# 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.

**Help Please. Math**

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 ...

**@Ms.Sue please help!!:(Math**

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.

**Mathematics-Geomeotry PLEASE HELPPPPPP!!!!!!!**

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.

**Mathematics-Geometry PLEASE HELP HELP!!!!!!**

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.

**Math please help me**

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