Wednesday

July 1, 2015

July 1, 2015

Total # Posts: 32,044

**AP Calc B/C**

we are told that p doubles every 35 years. That means p = Po * 2^(t/35) Each year it increases by a factor of 2^(1/35) = 1.02, or 2% per year.
*April 27, 2015*

**Calculus**

The curves intersect at x = ±1.18. So, using symmetry, the area is a = ∫[0,1.18] cos(x) - (x^2-1) dx Those are all easy integrals, so just plug and chug.
*April 27, 2015*

**Algebra**

If the classes are a,b,c,d, then b = a+60 c = 2a-50 d = 3a a+b+c+d = 1424 a+(a+60)+(2a-50)+(3a) = 1424 7a + 10 = 1424 7a = 1414 a = 202
*April 27, 2015*

**math**

2-2sin^2-sin-1=0 2sin^2+sin-1 = 0 (2sin-1)(sin+1) = 0 sin = 1/2 or -1 sin = 1/2 at pi/6 and 5pi/6 sin = -1 at 3pi/2 Better check your factoring. Your angles are correct for the values you obtained, but those values had the wrong signs, and hence, the wrong sines :-)
*April 27, 2015*

**Calculus**

y = √(x^3+2) y' = 3x^2/(2√(x^3+2)) y' = (6x)(2√(x^3+2)) - (3x^2)(6x^2)/(2√(x^3+2)) ----------------------------------------- 4(x^3+2) 24x(x^3+2)-18x^4 ------------------------ 8(x^3+2)^(3/2) 3x(8(x^3+2)-6x^3) ---------------- 8(x^3+2)^(3/2) 3x(...
*April 27, 2015*

**Precalc-functions**

y = 2/3 x + 5 y-5 = 2/3 x x = 3/2 (y-5) so, f^-1(x) = 3/2 (x-5) as with all polynomials, the domain is all real numbers.
*April 27, 2015*

**Math 222**

why not just use conventional math notation? 5x^2-9x+2 = 0 the discriminant is b^2-4ac = 81-40 = 41 Now just use that in the quadratic formula. Since the discriminant is positive, there are two real solutions: x = (9±√41)/10
*April 27, 2015*

**Math**

I think it is (C) Roll that rectangle along the horizontal axis, and fold over the ends.
*April 27, 2015*

**Math**

looks correct to me.
*April 27, 2015*

**Algebra 222**

{w|w > 1} = (1,∞) {w|w greater than less than 5} = {w|w≠5} = {-∞,5)U{5,+∞)
*April 27, 2015*

**geometry**

the distance out from the dock for an angle θ is just d = 14 cosθ So, the change in distance is just 14(cos 9.2° - cos 14.1°)
*April 27, 2015*

**math**

the arc length of the sector is s = rθ = 36(210/360)(2π) = 132 That becomes the circumference of the base of the cone. So, the radius of the base of the cone is 132/(2π) = 21 Now look at the cone from the side. Its height h is given by 21^2 + h^2 = 36^2 Now you ...
*April 27, 2015*

**math**

do you not have a simple formula for loan amortization? Just plug in the numbers...
*April 27, 2015*

**math**

p + 3p = 48 now you can find pink, and red = 3 times that.
*April 27, 2015*

**algebra**

4/1800 = x/8400 m/k = m/n * n/k = 1/5
*April 27, 2015*

**Chemisrty**

you are reducing the concentration by a factor of .2/6 So, you need to increase the volume by a factor of 6/.2
*April 27, 2015*

**PCM**

you can't get offline stuff from any web site...
*April 27, 2015*

**Business Math**

You want x, where 2200(1+.08x) = 1960(1+.10x)
*April 27, 2015*

**Math**

yes. Since sinθ is negative in QIII and QIV, that should make it clear that they only want one solution. sin(.2054) = .204, you want QIII, so θ=π+.2054
*April 26, 2015*

**math**

correct.
*April 26, 2015*

**math**

since the horizontal distance is the same for all three streets, the steepest street will be the one with the greatest vertical rise. steepness = vertical/horizontal change
*April 26, 2015*

**Math**

well, since 3 cm is 3 times as big a 1 cm, the real length will be 3 times as big as well.
*April 26, 2015*

**Math**

For a fair coin, the number would be 150 * 1/2 = 75 But, we just went through a lot of work to discover that our coin has a probability of 17/30 for getting heads. So, with that coin, 150 tosses should come up with 150 * 17/30 = 85 heads, not 75.
*April 26, 2015*

**Math**

f=w+6 p=w-3 f+p+w = 27 (w+6)+(w-3)+w = 27 3w+3 = 27 3w = 24 w = 8 So, the wrist has 8 bones. Now find the values for the other two parts of the hand.
*April 26, 2015*

**calculus**

4x dx/dt + 3y^2 dy/dt = 0 Now just plug in your values and solve for dx/dt.
*April 26, 2015*

**Maths**

well, you didn't say which side is the base, and which are the two equal sides. I will assume A is the vertex between the two equal sides. In that case, AB = AC, so (A): (4x-3)/15 = (x+3)/5 I assume you can now work out the rest. If I got the two equal sides wrong, then ...
*April 26, 2015*

**math**

Let y = x+28 Then xy=7 x(x+28) = 7 x^2+28x-7 = 0 x = -14±√203 I suspect a typo.
*April 26, 2015*

**math**

s(t) = t^3/3 - 12t^2/2 + 36t Strange, saying 12t^2/2 rather than 6t^2, but hey ... Assuming "moving to the right" means s(t) is increasing, then we just have to find when ds/dt > 0. That is, t^2 - 12t + 36 > 0 Hmmm. Something tells me you had a typo, and it was...
*April 26, 2015*

**math**

do you have some graph paper? pick any value for x, and calculate y. The easiest value to use is x=0. Then y=3. So, the point (0,3) is on the graph. For y, if you pick and even value, you will have a fraction for x, so pick any odd value, say, y=1. Then x=1, and (1,1) is on ...
*April 26, 2015*

**Math**

42 cm^2
*April 26, 2015*

**math**

so, did you use the product rule as advised? v = x^2 h dv/dt = 2xh dx/dt + x^2 dh/dt Now just plug in your values to find dv/dt. Note that dv/dt does not actually depend on h at all, just dh/dt.
*April 26, 2015*

**Math 222**

F = mv^2/r Fr = mv^2 Fr/v^2 = m
*April 26, 2015*

**Math**

-(3a-7b)/(2a) + (5a-4b)/(3a) + 3 The common denominator is (2a)(3a), so we have for the numerator -(3a-7b)(3a) + (5a-4b)(2a) + 3(2a)(3a) = 19a^2+13ab so, our final fraction is (19a^2+13ab)/6a^2 = (19a+13b)/(6a)
*April 26, 2015*

**Calculus 2**

I think it is (n^(1/n)-1)^n Using the ratio test, T(n+1)/Tn = ((n+1)^(1/(n+1))-1)^(n+1)/(n^(1/n)-1)^n If you can show that that ratio is less than 1, you're home free.
*April 26, 2015*

**Calculus**

the average value of f(x) on [a,b] is ∫[a,b] f(x) dx ----------------------- b-a f' = 3x^2-6x f = x^3-3x^2+4 so, you want ∫[-1,3] x^3-3x^2+4 dx -------------------------- 3 - (-1) which I'm sure you can do.
*April 26, 2015*

**Business Math Please Help me**

there is a nice discussion here. It should enable you to come up with the result you want. http://en.wikipedia.org/wiki/Economic_surplus
*April 26, 2015*

**Physics**

The two components of the distance between them is x = 4√3 y = 4 (a) if the velocity is v, then the two components are (using g=10 for simplicity) x(t) = vt y(t) = -5t^2 so, for the apple to travel 4√3 meters out, it will take 4√3/v seconds. It must fall 4 ...
*April 26, 2015*

**Calculus**

y'=x∛y The tangent line at (2,8) is thus y-8 = 4(x-2) y'=x∛y y^(-1/3) dy = x dx 3/2 y^(2/3) = 1/2 x^2 + c y^(2/3) = 1/3 x^2 + c since y(2) = 8, 4 = 4/3 + c c = 8/3 y^(2/3) = (x^2 + 8)/3 The domain is clearly where x^2+8 >= 0, or all reals. for minimum ...
*April 26, 2015*

**math**

what difference does it make how much water is in the tank? Are we to assume maximal packing, or just a rectangular lattice of spheres?
*April 26, 2015*

**Engineering Physics 1**

you know the height function, so set them equal. Let's use g=10 for convenience: 10t-5t^2 = 15(t-1)-5(t-1)^2 t = 4/3 that is, 1/3 s after the 2nd arrow takes off. You can figure the height, I guess.
*April 26, 2015*

**integral**

let u = 4+x du = dx x = u-4 3x/√(4+x) dx = 3(u-4)/√u du = (3u-12)/√u du = 3u^(1/2) - 12u^(-1/2) du Now it's just easy power rule stuff.
*April 26, 2015*

**math**

45 * (8/3)^2
*April 26, 2015*

**Calculus 2**

well sum of 1/n^2 converges, and cos^2(n)/n^2 <= 1/n^2
*April 25, 2015*

**Math**

18sin^2x + 3cosx - 17 0 18 - 18cos^2x + 3cosx - 17 = 0 18cos^2x - 3cosx - 1 = 0 (6cosx-1)(3cosx+1) = 0 cosx = 1/6 or cosx = -1/3 x = 1.40 or 2π-1.40 x = π-1.23 or π+1.23
*April 25, 2015*

**Calculus 2**

huh? They told you that p=e
*April 25, 2015*

**Calculus 2**

since 1/n^p converges if p > 1, 1/n^e converges
*April 25, 2015*

**Math**

my other thought was stratified because of the grouping and selection
*April 25, 2015*

**Math**

You want to find out what the favorite hot lunch in the school cafeteria is among the high school students. At an assembly for the whole school, you decide to survey all students who are sitting on the end of their rows in the auditorium. What type of survey are you conducting...
*April 25, 2015*

**Math**

in order to find out what the best restaurant is in town you call 25 of your friends at random. what type of survey is this? a- biased b-random c-systematic d-stratified my answer is biased because you may pick at random but they are all still your friends.
*April 25, 2015*

**Pre-Calculus**

nope. consider the fraction of the job done in one hour: 1/3 + 1/b = 1/2 In this case you got lucky, since indeed 1/3 + 1/6 = 1/2.
*April 25, 2015*

**Calculus**

circle: a = pi r^2 da/dt = 2pi r dr/dt = 2pi * 10/(pi-1) square: a = s^2 da/dt = 2s ds/dt = 2*10/(pi-1) da(circle)/dt - da(square)/dt = 20/(pi-1) (pi-1) = 20 in^2/s
*April 25, 2015*

**Calculus**

f is decreasing when f' < 0 f' = 2x-27 so, when is 2x-27 < 0?
*April 25, 2015*

**Math**

no you haven't the 274 is 275 though
*April 25, 2015*

**Math**

218 sorry
*April 25, 2015*

**Math**

What is the 1st quartile of the following data set? 2745,257,301,218,265242,201 a-201 b-218 c-257 d-270 my answer would be 28 because 201 would be the minimum.
*April 25, 2015*

**Calculus**

y = 3x^2 + 1 dy/dt = 6x dx/dt so, 3 = 6*2 dx/dt dx/dt = 1/4
*April 25, 2015*

**trigonometry pls help**

you saying that you cannot evaluate 18tanθ when given θ? the area is 1/2 bh = 18y since tanθ is increasing, the area will be greater when θ is greater.
*April 25, 2015*

**physics**

the acceleration was (-24m/s)/6s = -4m/s^2 s = vt - 1/2 at^2 = 24*6 - 2*6^2 = 144-72 = 72 m
*April 25, 2015*

**PHYSICS answer check**

a is usually used for acceleration, which is m/s^2. You ought to have used v, which in fact was entirely omitted: 50000*3 = 60000v v = 2.5 m/s Your answer was correct, but the "a" was misleading.
*April 25, 2015*

**Physics - eh?**

360 what?
*April 24, 2015*

**Algebra**

a=7 b = -5 c=0, so x = (5±√(25-0))/14 = (5±5)/14 x = 0 or 5/7 But then, we saw that just by factoring: x(7x-5) = 0
*April 24, 2015*

**Algebra**

(x^2-4)(x^2-25) = 0 (x-2)(x+2)(x-5)(x+5) = 0 . . .
*April 24, 2015*

**Intermediate Algebra**

150/1000 = x/5280
*April 24, 2015*

**Algebra**

2(9a-6)^2=11(9a-6)+63 2(9a-6)^2-11(9a-6)-63 = 0 If you rewrite this using u = 9a-6, you just have 2u^2-11u-63 = 0 (2u+7)(u-9) = 0 u = -7/2 or u=9 That means 9a-6 = -7/2 18a-12 = -7 18a = 5 a = 5/18 9a-6 = 9 9a = 15 a = 5/3
*April 24, 2015*

**geometry**

Nothing is wrong. The chord is 12 units from the center of the circle.
*April 24, 2015*

**geometry**

No idea where B is. arc length is radius time central angle. m<DEC = 2*m<DAC m<DEA = 180 - m<DEC = 180-124 = 56 arc AD = radius * 56 = AC*28*(pi/180)
*April 24, 2015*

**Calc 1**

basic stuff. f(x) = lnx/x^-6 limit is same as (1/x)/-6x^-5 = -x^4/6 = 0 f'(x) = x^6 * 1/x + 6x^5 lnx = x^5 (1+6lnx) so, f'=0 when x = e^(-1/6) f" = x^4 (11+30lnx) x^4 is always positive, so, f" > 0 when lnx > -11/30, or x > e^(-11/30) I expect you ...
*April 24, 2015*

**Calc 1**

take derivatives, and the limit is the same as that for (1/x)/(5π cos5πx) = (1/1)/(5π*-1) = -1/(5π)
*April 24, 2015*

**Math**

8C5 = 8C3 = 8*7*6 / 1*2*3 = 56 Note that 8C5 = 8C3 because 8*7*6*5*4/1*2*3*4*5 = 8*7*6/1*2*3 The extra two factors 4*5 cancel out.
*April 24, 2015*

**Math**

The first 4 are all ok. The last one is the Fibonacci Sequence, where each term is the sum of the two previous terms. It goes 1,1,2,3,5,8,13,21,34,55,... Your constant d=3 is just based on the terms 5,8 but clearly does not hold for the earlier terms.
*April 24, 2015*

**calculus**

Since the region is roughly triangular, with a horizontal top but a vertex underneath, either way the region has to be broken into two parts. Either ∫[0,1] 3-(2-x/2) dx + ∫[1,4] 3-(3/2 √x) dx or ∫[3/2,2] (4/9 y^2)-(4-2y) dy + ∫[2,3] 4/9 y^2 dy
*April 24, 2015*

**Math 222**

the Quadratic Formula cannot be factored. It is a formula You have used the formula correctly to find the roots of the given quadratic equation, but have come up with the wrong values. x = (3±√29)/2 = (3±5.385)/2 = 8.385/2 or -2.385/2 = 4.19 or -1.19 Not ...
*April 24, 2015*

**Prealgebra**

the three long faces are 8.1x12 rectangles, so the total area is 3 faces + 2 bases: 3*8.1*12 + 2*28.4 = 291.6 Looks like C to me.
*April 24, 2015*

**mathematics**

10-8 < x < 10+8
*April 24, 2015*

**Algebra**

The subject says algebra, but since you apparently have had some trig, just use the law of cosines: x^2 = 165^2 + 245^2 - 2(165)(245)cos115°
*April 24, 2015*

**Algebra**

the first leg moves the plane in th x- and y-directions by (165*cos40°,-165*sin40°) Now figure the amounts of the second leg, add them up, and then use the distance formula.
*April 24, 2015*

**math**

4 red,4 white,4 blue at random take out 2 , say 1 red 1 white do not replace. what is probarbility of now picking a)blue b)red c)white so red=1/12 or 1/4 white=1/12 or 1/4 now only 10 left red 3/10,white3/10, blue 4/10
*April 24, 2015*

**Math help?**

P(red&red) = 8/18 * 10/13 = 40/117 why does 18/31 make sense? That would be the probability if all the balls were in a single urn, but you have two separate events from two separate urns.
*April 23, 2015*

**Algebra**

which fraction can be reduced to 1/4? That is, can you divide the numerator and denominator by the same number, leaving 1/4 as the result?
*April 23, 2015*

**Math**

#1 Permutations, since each office is unique. If just selecting a committee, then it'd be combinations, since the member roles are all the same. #2,3 ok #4 is also permutations, since the order in line matters.
*April 23, 2015*

**Math**

still wrong. see earlier post.
*April 23, 2015*

**Math**

the line joining the points has slope (6-4)/(2-3) = -2
*April 23, 2015*

**math**

since (t-4), he delayed 4 seconds The timer started at t=0. His dive time is t-4. at t=4, he jumped. plug in t=4, and you are left with 2500.
*April 23, 2015*

**math**

5 faces, each with (20/5)^2 tiles
*April 23, 2015*

**Geometry**

area: 4^2/17^2 volume: 4^3/17^3 Now what is your answer?
*April 23, 2015*

**algebra**

since time = distance/speed, 325/(59+x) = 265/(59-x)
*April 23, 2015*

**Math 7th Grade**

#1 ok #2 Nope. Check your signs carefully. #3 Nope. Plot a point and fold the paper along the x-axis. Where does the point go?
*April 23, 2015*

**Analytic Geometry**

Since all three medians meet there, finding where two intersect is enough. AB has midpoint P at (5,0) BC has midpoint Q at (6,5) The lines CP and AQ are y-6 = (0-6)/(5-2) (x-2) y+4 = (5+4)/(6-4) (x-4) Now just find where the two lines intersect.
*April 23, 2015*

**Geometry**

conguent refers to two objects length describes properties of the objects two polygons are congruent if all their sides and angles have equal measure. Saying two objects are congruent is a short way of saying that all their properties are the same. Saves a lot of time.
*April 23, 2015*

**algebra**

Man, that's over three years ago!
*April 23, 2015*

**algebra**

well, just plug in the values! g(9) = |9-1| = 8 f(g(9)) = f(8) = √(8+1) = 3
*April 23, 2015*

**College algebra**

The fonts mangled your symbols, but if I manage to get it right, we have i = √-1 (i+1)(6i-10) = 6i^2 + 6i - 10i - 10 = -6+6i-10i-10 = -16-4i
*April 23, 2015*

**College Algebra**

just solve -0.01x^2 + 0.27x + 8.60 = 9.7 -0.01x^2 + 0.27x - 1.1 = 0 -0.01(x^2-27x+110) = 0 -0.01(x-5)(x-22) = 0
*April 23, 2015*

**College math**

depending on the precision of the numbers involved, your result might more correctly be written as 8.300
*April 23, 2015*

**History**

Viisit: http://www.myassignmenthelp.net/history-assignment-help.php You will get your all answers
*April 23, 2015*

**precalculus**

well, what are the discriminants? That is easy, and we can go from there.
*April 22, 2015*

**Calc 2 Indefinite Integrals**

Hmmm. dx/(x√(x^3-1)) as you suggest, let x^3 = 1/z^2 x = z^(-2/3) dx = -2/3 z^(-5/3) dz 1/x = z^(2/3) √(x^3-1) = √(1/z^2 - 1) = √(1-z^2)/z so now you have an integrand of -2/3 z^(-5/3) dz z^(2/3) z ------------------------------- √(1-z^2) = -2/3 ...
*April 22, 2015*

**Calculus Indefinite Intergral**

#1 makes no sense #2 dx/(1-x^2)^(3/2) If you let x = 1/z dx = -1/z^2 dz 1-x^2 = 1 - 1/z^2 = (z^2-1)/z^2 Now you have -dz/z^2 z^3/(z^2-1)^(3/2) = -z dz/(z^2-1)^(3/2) and the integral is simply 1/√(z^2-1) = x/√(1-x^2)
*April 22, 2015*

**precalculus**

The general form is r = k/(1 ± e cosθ) or r = k/(1 ± e sinθ) so, you need to massage what you have into that form 2/(3+2sinθ) = (2/3)/(1 + 2/3 sinθ) so, e = 2/3 do the others in like wise.
*April 22, 2015*

**math**

If we started with a in A and b in B, then we are told: a-25 = b+25 (a+100)/(b-100) = 7/2 a = 250
*April 22, 2015*