Friday

October 31, 2014

October 31, 2014

Total # Posts: 26,258

**Stats**

you can play around with Z table stuff at http://davidmlane.com/hyperstat/z_table.html It has very nice input forms so you can specify tails or middle regions.
*October 27, 2014*

**Math**

Since v = 1/3 pi r^2 h I get v = (1/3)(3.14)(9)(8) = 75.36 How did you get your answer?
*October 27, 2014*

**physics**

A small block of mass m = 225 g is released from rest at point circled A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 27.5 cm as shown in the figure below. (a) Calculate the gravitational potential energy of the block-Earth ...
*October 27, 2014*

**Algebra, help please!!!**

well, clearly, if you want values between 4.30000000000000... and 4.33333333333333... All you need to do is replace the string of zeroes with something less than 33333333333333... So, you could have 4.3123456789 4.3010101010 4.33333322222 Similarly for the other one
*October 27, 2014*

**algebra. help!**

|110-30t| = 40 so, either 110-30t = 40 30t = 70 t = 7/3 = 2.3 or -(110-30t) = 40 -110+30t = 40 30t = 150 t = 5 So, (A) is correct
*October 27, 2014*

**algebra. help!**

3x+10 >= 4 3x >= -6 x >= -2 (A) 2.25x >= 180 x >= 80 you are correct
*October 27, 2014*

**Calculus: need clarification to where the #'s go**

y = 2√(4x+4) = 4√(x+1) THE DISTANCE D IS FOUND USING d^2 = x^2+y^2 = x^2+16(x+1) = x^2+16x+16 when x=3, d=√73 so, 2d dd/dt = (2x+16) dx/dt d dd/dt = (x+8) dx/dt so, now for the numbers: √73 dd/dt = 11(5) dd/dt = 55/√73
*October 27, 2014*

**Pre-Calc/Trig**

it depends. as a function of time, or of distance? wikipedia's article on trajectory has many useful equations.
*October 27, 2014*

**Algebra**

(2a-b)^3 = (2a)^3 - 3(2a)^2b + 3(2a)b^2 - b^3 = 8a^3 - 12a^2b + 6ab^2 - b^3 better review your binomial coefficients. I think you used 2 instead of 3. the other is correct
*October 27, 2014*

**Pre-Calc/Trig**

If the side parallel to the river is x and the other side is y, then we have x+2y = 600 The area is a = xy = (600-2y)y = 600y-2y^2 This is just a parabola, where its maximum is at the vertex, which occurs at y=150 So, x = 300 and the area is 300*150 = 45000 ft^2 As is usually ...
*October 27, 2014*

**Math Homework**

3x^2+2x > 2x^2-3 x^2 + 2x + 3 > 0 since the discriminant is negative (4-12), this has no real roots. That is, it does not cross the x-axis. Since f(0) = 3 > 0, f(x) is always > 0, so the solution is all real numbers. 4x^2 - x < 0 x(4x-1) < 0 Recall what you ...
*October 27, 2014*

**algebra**

Since the domain of f is all x, the range of f(6x) is the same as the range of f(x). The graph is just contracted horizontally by a factor of 6. The range is unaffected. So, the range of g(x) = f(6x)+1 is just [-11,3]+1 = [-10,4] Things would have been stickier had the domain ...
*October 27, 2014*

**Calculus: need clarification to where the #'s go**

just use the formula v = 4/3 pi r^3 a = 4pi r^2, so dv/dt = 4pi r^2 dr/dt = a dr/dt da/dt = 8pi r dr/dt v(14) = 11494 a(14) = 2463 So, we have 80 = 2463 dr/dt da/dt = 8pi(14)(80/2463) = 11.43
*October 27, 2014*

**Algebra**

Hmm. I get 2x(3x^2-4xy+2y^2) + 5y(3x^2-4xy+2y^2) = 6x^3 - 8x^2y + 4xy^2 + 15x^2y - 20xy^2 + 10y^3 = 6x^3 + 7x^2y - 16xy^2 + 10y^3
*October 27, 2014*

**algebra--1 question**

Just use >= for greater than or equal Your solution is correct. x < -3 or x >= 5
*October 27, 2014*

**algebra 2**

He only falls 1500 ft before the chute opens. So, change d to 1500 in the above solution and then rework the numbers.
*October 27, 2014*

**Calculus: need clarification to where the #'s go**

you have your formula: pv^1.4 = c, so you get v^1.4 dp/dt + 1.4pv^.4 dv/dt = 0 Now just plug in your numbers: (330^1.4)(-10) + (1.4)(330^.4)(79) dv/dt = 0 -33567.54 + 1125.02 dv/dt = 0 dv/dt = +29.84
*October 27, 2014*

**Calculus**

To find the rate of increase, note that the distance x after t hours is given by x^2 = (50+19t)^2 + (24t)^2 At 3 pm, t=3, and we have x^2 = 117^2 + 72^2 = 16633 x = 137.4 Now we have to find the derivative: 2x dx/dt = 38(50+19t)+48(24t) = 1874t + 1900 So, at t=3, we have 2(137...
*October 27, 2014*

**science-gen chem**

.969g CaCl2 = 0.00873 moles 1.111g KIO3 = 0.00519 moles Each mole of CaCl2 requires 2 moles of KIO3, so clearly the KIO3 is the limiting reagent. So, since each mole of KIO3 produces 1/2 mole of Ca(IO3)2, we get 0.00269 moles of Ca(IO3)2 = 1.051g
*October 27, 2014*

**Calculus - PLEASE HELP!**

3√(x^2+y^2) * (x + yy') = (x+yy')/√(x^2+y^2) + 1 x+yy'(3√(x^2+y^2) - 1/√(x^2+y^2) = 1 y' = (1/(3√(x^2+y^2) - 1/√(x^2+y^2) - x)/y y' = [x+√(x^2+y^2) - 3x(x^2+y^2)]/(y(3(x^2+y^2)-1)) So, we want either the numerator ...
*October 27, 2014*

**Calculus**

draw a diagram. Let the distance from the pole be x, and the shadow's length be s. Using similar triangles, we have s/6 = (x+s)/15 or, 5s = 2x+2s 3s = 2x so, ds/dt = 2/3 dx/dt plug and chug
*October 27, 2014*

**Calculus**

16/(x^2+16)^(3/2)
*October 27, 2014*

**Calculus**

dv/dt = 4pi r^2 dr/dt plug and chug
*October 27, 2014*

**Calculus**

using implicit derivatives, (x+2y)y'=2x-y (1+2y')y' + (x+2y)y" = 2-y' Now plug in the numbers: y'(3) = 2, so (1+4)(2) + (3)y" = 2-2 10+3y" = 0 y" = -10/3
*October 27, 2014*

**Math**

Now, if you meant x^2-31x+108 = 0 then (x-4)(x-27) = 0 x^x is a bit nastier
*October 27, 2014*

**Calculus**

df/dx = df/dg * dg/dx = g/√(g^2-4) * (3) = 3(3x-2)/√((3x-2)^2-4) = 3(3x-2)/√(9x^2-12x) at x=3, that's 3(7)/√45 = 21 / 3√5 = 7/√5
*October 27, 2014*

**math**

Since when do Alonso and Italian chicken sound Chinese? and that's "whoever deciphers this." The phrase acts as the object of the preposition, not just "who".
*October 27, 2014*

**Calculus**

from you formula for v, dv/dt = 4pi r^2 dr/dt now plug in the numbers.
*October 26, 2014*

**Calculus**

I think you have a typo. I'm sure you meant y = 2sinx + (sinx)^2 since y(π/6) = 2(1/2) + 1/4 = 5/4 So, y' = 2cosx + 2 sinx cosx at x=π/6, y' = 2(√3/2) + 2(1/2)(√3/2) = √3 + √3/2 = 3√3/2 Now you have a point and a slope. Dig ...
*October 26, 2014*

**Fractions**

Starting with x peaches, a = x-(x/2 + 1/2) b = a - (a/3 + 1/3) c = b - (b/4 + 3/4) c - (c/5 + 1/5) = 19 now start substituting in and you wind up with x = 101
*October 26, 2014*

**Calculus**

f'(x) = -4x-2 f'(-1) = 2 point-slope form: y+2 = 2(x+1)
*October 26, 2014*

**Calculas**

y = secx y = (cosx)^-1 y' = -1 (cosx)^-2 (-sinx) = sinx/cos^2x = secx tanx similarly for tanx, using the quotient rule
*October 26, 2014*

**calculus**

a square has minimum perimeter for a given area
*October 26, 2014*

**Algebra**

it's easiest to use the polar form if z = r cisθ z^2 = r^2 cis2θ so, r^2 cis2θ = 2 cis π/2 r = ±√2 θ = π/4 So, we have √2 cis π/4 -√2 cis π/4 √2/√2 + √2/√2 i = 1+i or -(1+i) check: (1+...
*October 26, 2014*

**Algebra**

That vertical asymptote at x = -1 means the range is (-∞,∞)
*October 26, 2014*

**Algebra**

for the root, we need x >= 0, or [0,∞) but, the denominator cannot be zero, so the domain is (0,∞)
*October 26, 2014*

**algebra**

we want f(2), so that means we need √(x+1) = 2 so, x=3 f(2) = 1/3
*October 26, 2014*

**Algebra**

well, where are the denominators zero? x-64 = 0 x^2-64 = (x-8)(x+8) = 0 x^3-64 = (x-4)(x^2+4x+16) = 0
*October 26, 2014*

**Calculus**

df/dx = 6(2x+1)^2 so, g' = 1/f' g'(1) = 1/(6*9) = 1/54 check: g(x) = (∛x-1)/2 g'(x) = 1/(6(∛x^2)) g'(1) = 1/(6*9) = 1/54
*October 26, 2014*

**Calculus 1**

f'(x) = -3/x e^(1-x^2) (2x^2 lnx - 1) f(1) = 0 f'(1) = -3(1)(-1) = 3 So, now you have a point (1,0) and a slope (3) and the line is y-0 = 3(x-1)
*October 26, 2014*

**calculus**

Wow. This one gets tricky, since we're dealing in 3D, rather than the usual 2D. If the bottom of the pole is at B, and the top is at T, and the point directly across the road is at Q, then if the man is x from Q, the diagonal distance across the road (BQ) is y, then y^2 = ...
*October 26, 2014*

**calculus**

y = 9√x y(16) = 36 y(25) = 45 So, the secant has slope (45-36)/(25-16) = 1 So, now use the point-slope form to get the equation for a line with slope 1 through (16,36) For the tangent line, the slope y' = 4.5/√x is 9/8 at x=16 Again, use the point-slope form to...
*October 26, 2014*

**Calculus 1**

If the hypotenuse has length z, then h^2+36 = z^2 when h=8, z=10 2h dh/dt = 2z dz/dt or, h dh/dt = z dz/dt So, when h=8, 8*2 = 10 dz/dt dz/dt = 1.6 m/s
*October 26, 2014*

**Algebra**

GanonTEK is correct. If you don't have ready access to a graphing utility, just use what you know about parabolas. Since this is one which opens downward, it is positive between the roots, where it crosses the x-axis.
*October 26, 2014*

**precalculus**

If the cuts are at a distance of x from the center, then since the area of a circular segment is a = 1/2 r^2(θ-sinθ) we want that to be 1/3 of the area, or π/3 r^2 So, 1/2 r^2(θ-sinθ) = π/3 r^2 3(θ-sinθ) = 2π Solve that using your ...
*October 26, 2014*

**Geometry**

You have to tell us what angles 1 and 2 are, in terms of the points A,B,C,D.
*October 26, 2014*

**Algebra**

since the domain of √u is u>=0, we need 2-x >= 0 That is, x <= 2 Next, we also need 1-√(2-x) >= 0, so √(2-x) <= 1 2-x <= 1 means x >= 1 -(2-x) <= 1 means x <= 3 So, we have a final domain of 1 <= x <= 2 The graph is at http://...
*October 26, 2014*

**Algebra**

the denominator cannot be zero. So, x^2+bx+8 must have no real roots. For that to happen, we need the discriminant to be negative: b^2-32 < 0 So |b| <= √32 5 is the greatest integer less than √32
*October 26, 2014*

**science (Need FAST FEEDBACK!!!)**

No idea. I'd have to google the terms. Oh, wait, you could have done that just as easily as asking here and waiting for a response.
*October 26, 2014*

**Math algebra**

for any radius r, the circumference C is C = 2πr So, just plug in the numbers you have for r
*October 26, 2014*

**math**

(4+x)(8+x) = 4*8 + 64 x^2+12x-64 = 0 (x-4)(x+16) = 0 x = 4 So, the rectangle is 8x12 with area 96 (32+64)
*October 26, 2014*

**MATHS**

If there were x 50-cent coins, x/6912 = 7/16 x = 3024 So, they collected 6912*.25 + 3024*.50 = $3240.00
*October 26, 2014*

**Calculus**

a square encloses the maximum area for a given perimeter.
*October 26, 2014*

**second degree equation word problem**

If there are now x rows, then each row has 720/x seats. The new configuration means that (x-4)(720/x + 6) = 720 (x-4)(6x+720) = 720x (x-4)(x+120) = 120x x^2-4x-480 = 0 (x+20)(x-24) = 0 ...
*October 25, 2014*

**Pre-calc**

cos(-7pi/6) = cos(7pi/6) = cos(pi+pi/6) = -cos(pi/6)
*October 25, 2014*

**math**

See http://www.wolframalpha.com/input/?i=f%3E%3D4 There's a number-line plot there.
*October 25, 2014*

**Precalculus**

using the law of cosines, the ship's distance x from the port is x^2 = 170^2 + 90^2 - 2*170*90 cos 14° Now just figure the coordinates of the ship and port to get the desired course. That assumes that there are no cross currents for the rest of the trip.
*October 25, 2014*

**Precalculus**

Hmm. Not having the diagram is a bit of a problem. In my diagram, extend the sea-level line TR to intersect PS at Q. Then, I have the angle of depression α = ∠PTR and β = ∠PQR. Apparently that is wrong, so please describe the diagram using only labeled ...
*October 25, 2014*

**Precalculus**

I don't see how ∠SPT can be 99°. ∠SPT = ∠SPR + ∠RPT = 51°+69° = 120° Did I miss something in the explanation?
*October 25, 2014*

**Pre-Calc/Trig**

you want two factors of 24 which add up to 10. List the factors: 1 2 3 4 6 8 12 24 Take them in pairs, working in from the ends. Note that 4 and 6 add up to 10, so you have (x+4)(x+6) = 0 x = -4 or -6
*October 25, 2014*

**Precalculus**

That is the angle between the vectors (7,10,0) and (7,0,3) So since (7,10,0)•(7,0,3) = √149√58 cosθ, cosθ = 49/√8642 = 0.527
*October 25, 2014*

**math**

f is increasing where f' is positive f' = 4-3x^2 So, where is 4-3x^2 > 0? similarly for decreasing. max/min where 4-3x^2 = 0
*October 25, 2014*

**math**

f' = (x+3) / 4x^(7/4) f' = 0 when x = -3. what about x=0?
*October 25, 2014*

**Math**

b = 3j m = 4(b+j) m = 4b+4j m = 4b+4(b/3) m = 16/3 b yes 3 = 1+2 3*8 = (1+2)*8 = 8+16 = 24 (12-4)*(2+1) = 8*3 = 24 12-4*2+1 = 12-8+1 = 5 in the absence of parens, multiplication is done before addition. That's why 3x+2 is (3*x)+2 and not 3*(x+2)
*October 25, 2014*

**math**

3*5 = 1/3 * x
*October 25, 2014*

**Algebra**

AB has slope 2/-6 = -1/3 So, (y+2)/(6-x) = -1/3 3y+6 = x-6 x=3y+12 So, pick any y and that will give you an x.
*October 25, 2014*

**Calc**

since v = 4/3 pi r^3 and a = 4 pi r^2, v = 1/3 ar so, dv/dt = 4pi r^2 dr/dt = a dr/dt so, dr/dt = (dv/dt)/a dv/dt = 1/3 (r da/dt + a dr/dt) = 1/3 (r da/dt + a*(dv/dt)/a) = 1/3 (r da/dt + dv/dt) = 1/2 r da/dt So, 10 = 1/2 (6.5) da/dt da/dt = 3.077 cm^2/s
*October 25, 2014*

**Maty**

3d+300m = 111 5d+600m = 198 Now just solve for d and m
*October 25, 2014*

**algebra**

check the related questions below
*October 25, 2014*

**Chemistry**

11.9g of am carb = 11.9/96.1 = 0.124 moles So, you get 0.248 moles of NH3 0.124 moles CO2 0.124 moles H2O Total moles of gas: 0.496 So, since 1 mole makes 22.4L at STP, adjust for your conditions.
*October 25, 2014*

**college algebra**

Using the point-slope form of the line, y-39000 = (64/39)(x-2000) So, when is y > 61000?
*October 25, 2014*

**Physics**

a = (36.6 m/s)/(8.22 s) = 4.45 m/s^2 F = ma
*October 24, 2014*

**Calc**

who cares about the volume? A = πd^2 dA/dt = 2πd dd/dt Now plug in your numbers to find dd/dt
*October 24, 2014*

**Secondary Math**

Hmmm. You have TR(-15x+3) = (-15x+3)(-2x+6) So, TopRight = -2x+6 BC = -3(-5x+1) = 15x-3 So, BottomCenter = 15x-3 That means BottomLeft = 2x-6 TopLeft = (-2x+6)/-3 = 2/3 x - 2 So, the filled-in square is (2/3 x - 2) (-3)(-2x+6) (3)(-5x+1)(-15x+3) (2x-6)(15x-3)(30x^2-96x+18)
*October 24, 2014*

**Algebra**

if the domain is all real numbers then the denominator can never be zero. That means it has no real roots. Thus, the discriminant is negative: b^2-32 < 0 So, if 0<=b<=5 x^2+bx+8 is never zero
*October 24, 2014*

**Physic**

a = 320km/hr / 6.5s * 1000m/1km * 1hr/3600s = 13.68 m/s^2 Since F = ma, the net force is 13.68m Since the net force is 1.52*10^4-5.2*10^3, divide that by a to get m.
*October 24, 2014*

**MATH HELP asap PLEASE**

the vectors to add are v = (550cos7°,550sin7°)+(85cos20°,85sin20°) so, plug in the numbers and get the sum. Then just get |v| and its direction
*October 24, 2014*

**math**

.4(6) = 2.4 = 24/10 = 12/5 or, .4 = 2/5, so .4(6) = (2/5)(6) = 12/5
*October 24, 2014*

**Math**

correct Though, to save us having to search back for your previous postings, it's better just to add a reply to them yourself. You can see where lots of posts have several replies.
*October 24, 2014*

**Algebra**

the denominator cannot be zero. So, its discriminant must be negative. That means 1-4c < 0
*October 24, 2014*

**Math**

Unfold the prism into a sheet, and draw a line from one point to the other. Then fold it back up again.
*October 24, 2014*

**math**

After Amy, .85 remained. After Beth, .80*.85 = .68 remained After Cathy, .88*.68 = .5984 remained So, 59.84% of the job remained undone. So, how did you arrive at 53%/ It's always good to show your work, so we can check to be sure your logic is correct.
*October 24, 2014*

**Algebra**

at x=1, r is undefined (∞) As x gets huge, r-> 0 The denominator is always positive, So, the range is (0,∞) See the graph at http://www.wolframalpha.com/input/?i=1%2F%281-x%29^2
*October 24, 2014*

**Calc**

A = πd^2 dA/dt = 2πd dd/dt Now plug in your numbers and solve for dd/dt
*October 24, 2014*

**algebra**

we want 23 + 1.75x < 7 + 3x Now just solve for x
*October 24, 2014*

**algebra**

that would be x miles, where 45.00+0.25x = 35.00+0.30x Now just solve for x
*October 24, 2014*

**Trig calc**

x = r cosθ y = r sinθ Now just plug in your numbers
*October 24, 2014*

**Trig**

x = r cosθ y = r sinθ Now just plug in your numbers
*October 24, 2014*

**Math Calc**

for any x, sin(x+180) = -sin(x) check the sum formula for sines
*October 24, 2014*

**Math**

f(-x) = -x^3+x^2+10x-24 2 sign changes, mean either 2 or 0 negative roots "A few" is not a very precise answer.
*October 24, 2014*

**Math**

1 sign change, so there is exactly 1 positive real root. The sign change is from +x^2 to -10x
*October 24, 2014*

**algebra**

2f-21 looks good to me. 2f is two times 2f-21 is 21 less than that. good work.
*October 24, 2014*

**math**

10+30*2.5
*October 24, 2014*

**Math Trig**

h/(120*3) = tan 40 now just evaluate
*October 24, 2014*

**Algebra**

we need |500m - 200t| <= 5 where m and t are the rates in $/minute
*October 24, 2014*

**math please help**

Not sure how the answer is 16 gallons. Better check my math.
*October 24, 2014*

**math please help**

from the Z table, 95% are below 1.645 std above the mean That means .39 + 1.645*.26 = .8177 So, there's a 95% chance that the average consumption is below .8177 gallons For 33 people , that means that there's a 95% chance that .8177*33 = 26.98 gallons will fed them all.
*October 24, 2014*

**maths**

a + ar = 28 ar^2 + ar^3 = 252 a(1+r) = 28 ar^2(1+r) = 252 ar^2(1+r)/a(1+r) = r^2 = 252/28 = 9 r = 3 or -3 a(1-3) = 28 ==> a = -14 a(1+3) = 28 ==> a = 7 So, now we need to know that ar^6 = 5103 7*3^6 = 5103 Unfortunately, that works whether r is 3 or -3 Using r = -3 7+(-...
*October 24, 2014*

**Algebra/math 117**

well, just count by 6's starting with a1=6: 6 12 18 ...
*October 24, 2014*

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