Tuesday

September 2, 2014

September 2, 2014

Total # Posts: 24,555

**algebra 1**

well, the triangle obviously has area = 12 So, the whole outfit has area 6x+12. So, 6x+12 = 150 6x = 138 x = 23
*July 1, 2014*

**math**

surely you can count days on a calendar...
*July 1, 2014*

**Math**

If you subtract the 2nd equation from the first, you get 4x-3y-4 = 0 4x+2y+5 = 0 ---------------- -5y-9 = 0 now add 9 to both sides: -5y = 9 now divide by -5: y = -9/5 Knowing that y = -9/5, just plug it into either of the original equations and then solve for x. The work is ...
*July 1, 2014*

**algebra**

20m
*July 1, 2014*

**math**

Just think about it. Every second it travels 1 meter. So, 5 seconds ago it was down 5 meters. now think forward in time for 120 meters above.
*July 1, 2014*

**math**

if now is time t=0, then the temperature at any time t is -2t. So, we want -2(-3) and -2(5)
*July 1, 2014*

**Math**

that would be 420/12 = ?
*July 1, 2014*

**math**

If they were laid out flat, they'd each be 4.8 units long. So, they must be more than that, and since 4.8 is not an acceptable answer, 4.9 is what we have to go with.
*July 1, 2014*

**math**

The midpoint of two points is (6,2). One of the points is (3,5). Since the two ends must be equidistant from the midpoint, 3 = 6-3, 9 = 6+3 5 = 2+3, -1 = 2-3 So, (9,-1) is at the other end.
*July 1, 2014*

**Physics**

it gained (18-12)=6 m/s in (4-2)=2 seconds, so that is 6/2 = 3 m/s^2 In the first 2 seconds, then, it gained 6 m/s, so it started at 6 m/s.
*July 1, 2014*

**geometry**

hmmm. you have the formula. What's the trouble? The same formula applies for ym: (y1+y2)/2 A value midway between two others is just their average: the sum/2.
*June 30, 2014*

**10 gr math**

ok. plug in the data and evaluate: the distance between (1,3) and (7,2) is √((7-1)^2 + (2-3)^2)
*June 30, 2014*

**Chemistry**

When reviewing the procedure the student found that 4.912 x 10-1 M barium nitrate solution had been used instead of that in the following procedure: A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 5.912 x 10-1 M barium nitrate, Ba...
*June 30, 2014*

**algebra**

A. No - the decrease in intensity is not a constant, but depends on the previous intensity. The intensity I of x meters of cable is I = 1500 * 0.964^x C yes you can probably write that one on your own.
*June 30, 2014*

**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
*June 30, 2014*

**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.
*June 30, 2014*

**Non**

just a note: that's mu sub-k, not "muse of k" µk
*June 30, 2014*

**Volume**

Just figure the tank's volume (66.75*132*10) and divide it by 231 in^3, the volume of a gallon.
*June 30, 2014*

**algebra**

x^2+8x = x(x+8) Pick any positive value for x and you get a rectangle.
*June 30, 2014*

**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
*June 30, 2014*

**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...
*June 30, 2014*

**science**

I may have messed up the sign change, so double check my math.
*June 30, 2014*

**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
*June 30, 2014*

**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...
*June 30, 2014*

**Math**

54° is 3/20 of a full circle. SO, the whole circumference 3/20 c = 9 c = 60
*June 30, 2014*

**ALGEBRA - eh?**

cannot parse the "8n 12" part
*June 30, 2014*

**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.
*June 30, 2014*

**trigonometry**

well, we have h/120 = tan 10°18' h = 21.8 ft
*June 30, 2014*

**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.
*June 30, 2014*

**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.
*June 30, 2014*

**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
*June 29, 2014*

**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
*June 29, 2014*

**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?
*June 29, 2014*

**Algebra**

y(x) = 3*2^x y(x+1) = 3*2^(x+1) = 3*2*2^x = 2y(x) So, each year y doubles.
*June 29, 2014*

**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 ...
*June 29, 2014*

**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.
*June 29, 2014*

**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
*June 29, 2014*

**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.
*June 29, 2014*

**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 ...
*June 29, 2014*

**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? ...
*June 29, 2014*

**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...
*June 29, 2014*

**Trigonometry**

the angle is x, where cos x = 15/20 the area is 15√(20^2-15^2)
*June 29, 2014*

**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
*June 28, 2014*

**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 ...
*June 28, 2014*

**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.
*June 28, 2014*

**math**

area of ABCD = ab area of ABCD+path = (a+2c)(b+2c) area of path = (a+2c)(b+2c)-ab
*June 27, 2014*

**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.
*June 27, 2014*

**Algebra**

C is correct
*June 27, 2014*

**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....
*June 27, 2014*

**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 ...
*June 27, 2014*

**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.
*June 27, 2014*

**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...
*June 27, 2014*

**maths**

40m north and 40m south cancel out, leaving only the 30m east.
*June 27, 2014*

**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
*June 27, 2014*

**maths**

-j x i = k
*June 27, 2014*

**maths**

zero
*June 27, 2014*

**maths**

|A+B|^2 = 2^2 + 2^2 - 2(2)(2)cos 120°
*June 27, 2014*

**math**

.327... = 327/999 = 109/333 1000x = 327.327327... x = 0.327327... 999x = 327
*June 27, 2014*

**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
*June 27, 2014*

**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.
*June 27, 2014*

**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.
*June 27, 2014*

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

that cannot possibly be how the question was presented. why not post it as it was given?
*June 27, 2014*

**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...
*June 27, 2014*

**mathematics**

the volume is just |u•v×w| That's pretty straightforward. What do you get?
*June 27, 2014*

**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...
*June 27, 2014*

**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.
*June 27, 2014*

**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?
*June 27, 2014*

**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.
*June 27, 2014*

**maths**

R(x) = 120x if x <= 25 100x is 25 < x <= 40
*June 27, 2014*

**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
*June 26, 2014*

**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
*June 26, 2014*

**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 ...
*June 26, 2014*

**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
*June 26, 2014*

**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
*June 26, 2014*

**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 + ...
*June 26, 2014*

**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
*June 26, 2014*

**math**

The 6.02 looks good, but you should have added exponents 6.02 * 10^(5 + -13) = 6.02 * 10^-8
*June 26, 2014*

**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)
*June 26, 2014*

**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
*June 26, 2014*

**Calc please help**

massage the form to 1/x - cosx/sinx = (cosx-x)/(x*sinx) and apply l'Hospital's Rule
*June 26, 2014*

**Math**

(1.9/9.5) * 10^(-6-4) = 0.2 * 10^-10 = 2.0 * 10^-11 How did you get 1.6?
*June 26, 2014*

**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...
*June 26, 2014*

**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.
*June 26, 2014*

**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 ...
*June 26, 2014*

**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
*June 26, 2014*

**math**

just multiply the two numbers. I assume you want the dollar value. 7.25kg is always the same, regardless of its price...
*June 26, 2014*

**not maths**

and probably not even homework
*June 26, 2014*

**maths**

24*12*6 = ?
*June 26, 2014*

**maths**

a cloud
*June 26, 2014*

**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.
*June 26, 2014*

**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 ...
*June 26, 2014*

**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>
*June 25, 2014*

**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
*June 25, 2014*

**science**

conserve momentum: .062*150 + 90*0 = (.062+90)v v = 0.103 m/s
*June 25, 2014*

**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 (...
*June 25, 2014*

**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)
*June 25, 2014*

**ALGEBRA**

well, let's see 900 lbs capacity 100 lbs luggage subtract to get ?? lbs for passengers
*June 25, 2014*

**math**

-11 + 7 + x = 15 -4 + x = 15 x = 19
*June 25, 2014*

**science**

163/(163+755) = 163/918 = .1776 or 18%
*June 25, 2014*

**ALGEBRA**

8x > 72
*June 25, 2014*

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