Friday

July 25, 2014

July 25, 2014

Total # Posts: 23,955

**Math**

if there are 8x tools, then there are 3x shovels and 5x rakes. So, 5x = 3x+10 x=5 so, there are 15 shovels and 25 rakes

**Math**

If you can find the midpoint of a line segment know that the slope of the perpendicular is -1/(slope of line segment) remember the point-slope form of a line then all of these problems are straighforward. Review those topics, give it a try, show where you get stuck, if you do.

**Non**

just a note: that's mu sub-k, not "muse of k" µk

**Volume**

Just figure the tank's volume (66.75*132*10) and divide it by 231 in^3, the volume of a gallon.

**algebra**

x^2+8x = x(x+8) Pick any positive value for x and you get a rectangle.

**math**

take a look at the graph of (1+.10/n)^(n) It shows the effective yield of 10% compounded n times per year. It approaches continuous compounding, e^.1 http://www.wolframalpha.com/input/?i=plot+y%3D%281%2B.1%2Fn%29^n+for+n+%3D+1..40%2C+y%3De^.1

**algebra**

solve it just as you would an equation: 42 = -6d -7 = d However, multiplying and dividing by negative values changes the direction of the inequality. So, you have 42 < -6d 7 < -d -7 > d or d < -7 Since you can always add and subtract without changing the direction,...

**science**

I may have messed up the sign change, so double check my math.

**science**

the position falling from a height of h feet is y = h - 16t^2 After t seconds, y=0, so h=16t^2 So, we have (h-16t^2) - (h-16(t-1)^2) = 11h/36 0 - (16t^2 - 16(t-1)^2) = 11(16t^2)/36 32t-16 = 44/9 t^2 11t^2 - 72t - 36 = 0 t = 7.01 So, h = 16t^2 = 786.24 ft

**calc**

first off, I'd use partial fractions: (x^3 -3x^2 -9)/(x^3 -3x^2) = 1 + 1/x + 3/x^2 - 1/(x-3) There are no discontinuities on [4,5], so it's straightforward. The integral is just x + logx - 3/x - log(x-3) so we get (5 + log5 - 3/5 - log2)-(4 + log4 - 3/4 - log1) = 23/20...

**Math**

54° is 3/20 of a full circle. SO, the whole circumference 3/20 c = 9 c = 60

**ALGEBRA - eh?**

cannot parse the "8n 12" part

**math**

The solution of this equation is so complex, I have to assume you're doing numerical methods or something. Take a peek at http://www.wolframalpha.com/input/?i=sin^6x+-sin^2+4x+%3D+sin2x+sin10x for further insight.

**trigonometry**

well, we have h/120 = tan 10°18' h = 21.8 ft

**Maths**

y = 1/x y' = -1/x^2 y'(1/4) = -1/(1/16) = -16 I think you lost a minus sign. And, of course, you have displayed the usual sloppiness with parentheses.

**math**

I don't see any way. Each odd number is of the form 2k+1 Add up 7 of those, and you have a sum of the form 2n+7 = 20 But 2n cannot be 13. The sum of an odd number of odd numbers cannot be even.

**Math**

well, ∫ lnx/x^2 dx = -(lnx + 1)/x recall that ln √x = 1/2 lnx let u = 3-5x^2 and du=-10x dx That gives you ∫ -1/10 2^u du which is just 2^u / ln u Now it's pretty straightforward

**Math**

well, we know that d/dx arccosh u = 1/√(u^2-1) du So just use the chain rule, and note that (2x-1)^2-1 = 4x^2-4x

**physics**

(i) the range for angle θ is R = v^2/g sin2θ (ii) Naturally max height is reached at θ = 45° (iii) given the range, and that horizontal velocity is constant, getting t is easy. So, what do you get?

**Algebra**

y(x) = 3*2^x y(x+1) = 3*2^(x+1) = 3*2*2^x = 2y(x) So, each year y doubles.

**Wood working**

Of course, if each stripe is that wide, there will be no space between them, so the board will just be solid white. Now you have to decide how much blank space to allocate between the stripes, and whether the end stripes will reach to the very ends of the board, or whether the...

**Math-Parabolas**

If you mean y = x^2+9 you can see that since x^2 can never be less than zero, the vertex is at x=0. Since y=9 there, the vertex is at (0,9) Or, recalling that the vertex of y = (x-h)^2 + k is at (h,k), note that we have h=0 and k=9.

**pre cal**

since complex roots come in pairs, we have (x-3)(x+13)(x-(5+4i))(x-(5-4i)) (x-3)(x+13)((x-5)-4i)((x-5)+4i) (x-3)(x+13)((x-5)^2+4^2) (x^2+10x-39)(x^2-10x+41) x^4 - 98x^2 + 800x - 1599

**Science**

Some of the CO2 was lost as a gas. Also, depending on how hot the mixture became, maybe some of the water was lost as steam.

**pre-cal**

However, assuming the chopper starts at (0,h) and the vehicle starts at (0,0), after t minutes the chopper and vehicle are at (2640t,h - 5/33 t) and (4224t,0) So, the line between them has slope (h-5/33 t)/(1584t) Now you have the angle tanθ = (h-5/33 t)/(1584t) so you ca...

**pre-cal - incomplete**

So the helicopter drops 400 feet. How high was it when it started its descent? what is "the vehicle"? Is it a car driving directly underneath the chopper? Did we start at t=0 when the chopper was directly above "the vehicle"? Which angle of depression? chop...

**maths (exponents**

Well, just using brute force, we have b/(b+ab+1) + c/(c+bc+1) + a/(a+ac+1) expand all that over a common denominator, and you get (a^2b^2c + a^bc^2 + 2a^2bc + a^2b + ab^2c^2 + 2ab^2c + 2abc^2 + 6abc + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c) / (a^2b^2c^2 + a^2b^2c + a^2bc^2 ...

**Trigonometry**

the angle is x, where cos x = 15/20 the area is 15√(20^2-15^2)

**Algebra**

1.2^-t = 1 / 1.2^t That is clearly not the same as 1 / 2^t recall that x^-a = 1/x^a

**Maths**

Since the diagonals of a rhombus are perpendicular, and all four sides are the same length, the rhombus can be divided into four right triangles with one leg of 5 and an hypotenuse of 8. So, the other leg is √39 So, the area of all four triangles is 10√39 m^2 Multi...

**maths**

I see a lot of perfect cubes there 8x^3 - 27y^3 = (2x-3y)(4x^2+6xy+9y^2) Hmmm. No joy there. wolframalpha.com factors it, but I can see no motivation for trying (2x-3y+5) as a factor.

**math**

area of ABCD = ab area of ABCD+path = (a+2c)(b+2c) area of path = (a+2c)(b+2c)-ab

**Calculus help**

for multiple periods per year, just divide the interest rate and multiply the number of years. (b) P = 2000(1+.057/12)^(12*10) and similarly for other divisions of the year into multiple parts.

**Algebra**

C is correct

**calc**

Note that since (x-3)^2 = x^2-6x+9, what you have is ∫√(25-(x-3)^2) dx If u = x-3, du = dx, and it's just ∫√(25-u^2) du which you should know. If not, just use the trig substitution u = 5sinθ The answer can be seen at http://www.wolframalpha.co...

**Algebra**

continuously: A(t) = 1000*e^.08t daily: A(t) = 1100(1+.08/365)^(365t) so, when are they equal? When 1000 e^.08t = 1100(1+.08/365)^(365t) t = 10,873 years. continuous compounding is so close to daily compounding (1.08328% vs 1.08327%) that it takes a long time to overcome the l...

**MATH**

look at the differences: 2,3,5,7 My guess is that the next difference will be 11, the next prime. If not for the 2, I'd have said odd numbers were the differences.

**maths**

The formula must have a typo. If not, you are saying that r(x) = 600x, or $600 per person. I suspect that the problem is incorrectly stated. It appears that the $20 discount is only being applied to those people in excess of the 25-passenger price break. That explains the -20x...

**maths**

40m north and 40m south cancel out, leaving only the 30m east.

**maths**

draw a diagram, and you will see a 3-4-5 right triangle. So, the final displacement is 5, the hypotenuse. The direction (measured from due east) is x, where tan(x) = 40/30

**maths**

-j x i = k

**maths**

zero

**maths**

|A+B|^2 = 2^2 + 2^2 - 2(2)(2)cos 120°

**math**

.327... = 327/999 = 109/333 1000x = 327.327327... x = 0.327327... 999x = 327

**Maths**

(i) 3*b = 2*3+b = 6+b b*5 = 2*b + 5 = 2b+5 so, 6+b = 2b+5 b = 1 (ii) 3*d = 6+d d*(3*d) = 2d+(6+d) = 3d+6 d*3 = 2d+3 (d*3)*d = 2(2d+3)+d = 5d+6 so, 3d+6 = 5d+6 d = 0

**math**

first, note that if u = 6x, you have 1/6 ∫ sin^5 u du = 1/6 ∫sin^4(u) sinu du = -1/6 ∫(1-cos^2 u)^2 (-sinu du) If v = cos u, then dv = -sinu du, and you have -1/6 ∫(1-v^2)^2 dv just expand that and it's a regular old polynomial.

**Physics . pls help me :)**

we need to find where the cars have gone the same distance. So, 0 + 0t + (1/2)(2)t^2 = 30 + 4t that is, t^2 - 4t - 30 = 0 t = 7.83 Just plug that value in for t to see where they met.

**math - eh? eh? eh?**

that cannot possibly be how the question was presented. why not post it as it was given?

**mathematics**

well, just plug and chug: u = ln(x^2+y^2) - ln(x+y) ∂u/∂x = 2x/(x^2+y^2) - 1/(x+y) ∂u/∂y = 2y/(x^2+y^2) - 1/(x+y) x ∂u/∂x + y ∂u/∂y = 2x^2/(x^2+y^2) - x/(x+y) + 2y^2/(x^2+y^2) - y/(x+y) = 2(x^2+y^2)/(x^2+y^2) - (x+y)/(x+y) = 3 so...

**mathematics**

the volume is just |u•v×w| That's pretty straightforward. What do you get?

**mathematics**

assuming you meant t=π/6 The unit tangent vector T(t) = r'/|r'| = cot(t) i - tan(t) j ------------------------- tan^2(t) + cot^2(t) The unit normal is T'/|T'| = -(csc^2(t) i + sec^2(t) j) ---------------------------------- csc^4(t) + sec^4(t) which at t = ...

**mathematics**

find a line in each plane the cross product of those two vectors is perpendicular to them both. Then translate the vector so it passes through P.

**mathematics**

for all x ≠ -3, (x^2-9)/(x+3) = x-3 So, the limit as x->0 is just -3. How did you get -6?

**shizo homework dump**

if you're going to just dump your homework, at least use a consistent name... Lots of problems, no evident effort displayed on your part.

**maths**

R(x) = 120x if x <= 25 100x is 25 < x <= 40

**Algebra**

B = 2A-8 B = C+32 A+B+C = 152 C = B-32 = 2A-8-32 A + 2A-8 + 2A-40 = 152 5A - 48 = 152 5A = 200 A=40 B=72 C=40

**Algebra**

(3+√18) / 4√12 = (3+3√2) / 8√3 = 3√3 (1+√2) / 24 = (1+√2)√3 / 8 = (√3+√6)/8 or several other ways to express it

**Calculus I**

f = ax e^(bx^2) f' = (a + ax(2bx)) e^(bx^2) = a(1+2bx^2) e^(bx^2) So, f achieves a max/min where f'=0 That is, where a(1+2bx^2) = 0 Or, where x^2 = -1/2b So, it looks like we need b < 0, since x^2 is always positive Not surprisingly, the value of a does not affect t...

**algebra**

I will assume the usual carelessness with parentheses, and that you mean (7y+5)/3 = -8/(y-3) Clear fractions (assuming y≠3) and we have (7y+5)(y-3) = -8(3) 7y^2 - 16y - 15 = -24 7y^2 - 16y + 9 = 0 (7y-9)(y-1) = 0 y = 1 or 9/7

**Integral question**

for ∫x^3/(x^4+1) dx let u = x^4+1, and so du = 4x^3 dx. So, your integral becomes ∫ 1/4 du/u which should look familiar

**Integral question**

∫[-1,1] 5sinx - 2tanx + 3x^5 dx If you look carefully, all three terms are odd functions, so the whole integrand is odd. So, the integral from -1 to 1 is zero. However, if you want to do the integration, note that since tanx = sinx/cosx = -d(cosx)/cosx, we have -5cosx + ...

**Derivatives**

x secy + y^5 = 5x^3 - 7 secy + x secy tany y' + 5y^4 y' = 15x^2 (x secy tany + 5y^4)y' = 15x^2 - secy So, y' = 15x^2 - secy ---------------------------- x secy tany + 5y^4

**math**

The 6.02 looks good, but you should have added exponents 6.02 * 10^(5 + -13) = 6.02 * 10^-8

**Calc please help**

just use the chain rule: f = u^4 u = csc(v) v = 2x^2+1 f' = 4u^3 u' u' = -csc(v) cot(v) v' v' = 4x So, f' = 4 (csc(2x^2+1)^3 * (-csc(2x^2+1)*cot(2x^2+1)) * (4x) = -16x csc^4(2x^2+1) cot(2x^2+1)

**Calc please help**

Oops. A typo. Should be (x*cosx-sinx)/(x*sinx) just take derivatives, top and bottom. The limit is the same as for (-x*sinx+cosx-cosx)/(sinx + xcosx) = (-x sinx)/(sinx + x cosx) still -> 0/0, so do it again: (-sinx - x*cosx)/(cosx + cosx - x*sinx) -> 0/2 = 0

**Calc please help**

massage the form to 1/x - cosx/sinx = (cosx-x)/(x*sinx) and apply l'Hospital's Rule

**Math**

(1.9/9.5) * 10^(-6-4) = 0.2 * 10^-10 = 2.0 * 10^-11 How did you get 1.6?

**Algebra**

so, write it clearly: limit as x->1 of (x-1)/(√x - 2x) I see no problem here, since √x - 2x -> -1, which is not zero. (x-1)/(√x-2x) -> 0 Apparently I have it wrong, but cannot see an easy way to massage it into something which -> 0/0 Anyway, try l...

**math**

millions could be 5,7,9 in any order No idea what "all the other digits are 21" means, but surely you can work it out now.

**math**

the beads recur in cycles of 5 colors. 85 is the largest multiple of 5 which is less than 90. So bead #85 is red. After that is yellow,blue, making 87 beads in all. I have no idea what "the first bead is yellow, and the last bead is blue for each color" means, since ...

**math**

in tiles, the pool is 50*10 x 20*10, or 500 x 200. So, how many tiles is that? Or, you can do it by dividing the areas: 50mx20m / .1mx.1m = 1000/.01 tiles

**math**

just multiply the two numbers. I assume you want the dollar value. 7.25kg is always the same, regardless of its price...

**not maths**

and probably not even homework

**maths**

24*12*6 = ?

**maths**

a cloud

**maths**

assuming you meant 16x^2 + 8x + 7, it's just a parabola, with its vertex at x = -b/2a = -8/32, or -1/4 So, just evaluate f(-1/4) to find the minimum. Since the coefficient of x^2 is positive, there is no maximum. Of course, there may be other typos hiding in there.

**Algebra**

for higher-degree polynomials, try some synthetic division to try and find rational roots. In this case, there are none, so you're stuck with graphical or numeric methods. A check on the graph will show that there is a single real root, near x = 2.7, but that's about i...

**science**

just conserve momentum: 2900<34,0,0> + 4800<-18,0,28> = (2900+4800)v <98600,0,0>+<-86400,0,134400> = 7700v <12200,0,134400> = 7700v v = <1.58,0,17.45>

**math**

Since the curve intersects the line at (-2,1) and (2,1), we can use vertical strips to get a = ∫[-2,2] 1 - (x^2-3) dx = 32/3 Or, we can use horizontal strips to get x = ±√(y+3) a = ∫[-3,1] √(y+3) - (-√(y+3)) dy = 32/3

**science**

conserve momentum: .062*150 + 90*0 = (.062+90)v v = 0.103 m/s

**Precalculus**

the diagonal of the rectangle is a diameter of the circle. Since that is 4, if the rectangle has one side of length x, then the other side has length √(16-x^2) So, the area is x√(16-x^2) Well, that is not a choice, so if the side is 2x, rather than x, the area is (...

**calculus**

use the chain rule. let v = -x^2 u = cot v y = 1/2 u^7 y' = 7/2 u^6 u' u' = -csc^2 v v' v' = -2x So, dy/dx = 7/2 cot^6(-x^2) * -csc^2(-x^2) * -2x = 7x cot^2(-x^2) csc^2(-x^2) or, since cot(-x) = -cotx csc(-x) = -csc(x) 7x cot^2(x^2) csc^2(x^2)

**ALGEBRA**

well, let's see 900 lbs capacity 100 lbs luggage subtract to get ?? lbs for passengers

**math**

-11 + 7 + x = 15 -4 + x = 15 x = 19

**science**

163/(163+755) = 163/918 = .1776 or 18%

**ALGEBRA**

8x > 72

**math**

any number rounded to the nearest hundred is a multiple of 100, and so cannot be odd. I suspect that is not the way the question was worded.

**Algebra**

do things just as you would with numbers: (2+5)(7-3) 2*7 + 5*7 - 2*3 - 5*3 You have (n^3+8)(n-4) n^3*n + 8*n - n^3*4 - 8*4 now simplify each term n^4 + 8n - 4n^3 - 32 and arrange in descending powers n^4 - 4n^3 + 8n - 32 Sometimes you may have several terms with the same power...

**Maths**

each side is now √425.5969 = 20.63 Reduce by .03 to 20.6

**Maths - eh?**

no idea what that sentence means. what "same" number?

**Math**

140mi/14g = 140/14 mi/g = 10 mi/g

**physics**

current speed is 2.6 sin 28.5° boat velocity across is 2.6 cos 28.5°

**math**

I'll answer in mg, since 1/2 pint is more than 1 fl oz. Since 1/2 pint = 8 fl oz, there are 8 * 1.2 = 49.6 mg of otylmethoxycinnamate.

**math**

.02x = .23 x = 11.50

**:-)**

Google is your friend. It gave several hits, and this one was indeed elegant and simple.

**MathMate Help!!!**

oops - make that 1/a + 1/b < 1/c < 1/a - 1/b 1/84 < 1/c < 13/84 84/13 < c < 84 So, 83 is the maximum integer side

**MathMate Help!!!**

We know that given sides A,B,C, with A>=B, A-B < C < A+B With a little algebra (as seen here): http://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.577996.html we know that if the altitudes are a,b,c then 1/a - 1/b < c < 1/a + 1/b So, for you...

**pre cal**

I see a whole bunch of posts, but no indication of your attempts to solve them. I'll do one of them here: f(g) = g^2-3 = (√(3+x))^2 - 3 = 3+x-3 = x g(f) = √(f+3) = √(x^2-3+3) = √x^2 = x However, note that √x^2 is not necessarily x. If x = -3, ...

**pre cal**

Ok - do you know what (f+g)(x) means? It is f(x)+g(x) Now just plug in your function definitions, and simplify the result. Just to make things easier, lose all the words. You have f(x) = √(5x+7) g(x) = √(5x-7)

**math**

I guess she gives 24/25 of a chocolate to each kid. Or, she can give 1/2 a chocolate to two of the kids, and a whole one to the others.

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