Monday

August 3, 2015

August 3, 2015

Total # Posts: 32,562

**Calculus**

well, 1-sin^2 = cos^2, so you really just have sin/cos^2 = sec tan which I'm sure you recognize.
*April 19, 2015*

**Calculus**

the graphs intersect at (0,1) and (1.136,1.764) A The cross-sections have base of f(x)-g(x) and height 5. So, we add them all up: v = ∫[0,1.136] 5(1+sin(2x)-e^(x/2)) dx = 5(x - 1/2 cos(2x) - 2e^(x/2)) [0,1.136] = 2.1455 B The same thing, but the height is the same as the...
*April 19, 2015*

**Chem**

If the reaction is Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3 then each mole of PbCl2 uses one mole of Pb(NO3)2. So, with 3.264/278.11 = 0.0117 moles of PbCl2 produced, the 2.00mL was .0117/.002 = 5.868M 5.868*80.7/100 = 4.735M original concentration
*April 19, 2015*

**Calculus**

If you mean f(x) = √(x-3) f is only defined for x >= 3 For that domain, it is clearly integrable. So, the interval is [3,∞)
*April 19, 2015*

**Calculus**

clearly the intervals have width 1. So, with right-hand endpoints, we need 1*(f(1)+f(2)+f(3)+f(4)) = 1+4+9+16 = 30 what was so hard about that?
*April 19, 2015*

**Science**

I'm not sure. There are various chemical reactions taking place all the time (plant and animal respiration, mineral formation, etc.) that consume or release water. So, not only is C probably true, but we don't know the long-term effects of the various processes taking ...
*April 19, 2015*

**statistics**

you want z such that P(Z>z) = .63/2 z = .896 So, the scores are ±0.896
*April 19, 2015*

**Math/Algebra**

By the end of 2012, that's 3 years, not 2. A: 4000*.0315*3 = 378.00 B: 4000(1.0315^3-1) = 390.03 Looks like C to me.
*April 19, 2015*

**Vectors**

unless the normals are parallel, the planes must intersect in a line.
*April 19, 2015*

**mass**

1.3 = .000081x x = 16049.38 So, subtract the 1.3g from that
*April 19, 2015*

**Math**

least perimeter will be a square. So, what are the sides of such a square?
*April 19, 2015*

**statistics**

P(Z>z) = 0.2 z = 70
*April 19, 2015*

**math**

If there are x lbs of peanuts, then there are (100-x) lbs of cashews. So work with the value of the mixture and its parts: 2.80x + 5.30(100-x) = 3.30(100)
*April 19, 2015*

**7TH MATH HELP**

#1. Nope: -12(3)-3 = -36-3 = -39 #2 ok #3 Nope: as x increases, y decreases #4 Nope: 8*2+5 = 16+5 = 21 #5 ok #6 Nope J(10, 5) -> (15,9) #7 Nope: you didn't change anything!
*April 19, 2015*

**math**

how are the cubes arranged? 1x6 or 2x3?
*April 19, 2015*

**math**

π r^2 h = 355 r^2 h = 113 the area a=2πrh+2πr^2 which we want to be small 113 is prime, so there will not be integer values. However, using integers to approximate, r h a/2π(r^2+rh) 3 13 48 4 7 44 5 4 45 6 3 54 Looks like r is near 4
*April 19, 2015*

**maths**

625^2+4 = 360629 = 157*2297
*April 18, 2015*

**Calculus**

as always, n ∑ 2/n f(5+2/n) k=1 Since you didn't specify the endpoints or the number of intervals, you can fill in the numbers.
*April 18, 2015*

**precalc**

just look up the information for x^2 = 4py Figure p, and you're halfway there.
*April 18, 2015*

**math**

you should recall that for y = ax^2+bx+c the vertex is at x = -b/2a So, figure that for x, then evaluate y there. Since this parabola opens downward, its vertex is the maximum value.
*April 18, 2015*

**AP Physics**

conserve momentum. The sum of the two momenta after the collision must be the same as the original momentum of the moving ball. No diagrams here, so there's no way to tell the direction of the first ball.
*April 18, 2015*

**Math**

clearly both those equations cannot be true, since 99≠57. The sum of T5 to T8 is S8 - S4 = 8/2(2a+7d) - 4/2(2a+3d) = 4a+22d = 114 Similarly, S15-S11 = 15/2(2a+14d) - 11/2(2a+10d) = 4a+50d = 198 Subtract those two to get 28d=84, or d=3 So, a=12 T21 = 12 + 20*3 = 72
*April 18, 2015*

**Math**

oops typo. I hit 2 instead of 3. I'm sure you noticed that, and can fix it.
*April 18, 2015*

**Math**

just get all the y stuff on one side of the equation. x = 2r + 3ty 3ty = x - 2r y = (x-2r) / 3t
*April 18, 2015*

**Math**

8x3 * 3 = 24x9 8x3 * 4 = 32x12
*April 18, 2015*

**maths**

If you're talking about exponents, 5x^-3 = 5/x^3 If not, please explain "index" to poor ignorant me.
*April 18, 2015*

**Physics**

it will be the same as if it were just dropped. So, how long does it take to drop 60m? technically, this is not quite true, since the aerodynamics of flight will probably keep it aloft longer than just dropping it.
*April 18, 2015*

**math**

d = (T12-T5)/(12-5) = (50-8)/7 = 6 T5 = a+4*6=8, so a = -16 Thus, the sequence is -16 -10 -4 2 8 ...
*April 18, 2015*

**Further maths**

f(x) = 2x-3 so, f(g) = 2g-3 since g = x^2+5, g(x-1) = (x-1)^2+5 so, f(g(x-1)) = 2((x-1)^2+5)-3 = 2x^2-4x+9
*April 18, 2015*

**trig**

You know that A(b) = 43+b so, if b=22, plug that in wherever there is a b: A(22) = 43+22
*April 18, 2015*

**Algebra**

(8a^3b^6)^(1/3) = 2ab^2 (16a^8b^-4)^(1/4) = 2a^2b^-1 2ab^2 / 2a^2b^-1 = a^-1b^3 = b^3/a
*April 18, 2015*

**calc**

marginal cost is dC/dx = 49(-3x+3lnx+28)/(28-3x)^2 so at x=7, dC/dx = (7 + 3 ln7)/49
*April 18, 2015*

**Physics**

since the volume is constant, 4^2 * 2 = .5^2 * v
*April 17, 2015*

**Math plane or spherical trig?**

which kind of trig are we using?
*April 17, 2015*

**geometry**

the length of AB is √45 = 3√5 the length of CD is √80 = 4√5 So, since dilations scale lengths linearly, CD/AB = 4/3 No idea what EF is supposed to be for...
*April 17, 2015*

**Math**

assuming you want integer ratios, the medium rectangle must be 3 times as large, or 24x9 Similarly, since 3 does not divide 32, and 8 does, the large one is 32x12
*April 17, 2015*

**math**

There is a nice discussion and a good example here: http://en.wikipedia.org/wiki/Frobenius_method where do you get stuck?
*April 17, 2015*

**Math**

h/30 = tan 24
*April 17, 2015*

**Math**

If x students transfer, (480-x)/16 = (192+x)/12 x = 96
*April 17, 2015*

**Math**

ok, I'll bite. what is 2* ?
*April 17, 2015*

**Math 222**

when you multiply, you add powers: a^3 = a*a*a a^5 = a*a*a*a a^3 * a^5 = (a*a*a)(a*a*a*a*a) = a^8 similarly, when you raise powers to powers, you multiply them (a^3)^2 = (a^3)(a^3) = a^(2*3) = a^6 (a^1/2 b)^(1/2) = a^(1/2 * 1/2) b^(1 * 1/2) = a^1/4 b^1/2 so, now you have a^1/4...
*April 17, 2015*

**Math**

clearly, a = 0.5 d = 0.4 so, S10 = 10/2 (2*0.5 + 9*0.4)
*April 17, 2015*

**Algebra**

(x+5)(x-5)(3x-2) = 0
*April 17, 2015*

**Algebra**

4t + 16t^2 = 72 4t^2 + t - 18 = 0 (t-2)(4t+9) = 0 t = 2
*April 17, 2015*

**mathematics**

DE = √(1^2+9^2) = √82 DF = √((x+3)^2+5^2) = √(x^2+6x+34) So, x^2+6x+34 = 82 check my math, and solve for x.
*April 17, 2015*

**algebra**

8/12 = 2/3 24/36 = 20/30 = 2/3 So, you can see how to adjust either a or b. 5/3 = 25/x
*April 17, 2015*

**pre-calc**

if n is even, so is n^2, so the sum is even if n is odd, so is n^2, and since the sum of odd+odd is even, so is the whole thing. But that's just logic. To prove it, take S1: 2 is a factor of 1^1+1-2 = 0 So, assume Sn. Sn+1: 2 is a factor of (n+1)^2 - (n+1) + 2 = n^2+2n+1...
*April 17, 2015*

**Math**

w <= 6/2
*April 17, 2015*

**Algebra**

#5: T1 = 50 Tn+1 = Tn+20 #6: T1 = 1452 Tn+1 = Tn * 2/3 I expect you can tell an AP from a GP
*April 16, 2015*

**5th grade (math)**

since 5+8 = 13, one obvious combination is 2.35 in 5-cent stamps, and one 8-cent stamp Since 5*8=40, decreasing the 5-cent count by 8 and increasing the 8-cent count by 5 will keep the total at 2.43 So, just build a table 5¢ 8¢ 47 1 39 6 ... 7 26
*April 16, 2015*

**Ap Calculus**

You have a problem. x^2+y^2 cannot ever be negative.
*April 16, 2015*

**Ap Calculus**

The curves intersect at (-1,1) and (1,1) So, using discs, and allowing for symmetry, v = 2∫[0,1] π(R^2-r^2) dx where R = 2+√(2-x^2) r = 2+x^2 using shells, and taking advantage of symmetry, v = 2∫[0,1] 2πrh dy + 2∫[1,2] 2πrh dy where r = ...
*April 16, 2015*

**Calculus check and help**

A and B look good. The curves intersect at (0.1366,-3.9814) and (2.3977,1.7490) For C, you can either use washers, as you did for the horizontal axis. They are of thickness dy: x = 1/2 e^y x = √(y+4) v = π∫[-3.9814,1.7490] (√(y+4))^2-(1/2 e^y)^2 dy or, ...
*April 16, 2015*

**Math**

since rate*time is constant, 5*20 = 4*t
*April 16, 2015*

**Calculus check**

A is OK B is v = ∫[0,4] 2πrh dx, where r = 4-x h = (16-2x)-√x^3 For C, you need to split the Region into two parts at y=8, because we have to integrate along dy. The rectangles have a base of x, which changes at y=8 v = ∫[0,8] ∛y^2 h(y) dy + ∫...
*April 16, 2015*

**Math**

after n breaths, the amount left is 500*.88^n
*April 16, 2015*

**math**

area changes by the square of the linear ratio so, the area becomes 729*3^2
*April 16, 2015*

**Integration by Parts**

since sin(t) = (e^-t - e^-it)/2, isin(t) = i/2 (e^it - e^-it) i sint e^it = i/2 (e^2it - 1) so, the integral is just i/2 ((1/2i) e^2it - t) = -1/4 e^2it - t/2 do that from 0 to 2π and you have [1/4 e^4πi - π]-[1/4] = 1/4 - π - 1/4 = -π I haven't ...
*April 16, 2015*

**calculus word prob**

if the cable goes from A along the 360-ft side for a distance x past the indent at A's corner to y from the indent at B's corner, then the cost will be 8(60+x) + 11√((300-x)^2+(240-y)^2) + 8(60+y) Now, I don't think you have studied multi-variable derivatives...
*April 16, 2015*

**calc**

that's where E' is a max E' = .004(4x^3-24x^2) = .008(x^3-6x^2) E' is max where E" = 0 E" = .008(3x^2-12x) = .024x(x-4) so, |E'| is max when x=4, as you can see at http://www.wolframalpha.com/input/?i=.004+%28x%5E4-8x%5E3%29%2B2
*April 16, 2015*

**math 8th!**

since the slope is 2, y changes by 2 whenever x changes by 1. So, pick a point 1 unit to the right or left of (-2,3) These would be (-3,?) and (4,?) Since y changes by 2 for every 1 in x, the two points would be 2 units above or below (-2,3): (-2,1) and (4,5) You could have ...
*April 16, 2015*

**Geometry**

the ant has gone 3 times the distances, or 30 inches along, and 24 inches up. Using your distance formula, the final displacement is √(30^2+24^2) = 6√(5^2+4^2) = 6√41
*April 16, 2015*

**pre-calc**

I'll get you started. But you clearly need to study the examples in your text. Or, just do a google and you will find lots of examples online. You start with your matrix of coefficients: 4 -1 3 | 12 1 4 6 | -32 5 3 9 | 20 You want to work things so that you have a final ...
*April 16, 2015*

**pre-calc**

well, when I do it, it says I'm right, so why don't you show your work here and we can see how things go?
*April 16, 2015*

**pre-calc**

Enter your coefficients here and see all the details: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
*April 16, 2015*

**algebra**

after t days, the amount left is (1/2)^(t/13) So, after 3 days, the amount left is (1/2)^(3/13) = .852 = 85.2%
*April 16, 2015*

**Math**

we see that the length of the beam is 8/sin40 = 12.45 ft At 60 degrees, the top is 12.45sin60 = 10.78 ft high So, the wire is 12.78 ft up. Since the beam is only 12.45 ft long, it will not touch the wires.
*April 16, 2015*

**MATH**

40*9.25 = 370.00 The deductions are .0765*370 = 28.31 .12*370 = 44.40 .08*370 = 29.60 But, what's discretionary? The taxes and slacks are mandatory, but the transportation and lunches could be adjusted if necessary. So, I'd say 247.74 Hmmm. Maybe transportation is also...
*April 16, 2015*

**calculus**

The volume is just a stack of discs, each of radius y and thickness x, so v = ∫[a,b] πr^2 dx y(1) = 0 y(e) = 1 so, v = ∫[1,e] πr^2 dx so, what's r? If we were revolving around y=0, r would just be y. But the axis is 3 units farther away than that, and...
*April 16, 2015*

**calc**

just figure f(1)...f(4) since the rectangles have width 1, your approximation is just the sum of the four values of f. The value will be a little high, since f is concave downward, so using the right ends of the intervals means the rectangles are a bit too tall.
*April 16, 2015*

**MATH HELP**

well, what is 3+5 ? If you even considered (B) you clearly have not studied the Fibonacci sequence too closely.
*April 16, 2015*

**calc(very urgent)**

something wrong with the solution I posted a few hours ago? http://www.jiskha.com/display.cgi?id=1429156887
*April 16, 2015*

**precalculus**

we know that the center is at (0,0) and the axis is vertical, so y^2/a^2 - x^2/b^2 = 1 If the distance between the vertices is 2a = 8, then 4^2 + b^2 = c^2 Since (3,-5) is on the graph, 25/a^2 - 9/b^2 = 1 a=4, so 25/16 - 9/b^2 = 1 b^2 = 16 y^2/16 - x^2/16 = 1 See the graph at ...
*April 16, 2015*

**pre-calc**

You have Sn, but you have a typo, or the copy/paste mangled it. It should be Sn: 1^2+...+(3n-2)^2 = n(6n^2-3n-1)/2 So, S1: 1^2 = 1(6*1^2-3(1)-1)/2 S2: 1^2+4^2 = 2(6*2^2-3*2-1)/2 S3: 1^2+4^2+7^2 = 3(6*3^2-3*3-1)/2 which are all true We have not shown it to be true for all n. To...
*April 16, 2015*

**pre-calc**

8x+7y >= 336 Unless we're discussing pirates, that's borders.
*April 16, 2015*

**math**

well, clearly, the marble has a height of 2r, so the water must be that deep to cover it. Since the cylinder is said to have a radius of 4, I think you want the radius of the marble that requires the most water to cover it. The volume of water is v = π*16(2r) - 4/3 π...
*April 16, 2015*

**pre - CALCULUS**

enter your values here to see all the details: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
*April 15, 2015*

**Pre-calculus**

vehicles and winds travel on a heading, not a bearing. Sadly, this distinction is ignored or unknown to many textbook writers and teachers. But on to the math. the Cessna in still air travels be at (180 sin35°,180 cos35°) = (103.24,147.45) The wind blows it an ...
*April 15, 2015*

**calculus (please help. URGENT!)**

v = π/3 h^2 (3r-h) = πrh^2 - π/3 h^3 dv/dt = 2πrh dh/dt - πh^2 dh/dt = πh(2r-h) dh/dt Pluggin in your numbers, then, we have π*5(20-5) dh/dt = 3 dh/dt = 3/(75π) = 0.021 m/min
*April 15, 2015*

**Math (ASAP)**

well, just plug and chug a1 = 32 a2 = 1/2 (32) = 16 a3 = 1/2 (16) = 8 ...
*April 15, 2015*

**math**

I got 111.32, so I guess it depends on how you want to round it.
*April 15, 2015*

**Math**

4321 = 29*149 we know that 149 divides the product, so the question is, is any other number in teh product a multiple of 29? There are only 21 consecutive factors, so it's not guaranteed, so you'll have to check whether there is a multiple of 29 between 140 and 160. ...
*April 15, 2015*

**Math**

oops. 10*42^2 = t*60^2
*April 15, 2015*

**Math**

t = k/v^2 so, tv^2 is constant. Therefore, 10/42^2 = t/60^2
*April 15, 2015*

**Science/Physics**

3.9 + 2.4 + (3.9+3.1) = 13.3 How did you arrive at 10.5?
*April 15, 2015*

**geometry**

the midpoint is the average of the endpoints. So, for BC, P=(h,k) Now just use your distance formula to get the required lengths.
*April 15, 2015*

**math**

plug in your numbers and play around with Z table stuff at http://davidmlane.com/hyperstat/z_table.html
*April 15, 2015*

**math**

.42
*April 15, 2015*

**even wurse spelling**

yet correct, withal.
*April 15, 2015*

**More math**

and yes, you are correct.
*April 15, 2015*

**maths**

after 6 jumps, he has descended 6 steps The 7th jump will take him to the 9th step.
*April 15, 2015*

**geometry**

The inner rectangle is 70x105 the inner circle has radius 35 the outer rectangle is 84x105 the outer circle has radius 42 now just figure the areas and subtract inner from outer
*April 15, 2015*

**trig table?**

Excuuuuuse me! How long ago did you take trig? No one uses tables any more. However, the trig book I used was titled "Trigonometry with Tables." I remember our library had huge volumes from the National Bureau of Standards with 12-place tables for trig, log, and ...
*April 15, 2015*

**maths**

PAD is not necessarily congruent to PBC. They have two sides congruent, but that is not enough.
*April 15, 2015*

**Precalc**

f(x) = (3x)^-4 + 3x^4 f' = -4(3x)^-5(3) + 12x^2 = -12/(3x)^5 + 12x^2 where is 15x^2-30x = 0? where is 6x^2+12x = 0?
*April 15, 2015*

**math**

60mph = 88ft/s, so you want k where 88-kt=0 so t = 88/k 88t-1/2 kt^2 = 100 88(88/k) - 1/2 k (88/k)^2 = 100 k = 38.72 ft/s^2 now you can solve the other part.
*April 15, 2015*

**algebra**

recall that the slope-intercept form of a line is n = mt+b where m is the slope. Here, that is -2.23 It means that the number of refineries is declining by 2.23 per year, or about 9 refineries every 4 years.
*April 15, 2015*

**math**

(20+14)(30+14)-(20)(30)
*April 14, 2015*

**math**

just using the measurements, the tiled area is 8*6 - 2*3 = 42 ft^2 11" is nearly a foot, so we can estimate that about 42 tiles will be needed. Naturally, since the tiles are a bit smaller, more of them will actually be used.
*April 14, 2015*

**geometry**

Assuming a logical arrangement of W,X,Y,Z, your answer appears correct.
*April 14, 2015*