Sunday

February 14, 2016
Total # Posts: 37,659

**MATHS**

Clearly the large cube is 6x6x6. So, how many 2x2x2 cubes is that?
*January 2, 2016*

**Algebra**

what? No ideas? The fee is constant.
*January 2, 2016*

**Calculus help plz**

I suspect a typo. Wolframalpha indicates that it cannot be done with elementary functions: http://www.wolframalpha.com/input/?i=integral+%28%28x%2B1%29%2F%28x-1%29%29^%281%2F3%29%2F%28x%2B1%29dx If I got it wrong, and only the (x+1) is in the cube root, it can then be done, ...
*January 2, 2016*

**math**

naturally, 1-.2 = .8
*January 2, 2016*

**physcis**

of course, you have to watch the units (2π√80 cm)/(10 m/s^2) mixes up cm and m. Convert cm to meters by (2π√80 cm)(1m/100cm)/(10 m/s^2) = π√80/500 s^2 = 0.056 s^2 An odd quantity, s^2
*January 1, 2016*

**math- HELP!**

Oww! My bad! Should have caught my typo. 26π*200 in/min The ridiculous speed of less than 1/2 inch per minute should have caught your eye!
*January 1, 2016*

**math- HELP!**

26π in/rev * 200rev/min = 26π/200 in/min w is not 200. It is 200*2π radians/min
*January 1, 2016*

**Math**

You have the equations x+y=−2 y=13x+2 You know it is a 2-D graph The lines intersect at (-2/7,-12/7), so none of the choices listing a different intersection can be right. That leaves only (A). Now you just have to check to see whether the lines pass through the named ...
*January 1, 2016*

**Math**

so, did you graph it? Which choice matches your findings?
*January 1, 2016*

**Math**

(x + 1/x)^2 = 3 x^2 + 2 + 1/x^2 = 3 x^4 - x^2 + 1 = 0 x = ±i^(1/3), ±i^(5/3) using i^(1/3), we have x^37 = i^12 i^(1/3) = i^(1/3) x^31 = i^10 i^(1/3) = -i^(1/3) x^47 = i^15 i^(2/3) = -i i^(2/3) x^26 = i^8 i^(2/3) = i^(2/3) x^63 = i^21 = i x^9 = i^3 = -i Looks ...
*January 1, 2016*

**Math**

5^3x = 65 3x log5 = log65 3x = log65/log5 x = log65/3log5 I assume you meant 2^(2x+3) = 80 (2x+3)log2 = log80 2x+3 = log80/log2 x = (log80/log2 - 3)/2
*January 1, 2016*

**science**

google is your friend. 1st hit solved the problem.
*January 1, 2016*

**computer programming**

use the function to calculate the nth fib #, and return its value. Then loop for p=1..5 and print each value returned.
*January 1, 2016*

**Calculus need help**

do you mean (x+x^3)/y or x + x^3/y ? It makes a difference
*January 1, 2016*

**Math**

(a) clearly, a = (5x+15)(31-2x) (b) I think you can graph a parabola (c) values for zero area (d) use what you know about parabolas (e) ditto (f) since this is a real-world situation, negative lengths are meaningless. So, D = [-3,31/2] R = [0,<value at vertex>]
*January 1, 2016*

**MATHS**

see related questions below.
*December 31, 2015*

**Algebra**

well, 1/12 of w is w/12 so, what's 4 more than that?
*December 31, 2015*

**Math**

y = (7/8)^x log(y) = x log(7/8) ... y=log12x 10^y = 12x ... and similarly for the last one.
*December 31, 2015*

**Calculus**

Each slice of water is a rectangle. Its length is 10. If the water has depth y, then the width of the rectangle is √(1 - (1-y)^2) = √(2y-y^2) Now you have to figure the water's depth (h), and the height to pump each slice of water is 2-y. So, you have to ...
*December 31, 2015*

**Math**

car 1 is 1/6 * 65 = 65/6 miles ahead when car 2 starts. car 2 is going 10 mi/hr faster, so it will take (65/6)/10 = 65/60 hr to catch up. In other words, 65 minutes.
*December 31, 2015*

**Math**

just add powers for multipliers, subtract for divisors. f^3 g^-1 / f^2g^3 = f^(3-2) g^(-1-3) = f^1 g^-4 = f/g^4 (k^−2m^3n)×(kmn^5)÷(k^− 1m^2 n) = k^(-2+1-(-1)) m^(3+1-2) n^(1+5-1) = k^0 m^2 n^5 = m^2 n^5
*December 31, 2015*

**Math**

domain is all reals since 2^x > 0, 2^x-7 > -7, so the range is (-7,∞) I expect you can do the intercepts, right? Just set x=0 or y=0.
*December 31, 2015*

**Maths need help trigonomentery**

If the two radii are x and y, then π(x^2+y^2) = 29π/2 2π(x+y) = 10π factoring out the π, x^2+y^2 = 29/2 x+y = 5 Now you should be able to solve for x and y
*December 31, 2015*

**Geometry**

check out the angle bisector theorem.
*December 31, 2015*

**Operating system**

Just a gut feeling, I'd say it is not. The MRU page is likely to be the next page accessed, don't you think? Swapping it out is likely to be a waste of time.
*December 31, 2015*

**geometry**

CD will be parallel to AB, and of the same length. B-A = (2,2) So, D = C+(2,2) = (0,8) or C-(2,2) = (-4,4)
*December 31, 2015*

**math**

r = (a+b-c)/2 = 2
*December 31, 2015*

**Maths**

If the digits are t and u, then the value of the number is 10t+u. So, we have t+u=7 10u+t = 10t+u-9 Now just solve for t and u. You can do this without any algebra. I mean, how many pairs of numbers add up to 7? Just try them.
*December 31, 2015*

**computer**

read x,y if y == x*x then { yes } else { no }
*December 31, 2015*

**maths**

well, 6000 is 5/12 std above the mean. Look that up in your Z table. You can play around with this stuff at http://davidmlane.com/hyperstat/z_table.html
*December 31, 2015*

**Math**

you want to solve for t in 400t - 16t^2 = 2400 Then find the difference in the two times.
*December 31, 2015*

**probability**

well, using brute force, just list the ways you could pair up the socks. How many of those ways fit the requirements?
*December 30, 2015*

**Physics**

for inital velocity v, we have h(t) = vt-(g/2)t^2 This reaches a max height at t=v/g, and h(v/g) = v(v/g) - (g/2)(v/g)^2 = v^2/g - v^2/2g = v^2/2g So, we see that max height is directly proportional to v^2 So, double v, and you quadruple the max height. M does not matter, as ...
*December 30, 2015*

**Math**

In the first G.P. the numbers are a, ar, ar^2 To form the A.P., the differences are constant, so ar-a = ar^2-4 - ar Then, forming a new G.P. with constant ratio, we have (ar-1)/a = (ar^2-4-1)/(ar-1) Solving for a and r, we have a=1, r=3 or a = 1/9, r=7 The 1st G.P. is 1, 3, 9 ...
*December 30, 2015*

**physics(optics)**

nothing.
*December 30, 2015*

**Physics**

well, the acceleration is 4 m/s^2, and F=ma.
*December 30, 2015*

**Math advanced function**

Since a quadratic can two solutions, and a trig function can two solutions per period, I expect that 4 solutions is as many as you can have. For example, sin^2(x) = 1/4 has 4 solutions in one period. sin^2(x) = 2 has none, sin^2(x) = 1 has two Maybe you can find some with an ...
*December 30, 2015*

**Math**

Draw the triangles. Recall that the sides are in the ratio 1:√3:2 Label the longest hypotenuse, and then work down to get the desired leg's size. I get 6. I am assuming there are only two triangles.
*December 29, 2015*

**Physics**

check your previous post.
*December 29, 2015*

**maths**

In that case, recall that one log is the reciprocal of the other. So, If you let u = log3x, then you have u + 1/u = 5/2 2u^2 - 5u + 2 = 0 (2u-1)(u-2) = 0 u = 1/2 or u=2 Thus, u = √3 or 9
*December 29, 2015*

**maths**

Or, did you mean log3x+logx3 = 2.5 ?
*December 29, 2015*

**math**

If the lawn has side x meters, then lawn area: x^2 total area: (x+1)^2 path area: (x+1)^2-x^2 = 2x+1 So, we have x^2 = 2x+1 + 23 Now just find x.
*December 29, 2015*

**Physics**

original: <0,-398> wind: <-104,0> result: <-104,-398> = 411 at 195° or 411 at S15°W To get back on course, you just need to negate the wind. So, you want <x,y> + <-104,0> = <0,-398> <x,y> = <104,-398> or 411 at S15°E
*December 29, 2015*

**stats**

9 dancers total 5 warmed and had no injuries (W & ~I) You want ~W or I = ~(W & ~I) So, your probability is 1 - 5/9 = 4/9
*December 29, 2015*

**maths**

well, 3000 is 20 times 150, so it will take 20 times as long, right?
*December 29, 2015*

**Maths**

which two factors of 15 differ by 2?
*December 29, 2015*

**Factorize**

a^3-b^3 = (a-b)(a^2+ab+b^2) so, x^3 - (x+y)^3 = (x-(x+y))(x^2+x(x+y)+(x+y)^2) Now just simplify that.
*December 29, 2015*

**Maths**

well, just substitute: √(1-x^2) = √(1-sin^2 θ) Now, what is your most basic trig identity?
*December 29, 2015*

**chemistry**

same as 1 mole of any compound: Avogadro's Number of molecules. Maybe you should review molar mass...
*December 29, 2015*

**Wave**

look at the units: f = 1/s speed = m/s So, wavelength (m) = speed/f
*December 29, 2015*

**Math**

w+d = 46 (w+4)/(d+4) = 7/2 Now just solve for w and d.
*December 29, 2015*

**St. Francis English medium school.**

If the price is each, then we have cost: 110 revenue: 110, with one left over On the other hand, he made a gain of 10% one each clip sold.
*December 28, 2015*

**algebra**

If the length is y, then the width w is 1 + y/4. So, 2(1 + y/4 + y) = 62 Now just solve for y, and you can find w.
*December 28, 2015*

**Math**

If you mean (2x+7)/(2x^4+21x^3+35x^2-37x+46) then that is approximately (2x+7)/((x+7.966203)(x+3.527066)(2x^2-1.98654x+1.63717)) Now you can just use partial fractions on that mess.
*December 28, 2015*

**mathematics**

And assuming the usual sloppiness in the numerator, we have (x²+x+4)/[(1-x)(x²+1)] = (2x+1)/(x²+1) + 3/(1-x) amazing how little difference it made, eh?
*December 28, 2015*

**maths**

for x in [0,10000), f(x) = 900 for x in [10000,15000), f(x) = 900+.02(x-10000) for x in [15000,25000] f(x) = 900+.02(x-10000) + 500 I think you can discuss the continuity. If not, draw the graph and it will become clear.
*December 28, 2015*

**mathematics**

this cannot be more straightforward, if you have read the relevant section in your text. Namely, the equation for a circle with center at (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2 Surely you can now see what to do...
*December 28, 2015*

**<b>͡° ͜ʖ ͡°</b>**

:-)
*December 28, 2015*

**pi -- NOT pie!**

grrr
*December 28, 2015*

**PHYSICS**

Resolve F1 and F2 into x- and y-components, and then just add them up.
*December 28, 2015*

**physics**

conserve momentum: 5(-10) + 6(12) = (5+6)v That will give the final velocity v. Then find KE (1/2 mv^2) for each body before and after the collision.
*December 27, 2015*

**physics**

since C=2πr, just multiply the circumference by the revs/min. That will give the speed in cm/min.
*December 27, 2015*

**Math**

I suggest that would be 3/5 * 3 = 9/5 mi
*December 27, 2015*

**Calculus**

Of course it works. Why don't you show your work and the reason why you rejected it? 2/((2-x)(x+2)^2) = (1/8) (1/(x+2) + 4/(x+2)^2 - 1/(x-2)) So, the integral is just (1/8) (ln(x+2) - 4/(x+2) - ln(x-2)) = (1/8) (ln((x+2)/(x-2)) - 4/(x+2)) + C
*December 27, 2015*

**math**

for 2 cm, r=2. So, just figure the circumference of the circle of radius 2... 100/3 rev/min * 2π*2 cm/rev = 400π/3 cm/min similarly for r=15.
*December 27, 2015*

**komenco**

Assuming you mean Geometric Sequence, we have For The AP, a=7, so we want d. The ratio between terms of a GP is constant, so we form the ratio between successive given terms of the AP: (7+d)/7 = (7+4d)/(7+d) d=14 or d=0 Now, d=0 is not very interesting: AP = 7,7,7,7,7... GP...
*December 27, 2015*

**mustapha**

well, first evaluate 10^(1/3) = 2.1544 Then just multiply as usual.
*December 27, 2015*

**Math**

the derivative of sec(u) is sec(u) tan(u) But, if u is a function of x, then we have to apply the chain rule. Since u=tanx, then du/dx = sec^2(x) So, the final damage is sec(tanx) tan(tanx) sec^2(x)
*December 27, 2015*

**math**

well, what is half the circumference of the circle? Then just multiply that by the cost per foot.
*December 27, 2015*

**Math**

I think you've both gone stray. See http://www.wolframalpha.com/input/?i=derivative+%28sinx%2F%281%2Bcosx%29%29 and http://www.wolframalpha.com/input/?i=derivative+%28sinx%2F%281%2Bcosx%29%29^2
*December 27, 2015*

**math**

Because y(0) = 0. It helps if you actually read what I write. Or maybe it has something to do with the mysterious "a".
*December 27, 2015*

**math**

well, f'(x) = cos(√2 x) - √2 x sin(√2 x) Not sure what a has to do with things, but if you meant x=0, then f(0) = 0 f'(0) = 1 So, you want the line with slope 1, which passes through (0,0). No too much of a challenge there, eh?
*December 27, 2015*

**math**

cancel what out? There are no common factors.
*December 27, 2015*

**math**

well, the 1st three terms are 2^2, 3^3, 4^4, so what's your best guess?
*December 26, 2015*

**physics**

As I recall, the force grows as the ratio of the areas of the pistons. That means, as the square of the diameter. So, since 6 = 3*2, the force would be 3^2 * 20 Review your text on this subject. I'm sure there are examples. And you can always rely on google for lots of ...
*December 26, 2015*

**Calculus**

rearranging things a bit, we have y' = y(t+1) + t+1 y' = (y+1)(t+1) dy/(y+1) = (t+1)/t^2 dt ln(y+1) = ln(t) - 1/t + c y+1 = t e^(-c/t) + c2 or, altering the c values, y = ct e^(-1/t) + c2 y(1) = 0, so 0 = c/e + c2, so c2 = -c/e y = c(t e^(-1/t) - 1) We can ignore the c...
*December 25, 2015*

**Math**

how about if we start by saying y-1 = 3sin((1/2)(x+pi)) and y+2 = -1/3 cos(2(x - pi/2)) the shifts and scalings should now be clearer.
*December 25, 2015*

**math**

There are infinitely many chords passing through any point within the circle. Two such chords, easy to describe are x=2 and y=3.
*December 25, 2015*

**Statistics**

18 rings total, so that would make the probability 10/18 * 9/17
*December 25, 2015*

**math**

There are infinitely many chords passing through any point within the circle. Two such chords, easy to describe are x=2 and y=3. ---------------------------------------- Clearly, the vertex of the triangle is at (0,5), so the altitude is 3. The base of the triangle goes ...
*December 25, 2015*

**math**

Draw diagonal BD Then triangles ABD and CBD are isosceles. Hence, their altitudes bisect their bases and their vertex angles.
*December 25, 2015*

**physics**

54.45 mph = 79.86 ft/s If the rock was thrown with an initial speed k, then k-32*2.75 = -79.6 k = 8.4 ft/s So, the rock was thrown upward at 8.4 ft/s from a height h: h - 32*2.75 - 16*2.75^2 = 0 h = 209 ft
*December 25, 2015*

**mathematics**

(x-1)(x-2)(x-k) = x^3 - (k+3)x^2 + (3k+2)x - 2k k+3 = p 3k+2 = 2 q = -2k Looks like k=0, so p=3 q=0 and we have x(x-1)(x-2) = x^3 - 3x^2 + 2x
*December 25, 2015*

**maths**

or, using partial fractions, 2 - (5/9)/(x+2) - (13/9)/(x-1) + (1/3)/(x-1)^2
*December 25, 2015*

**Trigonometry- ??????**

oh come on. If ab=0, then either a=0 or b=0 So, we have tanx = 0 --> x is a multiple of pi or sin^2(x) = 2 --> no solutions there.
*December 24, 2015*

**Trigonometry- ??????**

tan(x) sin^2(x)=2tan(x) tanx(sin^2(x)-2) = 0 How about now?
*December 24, 2015*

**Algebra 1**

39.80x + 49.70y <= 150,000
*December 24, 2015*

**Math**

looks good to me. or, you could have noted the 41-22 = 19%, and that would be 76.
*December 24, 2015*

**math**

the units would be dollars/yd^2 The rate of change is always ∆y/∆x, so the associated units just follow along. y might be the cost (dollars) of some quantity x (yd^2) of carpet.
*December 24, 2015*

**Calculus**

u = sinx du = cosx dx so, dx = du/cosx = du/√(1-u^2) sinx(cosx)^2 dx = u(1-u^2)/√(1-u^2) du = u/√(1-u^2) du Now, let v = √(1-u^2) dv = -u/√(1-u^2) du and you have -v dv integrate that to get -1/2 v^2 = -1/2(1-u^2) = -1/2 (1-sin^2(x)) = -1/2 cos^2(...
*December 23, 2015*

**Calculus**

because now we are using u, not x. If, after integration on u, you substitute back into terms of x, then you would use the limits [0,2]. When x=0, u=3 and when x=2, u=5.
*December 23, 2015*

**Calculus**

∫[0,2] x√(5-√(4-x^2)) dx Let u = 5-√(4-x^2) du = x/√(4-x^2) dx so, x dx = (5-u) du and you now have ∫[3,5] √u (5-u) du and I'm sure you can handle that, eh?
*December 23, 2015*

**Math**

n is starting population. t is time period k indicates how fast it grows. To use it, um, just plug in your numbers... Surely, knowing the formula, you have examples of its use. If not, google is your friend.
*December 23, 2015*

**Algebra 2**

you have the sum of two cubes, so it factors into (3x+5)((3x)^2-(3x)(5)+5^2) = (3x+5)(9x^2-15x+25) It's easy to check that the quadratic has no real solutions. Just use the quadratic formula, learned long ago in Algebra I...
*December 23, 2015*

**MATH Please help!**

if the dimensions increase by a factor of 3, the area increases by a factor of 3^2, so A is correct. 13^2 * 3^2 = (13*3)^2 = 39^2
*December 23, 2015*

**math**

domain is positive integers range is (0,N] where N is the maximum time for a single painter. Further restrictions depending on the maximum number of painters, and other information on their required times.
*December 22, 2015*

**math**

clearly, the mass depends on the number of candies, which is an easily controlled integer value.
*December 22, 2015*

**Gwu**

I think that is supposed to read and spends it takes .75mi * 1hr/2mi = .375hr = 22.5 min to travel between stops. So, assuming we start counting as the train arrives at the 1st intersection, it takes (in minutes) 12*.75 + 11*22.5 = 256.5 minutes before it leaves the final ...
*December 22, 2015*

**SCIENCE**

surely it would have taken less time just to look in your text, or google, rather than post here and wait.
*December 22, 2015*