sum of angles formula for tan recall that tan(180) = 0
*November 8, 2014*

**Math Question**

QR = 7.1 tan58° Area = (1/2)(PR)(QR)
*November 8, 2014*

**math**

your math is ok, freddy, but the spelling, not so much...
*November 8, 2014*

**math**

Hmmm. How can someone savvy enough to use a computer, not know what 1+1 is? I suspect either a test posting, or someone really bored.
*November 8, 2014*

**Algebra 2**

You always want to set things equal to zero: 6x^2-13x+6 = 0 Now, just have to have some practice factoring to get to (3x-2)(2x-3) = 0 Now, since if the product of two numbers is zero, one of the numbers must be zero. So, either 3x-2 = 0 and x = 2/3 or 2x-3 = 0 and x = 3/2 You ...
*November 8, 2014*

**Calculus (math)**

Draw a diagram, looking from the side. If the observer is at distance x, then The angle Ø from the eye to the bottom of the picture is such that tanØ = d/x The angle θ subtending the picture is tan(Ø+θ) = (d+h)/x We want to determine x for ...
*November 8, 2014*

**math**

(3x-2)/8 + (2-x)/ 4 = - 1/2 easy way: clear fractions by multiplying by 8: (3x-2) + 2(2-x) = (-1/2)(4) 3x-2+4-2x = -2 x+2 = -2 x = -6 check: (3(-6)-2)/8 + (2-(-6))/4 = -20/8 + 8/4 = -5/2 + 2 = -1/2
*November 8, 2014*

**mathematics**

a + a+2d = 12 (a)(a+d) = 24 a=4, d=2 check: 4,6,8,10,... 4+8=12 4*6=24
*November 8, 2014*

**Mustapha**

ever try google? Here's one discussion: http://www.meritnation.com/ask-answer/question/determine-the-oxidation-number-of-cr-in-k2cr2o7/redox-reactions/1515787
*November 8, 2014*

**algebra**

#1 If you mean f(n) = 3(n-3) then f(5) = 3(5-3) = 6 f(n) = 3^(n-3) then f(5) = 3^2 = 9 f(n) = 3^n - 3 then f(5) = 3^5-3 = 240 How did you come up with 279? #2 f(n) = 2f(n-1)+2 f(1) = 4 f(2) = 2(4)+2 = 10 f(3) = 2(10)+2 = 22 #3 f(n) = (4-n)/(n+3) f(3) = (4-3)/(3+3) = 1/6 #4 I ...
*November 8, 2014*

**Maths**

C = pi * d so, the turns is 44/C where C is in meters
*November 8, 2014*

**math - typo**

3:1:15 requires three quantities.
*November 8, 2014*

**calculus**

d/dx u^n = nu^(n-1) u' d/dx arctan(u) = 1/(1+u^2) u' d/dx ln(u) = 1/u u' So, we have y' = 2(3/2) √(22x+31) * (22) + 1/(1+(22x+21)) * (1/2) * 1/√(22x+21) * 22 + 1/(11x+16) * 11 - 22 = 66/√(22x+31) + 11/((22x+22)√(22x+21)) + 11/(11x+16) - 22
*November 8, 2014*

**calculus**

That would be 4(cos(6x-1) - 6xsin(6x-1)) ------------------------------ 3cos(3x) - 3 at x=0 that is 4(cos(-1)-0) ---------------- 3-3 -> -∞
*November 8, 2014*

**Calculus**

See the related questions below
*November 8, 2014*

**Trig**

Extend AB to meet the vertical from C at D. It is easy to determine that in triangle ABC ∠C = 47° ∠B = 88° Now just use the law of sines to get AB.
*November 8, 2014*

**Intermediate Algebra**

well, let's just assume that there is a t. After all, if there is none, the B is just a constant: 604.47 So, since polynomials are normally written in descending power order, I'd guess B = 18.75t + 585.72 So, you want to solve 18.75t + 585.72 > 1050 18.75t > 464....
*November 8, 2014*

**Intermediate Algebra**

B=18.75+585.72 where is the t?
*November 8, 2014*

**Calculus (math)**

See the related questions below
*November 8, 2014*

**Math**

cos is negative in QII, QIII You reference angle is 0.927 So, x = pi±0.927
*November 8, 2014*

**calculus**

V = IR dV/dt = R dI/dt + I dR/dt Now just plug in your numbers.
*November 8, 2014*

**Calculus**

I guess you should have checked my math. The actual formula for the volume is (π/3)((18+y)(3+y/6)^2-18*3^2) = π(y^3/108 + y^2/2 + 9y) because the water is in the bottom of the cone dv/dt = π/36(y+18)^2 dy/dt 4.9 = π/36*20^2 dy/dt dy/dt = .14 m/min or 14.0 ...
*November 7, 2014*

**Calculus**

When the water has depth y, the radius of the surface is 3 + (y/12)(5-3) = 3 + y/6 Its volume at that point is v = 1/3 (R^2-r^2) h = (1/3)((3+y/6)^2-3^2)y = y^3/108 + y^2/3 Now we are ready to begin. When y=2, v = 8/108 + 4/3 = 38/27 dv/dt = (y^2/36 + 2y/3) dy/dt 4.9 = (4/36...
*November 7, 2014*

**Calculus (math)**

as usual, draw a diagram. The distance z between the boats at time t hrs after 2:00 is z^2 = (15t)^2 + (20-20t)^2 2z dz/dt = 2(15t)(15) + 2(20-20t)(-20) z dz/dt = 1250t - 800 clearly dz/dt=0 when t=8/1250 hours Now just change that to a time of day.
*November 7, 2014*

**AP CALC. AB**

the .75 comes from the fact that the diameter is increasing 1t 1.5in/min. The radius is half that. and yes, he meant to solve for dh/dt, since that was the question asked. Part 1 just uses the fact that the volume of a box is length*width*height. can you say duh?
*November 7, 2014*

**helpppppppp**

7c+3d = 4m 1m = 1c+1d So, 4m = 4c+4d 7c+3d = 4c+4d 3c = d
*November 7, 2014*

**BRAINMATH**

If the family has b boys and g girls, b = g+4 So, b = (g-1)+5 so, each sister has 5 more brothers than sisters
*November 7, 2014*

**Pre-Calc/Trig**

numerator will be 1 or 2 denominator will be 1 or 3 all can be positive or negative
*November 7, 2014*

**Math**

cos pi/3 = 1/2 so, t = pi/3 That's the reference angle. cos is positive in QI and QIV, so the other solution is t = 2pi - pi/3
*November 7, 2014*

**Physics**

the vertical speed is 200 sin30 = 100 m/s So, the height is h(t) = 100t - 4.9t^2 As you recall from algebra I, the vertex of the parabola is at t = -b/2a, which in this case is 100/9.8 So, plug that in to get the max height. The ball is in the air twice that long, and you know...
*November 7, 2014*

**Algebra**

I'll do one, and you try the other. 2x^2+7x = 4 (x^2 + 7/2 x) = 2 (x^2 + 7/2 x + (7/4)^2) = 2 + (7/4)^2 (x + 7/4)^2 = 81/16 x + 7/4 = ±9/4 x = -7/4 ± 9/4 x = -4 or 1/2 Just recall that (x+a)^2 = x^2 + 2ax + a^2 That's why I added (7/4)^2 to both sides. ...
*November 7, 2014*

**science Help 911!!**

Looks good to me.
*November 7, 2014*

**ALakh narayan sing high school Ekma saran**

1/6(2x+3y)+x/3=8 1/2(7y-3x)-y=11 Rearranging things a bit to get rid of the bothersome fractions, we have 4x+3y = 48 -3x+5y = 22 Since y = (48-4x)/3, -3x+5(48-4x)/3 = 22 Or, -9x+240-20x = 66 29x = 174 x = 6 Now you can easily find y.
*November 7, 2014*

**algebra - incomplete**

I think you are missing a t in the formula.
*November 7, 2014*

**multiplying integers**

10^2 = 10*10 = 100 So, -10^2 = -100 Powers are done before multiplication. SO, the expressions is evaluated as -(10^2), NOT (-10)^2, which is (-10)(-10) = +100
*November 7, 2014*

**algebra**

no, it would never be positive. multiplying two negatives always gives a positive. That third negative then yields a negative product.
*November 7, 2014*

**MATH**

#1 -15 - (-5.6) = -9.4 #2 ...
*November 6, 2014*

**Programming (Integer generator)**

Ignoring the syntax error of the missing close brace, all you do is set the text to the latest string displayed. All the other strings are replaced. You need to append a new line for each value of i, not just do a clean reset of the text area each time.
*November 6, 2014*

**algebra help!**

The domain is the set of 1st values from the pairs: (b) In #2 you are correct.
*November 6, 2014*

**Algebra**

The slope of the line is 2, so y-a^2 = 2(x-a) y = 2x+(a^2-2a) 2(-3/2) + a^2-a = 0 a^2-a-2 = 0 2(0) + b^2-b = 3 b^2-b-3 = 0 Now you can find a and b, and pick the ones that work with the line.
*November 6, 2014*

**Algebra**

well, that would be where x^3-x^2+x+1 = x^3+x^2+x-1 2x^2-2 = 0 x = ±1 See http://www.wolframalpha.com/input/?i=solve+x^3-x^2%2Bx%2B1+%3D+%28+x^3%2Bx^2%2Bx-1%29
*November 6, 2014*

**algebra**

you have slope=3, so using the point-slope form, the line can be written as y-3 = 3(x-0) or, y = 3x+3
*November 6, 2014*

**Calculus**

If v is viewed as a piecewise linear function, then the slopes of the various segments are 154/7 = 21 (374-154)/(17-7) = 22 (418-374)/(19-17) = 22 (770-418)/(35-19) = 22 (1320-770)/(60-35) = 22 (1508-1320)/(64-60) = 47 (3905-1508)/(115-64) = 47 Looks like the 2nd stage kicked ...
*November 6, 2014*

**Intermediate Algebra**

If you are talking about powers, 9^(1/2) = 3 x^(1/2) = √x so you are talking about square roots. (-9)^(1/2) is not real, because negative numbers do not have real square roots. That is because if x = √(-9) then x^2 = -9. But all numbers are positive when squared.
*November 6, 2014*

**Algebra help**

ok ok v = 243 - 0.2s
*November 6, 2014*

**Physics**

Consider m3 to be at (0,0) m1 exerts its force at an angle of 30° m2 exerts its force at an angle of -30° As you say, each force is F=GMm/r^2 The resultant force will be at 0°, with a magnitude of F√3, the sum of the x-components of the two forces.
*November 6, 2014*

**Calculus**

y = (3x^2+2√x)/x = 3x + 2/√x y' = 3 - 1/x^3/2 You are correct
*November 6, 2014*

**Trigonometry**

secx = 2.1754 x = 0.4776 SInce secx is positive in QI and QIV, the other angle is 2π-0.4776 arcsin(2x) makes no sense, since x is an angle. That's if we are still referring to the first question. Now, in general, if sinθ = 2x, then cosθ = √(1-sin^2&#...
*November 6, 2014*

**Trigonometry**

c = 2πr, so one revolution of the plane travels 4π meters So, now we know that 5 min * 60s/min * 0.4rev/s * 4πm/rev = (5*60*4π)/(0.4) = 300π meters
*November 6, 2014*

**Math**

There is a good discussion of this problem at http://mathforum.org/library/drmath/view/65003.html This is, of course, a case involving Fermat's Last Theorem.
*November 6, 2014*

**Math- Had Sub teacher, need help**

apparently you stopped reading before finishing things. Read on past where it says So, the revenue will be ...
*November 6, 2014*

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