Posts by Steve
Total # Posts: 51,772
1/3 x + 6 = 5/6 x 6 = 1/2 x x = 12 Check: the 1/3-full tank has 4 gallons. Adding 6 makes 10 gallons, or 5/6 of 12
Surely not 3, since it does not appear in both top and bottom 8/15 * 5/16 = 8*5 / 15*16 = 40/240 Now, 2 can be factored out to give 20/120, but that is not fully simplified Same for 5 and 8, done individually. However, if 5 and 8 are both factored out, the resulting 1/6 is as ...
for the parallel line, you have a slope and a point, so use the point-slope form to get y+1 = 3(x+1) The perp. line has slope -1/3, so it is y+1 = -1/3 (x+1) Now massage those into slope-intercept form.
MATH HELP ASAP PLEASE!!!!!
You and "Naomi" should get together. can't read you attempts at fractions. Try something like 11 1/4 or 10 1/2 Consider using a regular pentagon with an apothem of 7", or the radius of the pizza. Look online to find the area of such a figure. It will be the ...
Just write algebra for the conditions. Let w = municipal bonds x = CD's y = T-Bills z = Stocks Then the conditions are w <= (w+x+y+z)/5 x <= w, x <= y, x <= z x+y >= 3/10 (w+x+y+z) x+y >= 1.2(w+z) Rearrange things a bit, and you want to maximize p = .085w...
find v(t), and then just take its derivative to find its maximum: v(t) = 3t^2-6t-9 v'(t) = 6t-6 clearly, v'=0 at t=1 However, since v' is just a parabola, its vertex is a minimum, not a maximum. That means that its maximum is either at t=0 or t=2. So, just compare ...
You have mangled the question, but if I parse the meaning correctly, you need to check the amount of alcohol before and after mixing in the water: .60(80) = .40(80+x)
the fine is just an arithmetic progression where the amounts (in thousands) are a = 4 d = 1 You want n where 4+1(n-1) <= 60
Take the cross product and the normal to both lines is <2,-2,1> The plane is then just 2(x-1)-2(y-5)+1(z-12) = 0 2x-2y+z = 4
since ((2x-1)^3)/6 = 4/3 x^3 - 2x^2 + x - 1/6 the two solutions differ by 1/6 Since C is an arbitrary constant, its value does not matter.
just add up the interest amounts: .05x + .10(30000-x) = 2500
to find n, just solve n/2 (9+105) = 741 Now find d, using 9+(n-1)d = 105
poke around here, and I think all will become clear: http://davidmlane.com/hyperstat/z_table.html
math revise @Damon
I think Damon went a bit overboard. The line has a slope, and it is zero. Not at all the same as having no slope at all. A vertical line's slope is undefined. Hmmm. Is that the same as not having a slope? Not sure...
series maths steve reiny bob reiny steve!!!
There are lots of sequences that fit those three numbers. Possibly the simplest is a quadratic, 10x^2-18x+16 So, p ? 10x^2-18x+16 = 2/3 p(5p^2-6p+13) k=1
I suspect you meant that they are draining. So, the two equations would be y = 75-7x and y = 115-3x for suitable domains of x.
You just need to remember the characteristics of parabolas. y^2?16y=12x?8^2 y^2-16y+64 = 12x-64+64 (y-8)^2 = 12x Now recall that for y^2 = 4px, we have vertex = (0,0) focus = (p,0) directrix is x = -p So, for your parabola, that means that it is shifted up 8, and p=3: vertex...
Algebra (Please help!)
I get 0.90 as the best match.
12+5+3 = 20 5/20 = 1/4 12:5:3 * 1/4 = 3 : 5/4 : 3/4
Looks like 45 and 450 to me.
what does the cylinder have to do with it? For the cone, v = ?/3 r^2 h = ?/3 (16)(3) = 16?
the line joining the points has slope (-3+1)/(2+2) = -1/2 What do you know about the slopes of perpendicular lines?
just divide salt/fresh: (3*10^20)/(6*10^18) = (3/6)*10^(20-18) = 0.5*10^2 = 50
@Reiny gets the save
Wow - I guess I mangled that one, eh?
Where's the problem? Just plug your numbers into the formula, which I assume you have handy. A = Pr(1+r)^n/((1+r)^n-1) = 1000*0.16(1+.16/4)/((1+.16/4)^20-1) = 139.70
I assume your question is which two numbers have a product of -168 and a sum of 21? Well, check the factors of 168: 2 84 3 56 4 42 8 21 6 28 12 14 Now, which two differ by 21? None. So, your numbers are not rational. I suspect a typo. Algebraically, xy = -168 x+y = 21 x + (-...
try poking around here, if you don't have a handy Z table: http://davidmlane.com/hyperstat/z_table.html
Could be any of them, depending on how much of each you have. And what is H20? Molecules of hydrogen are just H2 :-)
218x - 36(30-x) = 5524
(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 ...
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
ever heard of google? It will provide lots of discussions on just what it is and what it is not.
It is defined that way. It makes possible the definition that n! = n(n-1)! for all integer n>0
OK. Now you have ln(y-1) = -1/x + 1 y-1 = e^(-1/x + 1) y = e^(1 - 1/x) + 1
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.
4/3 the speed means 3/4 the time: 7.5 min.
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?
Looks good to me.
a good place to check your work is http://calc101.com/webMathematica/long-divide.jsp
4:1 = 64:16
1/2 of all that
(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...
well, it will be 5/2 as long as NL
well, it will be exactly the same as any other three specific tosses. Each toss has a 1/2 chance of success.
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
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?
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)
the 17's cancel, so it's just 93/97
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
Out of every 8 students, there are 5 boys and 3 girls. So, b/g = 5/3 g = 240 so, b = 400
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
cool. So, what?
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
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
if you say so.
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
x - x/2 - x/4 - x/8 = 3 x = 24
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
see related questions below
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
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
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%...
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
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...
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.
Just expand the polynomial and integrate all the terms: 4x^2-4x+1 dx
well, 363 = µ - 2? what does the rule have to say about that?
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
(s/2)^3 / s^3 = (s^3/8) / s^3 = 1/8
there are 26 choices for each letter. So, multiply.
500 - 8*45 = ?
looks good to me
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
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
I get S-S6 = 5.93*10^-6 < 1.0*10^-5
I disagree. Why are you subtracting the last terms?
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'
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
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
add 30 ft to each dimension (15 ft on each side) then do perimeter as usual
vertical shrink by 4 shift down 12
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)+%...
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 ...
Of course it's possible. If 18 students get 100%, they all score above the average, which is less than 100.