# Posts by steve

Total # Posts: 51,743

maths geometry
(a) ?r^2h = 450 a = 2?r^2+2?rh = 2?r^2+2?r(450/?r^2) = 2?r^2 + 900/r Now just find r when da/dr=0 (b) is similar... ?3/4 s^2h = 450 a = 2(?3/4 s^2) + 3sh = ?3/2 s^2 + 3s(1800/?3 s^2) = ?3/2 s^2 + 1800?3/s ...

maths geometry
I object!

trigonometry and geometry
2x+y = 2000 a = xy = x(2000-2x) da/dx = 2000-4x da/dx=0 when x=500 So, the maximum area is 500*1000 m^2 Note that this is achieved when the available fence is equally divided between lengths and widths.

No, your equation cuts the amount in half every time t increases by 1. Since the half-life is 5730 years, you want the amount to be cut by half only that often. So, A9t) = 600(0.5)^(t/5730) To do a sanity check on your answer, note that 40000 years is about 7 half-lives, so ...

I = k/r^2 Now replace r with 5r, and you have I' = k/(5r)^2 = k/25r^2 = (1/25)(k/r^2) If r is replaced with nr, then I gets scaled down by a factor of 1/n^2

Maths
ever heard of google? It will provide lots of discussions on just what it is and what it is not.

Maths
It is defined that way. It makes possible the definition that n! = n(n-1)! for all integer n>0

Calculus
OK. Now you have ln(y-1) = -1/x + 1 y-1 = e^(-1/x + 1) y = e^(1 - 1/x) + 1

math
Maybe the sequence is squares of primes, since 4^2 and 6^2 are both missing.

The prism is regular, meaning its bases are squares, of side s. Now you know that h=13 4sh = 624 So, s = 12 Now you can work out the rest.

the area of each base is just 10*3=30 Then you have four rectangles. Two that are 10x12 and two that are 6x12 Now just add up all those six areas.

Maths
4/3 the speed means 3/4 the time: 7.5 min.

Differential Equations
I know that sometimes finding the solution is tricky, but surely just taking the derivatives is easy, no? y = c1-cos(x+c2) y' = -sin(x+c2) y" = -cos(x+c2) (y")^2 + (y')^2 - 1 = 0 sin^2+cos^2 = 1, right?

algebra
Looks good to me.

algebra
a good place to check your work is http://calc101.com/webMathematica/long-divide.jsp

math
420(1+log4204)

math
4:1 = 64:16

algebra
1/2 of all that

algebra
4-2(-3)^k.....-8+8(-3)^k 1-(-3)^k......-2+4(-3)^k

Calculus
(a) x=2 2y^2-4y = 6 y^2-2y-3 = 0 (y-3)(y+1) = 0 So, the points are (2,3) and (2,-1) (b) xy^2-x^2y = 6 y^2 + 2xyy' - 2xy - x^2y' = 0 y'(2xy-x^2) = 2xy-y^2 y' = (2xy-y^2)/(2xy-x^2) (c) y'(2,3) = 3/8 y'(2,-1) = 5/8 check: http://www.wolframalpha.com/input...

Math
well, it will be 5/2 as long as NL

Math
well, it will be exactly the same as any other three specific tosses. Each toss has a 1/2 chance of success.

Caclulu
what's the trouble? Just use the power rule. f(x) = 9x^-3 - 10x^-7 F(x) = -9/2 x^-2 + 10/6 x^-6 + C Now use F(1)=0 to find C

Caclulus
how about a question here? OK, I'll ask one. Is that function the position or the velocity? We don't mind helping with the answers, but asking us to come up with the question as well is bad form, no?

math
start off with u^2 = 1+sin(2x) 2u du = 2cos(2x) dx See where you get with that, and come on back with your work if you get stuck. recall that cos(2x) = 1-2sin^2(x)

math
the 17's cancel, so it's just 93/97

Physics
If the speeds change by v, then they must balance out. 4.90+v = 6.20-v 2v = 1.30 v = 0.65 So, the leading car ends up at 5.55 m/s the trailing car ends up at 5.55 m/s

value
Out of every 8 students, there are 5 boys and 3 girls. So, b/g = 5/3 g = 240 so, b = 400

math
p = kq^2/r, so pr/q^2 = k, a constant. So, you want to find q such that 200*2/q^2 = 36*4/3^2 400/q^2 = 16 q^2 = 25 q = 5

chemistry
cool. So, what?

calculus
y' = y^3/2 - 1/(2y^3) 1+y'^2 = (y^6+1)^2/4y^6 so, the arc length is s = 1/2 ?[1,2] y^3 + 1/y^3 dy = 33/16

Wayne C C C
W = PE = mgh

pre calc
what about when sin=-1 and cos=0? to make sure, let's work it out. Note that cos ?/4 = sin ?/4 = 1/?2 cos? - sin? 1/?2 cos? - 1/?2 sin? = 1/?2 cos(?+?/4) = 1/?2 ?+?/4 = ?/4 or 7?/4 ? = 0 or 3?/2

Chemistry
seems reasonable.

Chemistry
if you say so.

Algebra
If J is Jill's age now, then J+6 is her age in 6 years. But Amanda will also be 6 years older: A+6

MAth
x - x/2 - x/4 - x/8 = 3 x = 24

Physics
conserve momentum. Note that the actual value of m does not matter, since we have the relative masses. 1.4m*5.7 + m*4.7 = 2.4m*v 7.98+4.7 = 2.4v v = 5.28 m/s

Math 120
see related questions below

Calculus
Oops. That's not quite right. Can you spot my error? Also, it is no better to take horizontal rather than vertical strips. For vertical strips a = ?[0,?2] 4x - x/4 dx + ?[?2,?32] 8/x - x/4 dx

Calculus
as for the area, it's best to use horizontal strips (why?) of thickness dy. Then the area is a = ?[0,?32] 8/y - y/4 dy

Calculus
you are taking calculus, and you cannot draw two lines and an hyperbola? There are lots of online graphing sites; maybe you should spend some time working on this, rather than just carping and waiting. Try this: http://www.wolframalpha.com/input/?i=plot+y%3Dx%2F4,+y%3D4x,+y%...

Algebra
f(1) = -1+8+2 = 9 f(4) = -16+32+2 = 18 So, water traveled 9 ft in 3 seconds. Looks like (C) to me

Math
well, does your solution work? 4*4-9 = 16-9 = 7 Nope 4x-y-9=0 4x-(x-3)-9=0 You messed up right here It would be easier just to use the two values of y to find x: 4x-9 = x-3 3x = 6 x = 2 y = -1 4*2-9 = -1 2-3 = -1 That's better...

Math (Integrals) (Basic Integration)
that is correct. There is no du to work with. I mean, I did all the algebra for you, and explained the 2nd example to show why it worked there.

Math (Integrals) (Basic Integration)
Nope. Not that easy. x^4+1 = x^4+2x^2+1 - 2x^2 = (x^2+1)^2 - (?2 x)^2 = (x^2+?2 x+1)(x^2-?2 x+1) 1/(x^4+1) = 1/[(x^2+?2 x+1)(x^2-?2 x+1)] = 1/(2?2) ((x+?2)/(x^2+?2 x+1) - (x-?2)/(x^2-?2 x+1)) = 1/(2?2) ((2x+?2)/(x^2+?2 x+1) - x/(x^2+?2 x+1) - (2x-?2)/(x^2-?2 x+1) + x/(x^2-?2 x...

Math
that depends on a number of values not given here, as I'm sure you can tell. Better research the topic a bit more.

just multiply all the numbers from 1 to 4: 4! = 1*2*3*4 = 24 5! = 1*2*3*4*5 = 120 and so on

tangential velocity physics
google is your friend. Or, consider that the moon is roughly 250,000 miles from earth It takes about 28 days to orbit once. So, its linear speed is 2pi*250000mi/28days Find out the actual values, in meters and seconds, and that will give you its tangential speed.

Math (Integrals)
Just expand the polynomial and integrate all the terms: 4x^2-4x+1 dx

Math
well, 363 = µ - 2? what does the rule have to say about that?

Mathematics
since cosine has a period of 2pi, either A=X or |A-X| = k*2pi Since both A and X are less than pi/2, A=X

Maths
(s/2)^3 / s^3 = (s^3/8) / s^3 = 1/8

Math
there are 26 choices for each letter. So, multiply.

Math
500 - 8*45 = ?

algebra
looks good to me

math
a = 3 Since the GP has a common ratio (a+3d)/(a) = (a+12d)/(a+3d) (3+3d)/3 = (3+12d)/(3+3d) 1+d = (1+4d)/(1+d) d = 2 a+9d = 3+9*2 = 21

MAth
950000 * 0.90^5

Math - Calc
see whether this helps http://www.wolframalpha.com/input/?i=plot+y%3Dx%2B6,+y%3Dx%5E3,+2y%2Bx%3D0

calculus
I get S-S6 = 5.93*10^-6 < 1.0*10^-5

Algebra
I disagree. Why are you subtracting the last terms?

Math
timothy's coins were 10 more than pearly's. So, if the new amounts were t' and p', 10 more than pearly = 10+p' t' = 10+p'

Math
t = 5/8 p t + 1/4 p = 10 + 3/4 p now just solve for t

Calculus indefinite integrals
(a) u = x^2-7 (b) u = 1/5 x^(5/3) + 2

Pre calc
looks like you need to review your basic identities and double-angle formulas: cos^4x-sin^4x = (cos^2x-sin^2x)(cos^2x+sin^2x) = cos^2x-sin^2x = cos2x (B) is clearly not true, since sin^2x/(1+cos^2x) has no asymptotes

Calculus - Integrals
The curves intersect at (-1,2) and (-1,-2) So, using horizontal strips of width dy, the area is a = ?[-2,2] (3-y^2)-(y^2-5) dy Using vertical strips, we have to split the region in two at the intersections, and then symmetry helps: a = ?[-5,-1] 2?(x+5)dx + ?[-1,3] 2?(3-x)dx

Math
add 30 ft to each dimension (15 ft on each side) then do perimeter as usual

algebra
vertical shrink by 4 shift down 12

Algebra
y = (3x^2+8x-10)/(x^2+7x+12) = (3x^2+8x-10) / (x+3)(x+4) no holes, since y is never 0/0 vertical asymptotes where the denominator is zero: x = -3,-4 as x gets huge, y -> 3x^2/x^2 = 3 so that is the horizontal asumptote http://www.wolframalpha.com/input/?i=(3x%5E2%2B8x-10)+%...

Math
The average lies somewhere between the lowest and highest values. There is no way that all the children in a group can be above the average for that group. However, it's possible that all the kids in Lake Wobegone are above the national average, or above the average of ...

Math
Of course it's possible. If 18 students get 100%, they all score above the average, which is less than 100.

Math
6C2 = (6P2)/2! = 6*5/2 = 15

science
C looks good to me.

Math
The wording is very strange. Maybe you mean there are thirteen 2/3-cup servings in a bag. In that case there are [13*(2/3)]/(1/2) = 52/3 = 17 1/3 half-cup servings

Calc 1
since e^(ln?) = ?, sin(?) is the first time the curve intersects the x-axis. So, the volume, using discs of thickness dx is v = ?[0,ln?] ?r^2 dx where r=y=sin(e^x) v = ??[0,ln?] sin^2(e^x) dx Now, sin(e^x) is not integrable using elementary functions. I guess you'll have ...

Geometry
Draw a horizontal line through M to intersect BT at P. Since MP is parallel to AC, angles TMP and TNC are congruent, making right triangles TMP and TNC similar. So, PT/PM = NT/NC since M is the midpoint of AB, P is the midpoint of BN. Thus, NC=2MP PT/MP = NT/2MP 2PT = NT Since...

math
10P4 = 10*9*8*7 = ? not sure what the 4- means

math 105
so, what is the rule? You have the mean and the std....

physics
correct

physics
T^2 is proportional to L/g. The mass has no effect. So, if A^2 = L/g B^2 = 2L/g B^2/A^2 = 2 B/A = ?2 ? 1.4 answer C

physics
Since the period T = 2??(m/k) y = 0.1 sin(?(k/m) t) = 0.1 sin(5t) the speed v = 0.5 cos(5t) which has a maximum value of 0.5

Math
4 6/12 - 3 8/12 = 3 18/12 - 3 8/12 = 10/12 so, 10" left or, 54"-44" = 10"

Math help plz
the area needed is ?r^2 + 2?rh = ?r(2h+r) = 6?(24+6) = 180? = 565

Math
#14 wants volume, not area the others look good.

math
For #1 and 2, type in your function at wolframalpha.com or any of many other fine graphing web sites. -4 = 4-8 7 = -4+11 so, (x,y) -> (x-8,y+11)

Pre calc
because I studied the double-angle formulas, as you need to do!

Pre calc
In QI, you have sinx = 5/13 cosx = 12/13 tanx = 5/12 sin2x = 2sinx cosx = 2(5/13)(12/13) = 120/169 similarly using the other double-angle formulas.

Pre calc
If this isn't even math to you, you clearly haven't been paying attention during the discussion of the double-angle formulas ... (1-cos2x)/sin2x = (1-(2cos^2x-1))/(2sinx cosx) = (2-2cos^2x)/(2sinx cosx) = 2sin^2x/(2sinx cosx) = sinx/cosx = tanx sin4x/sinx = (2sin2x ...

Urgent math help!!
Huh? You are in trig, and cannot tell what the x- and y- coordinates are? The point (-3,4) is at x = -3 y = 4 since r^2=x^2+y^2, r=5. sin? = y/r = 4/5 You got the right answer; was it just a guess?

Urgent math help!!
Draw your triangle in standard position. It is clearly a 3-4-5 right triangle. sin? = y/r

Trig
well, csc^2x-1 = cot^2x cot*cos/cot^2 = cos/sin * cos * sin^2/cos^2 = sin

Math
she kept 3/8 * 25 = ? kg

Math
you have to keep writing an equation; you just dropped off the right side. In any case, remember from your Algebra I that you usually need to set stuff to zero to solve. Now, using the fact that sin^2x+cos^2x=1, 2cos^2x+sinx-1 = 0 2-2sin^2x + sinx-1 = 0 2sin^2x-sinx-1 = 0 (...

Trig
oops, sorry Reiny...

Trig
Remember your double-angle formulas. You have tan(2*30) = ?3 Reiny dropped the 2 in the numerator...

Trig
(1+tan^2x)/tan^2x = 1/tan^2x + 1 = cot^2x + 1 = csc^2x

laplace transformation:help damon or steve or rein
In case you don't have an online calculator handy, L{1} = 1/s L{e^t} = 1/(s-1) = F(s) L{t^2 e^t} = (-1)^2 F"(s) = 2/(s-1)^3 L{y} = f(s) L{y'} = sf(s)-f(0) = sf(s)-1 L{y"} = s^2f(s)-sf(0)-f'(0) = s^2f(s)-s+2 L{y'"} = s^3f(s)-s^2f(0)-sf'(0)-f&...

Bachelor of computer science
I'm sure you can do the I/O. The calculation is if hours <= 0   print "error - no hours entered" else if hours <= 40    pay = rate * hours else   pay = rate*40 + rate*1.5*(hours-40) endif

math
make up your mind -- BASIC or FORTRAN ? in any case, find the prime factors of each number, and select all those common to both. Their product is the HCF.

algebra
I don't see an inequality either. But, since the cost is not zero, you know that profit < earnings

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