Monday

July 6, 2015

July 6, 2015

Total # Posts: 32,118

**math**

um, how about 16*3.5 ?
*May 7, 2015*

**Math**

If I get past the font gibberish, I sense that you want to prove that 4 sinθ sin(60-θ) sin(60θ) = sin 3θ That isn't so, so try reposting using sin^3 θ for cube of sinθ if that's what you mean.
*May 7, 2015*

**Calculus**

the curve is concave down. So, (c)
*May 7, 2015*

**trigonometry**

what, forgotten your algebra I, now that you're taking trig? -3sin(t)=15cos(t)sin(t) 15cos(t)sin(t) + 3sin(t) = 0 3sin(t)(5cos(t)+1) = 0 sin(t) = 0 or cos(t) = -1/5 So, find the 4 values of t which do that. 8cos^2(t)=3-2cos(t) 8cos^2(t)+2cos(t)-3 = 0 (4cos(t)+3)(2cos(t)-1...
*May 7, 2015*

**Math**

The horizontal parabola y^2 = 4px has directrix p units from the vertex. So, since our directrix is 3 units from the vertex, we start with y^2 = 12x But, that's with a vertex of (0,0). So, our parabola is (y-1)^2 = 12(x+2) But, that opens to the right. Our vertex is to the...
*May 7, 2015*

**Calc/Precalc**

just do the same stuff: xy = cot(xy) y + xy' = -csc^2(xy) (y + xy') y'(x + xcsc^2(xy)) = -ycsc^2(xy)-y xy' (1+csc^2(xy)) = -y(1+csc^2(xy)) xy' = -y y' = -y/x The other is just a simple chain rule. y = cos(u), so y' = -sin(u) u'
*May 7, 2015*

**Poly**

42
*May 7, 2015*

**math**

4 of each length...
*May 7, 2015*

**Math**

can't happen. The angles must add up to 360.
*May 7, 2015*

**Math**

hint: r = 3
*May 7, 2015*

**Math**

That would of course be 6C2 (3x)^4 (-y)^2 = 15(81x^4)(y^2) = 1215x^4y^2
*May 7, 2015*

**Trig**

no, no. Draw your triangles: if cos(s) = 1/5, then sin(s) = √24/5 = 2√6/5 if sin(t) = 3/5, then cos(t) = 4/5 sin(s+t) = (2√6/5)(4/5) + (1/5)(3/5) = (8√6+3)/25 and similarly for sin(s-t)
*May 7, 2015*

**maths**

you want km/L. So, use what you have: (44km)/(11/4 L) = 44km * 4/11L = 16 km/L
*May 7, 2015*

**Calculus**

that would be ∫[0,2] f(x) dx ------------------- 2-0
*May 7, 2015*

**Algebra**

15+12x+40 = 127
*May 7, 2015*

**matha**

250500 * 1/5 * 2/5 = ?
*May 7, 2015*

**Calculus**

V = ∫[-1,1] π(e^x)^2 dx or V = ∫[0,1/e] 2π(1/e)(1-(-1)) dy + ∫[1/e,e] 2π(y)(1-lny) dy
*May 7, 2015*

**maths**

3/4 of 4L = 3L 1 - 3/4 = 1/4
*May 7, 2015*

**Calculus**

we can check using shells. V = ∫[0,1] 2πrh dy where r = 1-y and h = y-y^3 V = 2π∫[0,1] (1-y)(y-y^3) dy = 2π∫[0,1] y^4-y^3-y^2+y dy = 2π(1/5 y^5 - 1/4 y^4 - 1/3 y^3 + 1/2 y^2) [0,1] = 2π(1/5 - 1/4 - 1/3 + 1/2) = 2π(7/60) = 7π...
*May 7, 2015*

**Calculus**

or, using shells, you can do V = ∫[0,9] 2πrh dy where r = y and h = 3-x = 2π∫[0,9] y(3-√y) dy = 2π(3/2 y^2 - 2/5 y^(5/2)) [0,9] = 2π(3/2 * 81 - 2/5 * 243) = 2π(243/10) = 243π/5
*May 7, 2015*

**Math**

x + 1/ x - 1 + x^2 -1 / x + 1 x+(1/x)-1+x^2-(1/x)+1 x-1+x^2+1 x+x^2 However, assuming the usual sloppiness with parentheses, I suspect you meant (x+1)/(x-1) + (x^2-1)/(x+1) (x+1)/(x-1) + (x-1)(x+1)/(x+1) (x+1)/(x-1) + (x-1) (x+1)/(x-1) + (x-1)^2/(x-1) (x+1 + (x-1)^2)/(x-1) (x^...
*May 7, 2015*

**Help plz on Calc**

Think of the shells as nested cylinders, starting 1 unit away from the y-axis, and extending to the end of the ellipse: V = ∫[1,3] 2πrh dx where r=x and h=y V = 2π∫[1,3] x(2√(1-x^2/9)) dx = 2π/3 ∫[1,3] 2x√(9-x^2) dx = 2π/3 (32/...
*May 6, 2015*

**intermediate algebra**

if the radiator already contains pure antifreeze, how would adding more antifreeze change anything?
*May 6, 2015*

**intermediate algebra**

better read what you posted . . .
*May 6, 2015*

**Algebra**

a = 100 r = -1/20 An = ar^(n-1) Sn = a (1-r^n)/(1-r) A4 = 100(-1/10)^3 = 100(-1/1000) = -1/10 S4 = 100(1-(-1/10)^4)/(1 - (-1/10)) = 100(9999/10000)/(11/10) = 909/10
*May 6, 2015*

**calculus**

so, find a table of integrals and look it up. You will probably just find some power reduction formulas, such as ∫ x^n sinx dx = -x^n cosx + n∫ x^(n-1) cosx dx You can see that you will have to use integration by parts 4 times to get rid of all the x^n terms. So, ...
*May 6, 2015*

**precalculus**

since the vertices are on the y-axis, we will have y^2/a^2 - x^2/b^2 = 1 the slope of the asymptotes is b/a, so y^2 - x^2/4 = 1 but that has vertices at (0,+/-1) so y^2/4 - x^2/16 = 1 To verify, see http://www.wolframalpha.com/input/?i=hyperbola+y%5E2%2F4+-+x%5E2%2F16+%3D+1
*May 6, 2015*

**geometry**

tanθ = 1/11
*May 6, 2015*

**math**

all are correct, with the following notes: #1 has a typo. The 3rd y value should be zero Why say "1 over 3" when real mathematical notation (1/3) is so much better?
*May 6, 2015*

**chemisrty**

The molar mass of Fe(OH)3 is 106.87 g/mol. How many moles of H2SO4 are needed to react completely with 5.419 g of Fe(OH)3?
*May 6, 2015*

**math**

well, on the RS, csc^2 - sec^2 = (cos^2-sin^2)/(sin^2 cos^2) That should help...
*May 6, 2015*

**mathematics - eh?**

what does "sin alpha 9" mean?
*May 6, 2015*

**Math**

that's cylindrical cup sheesh. the area of the circular base, is just the area of a circle. pi r^2 the curved surface is just the circumference times the height: 2pi r h Now just plug in your values for r and h
*May 6, 2015*

**Math**

well, just offhand, I'd say that's 80% of 64. Or, .80 * 64 = ? Hmmm. Since 64 is not a multiple of 5, how could 80% (4/5) do it?
*May 6, 2015*

**GEOMETRY**

Oh come on. Do you have a triangle? parallel lines? intersecting lines? Geez, just describe the arrangement of the points and lines, fer cryin out loud! What is the relationship between MO and NA? Perpendicular? Parallel? Pretend I can't see the diagram. Tell me what to do...
*May 6, 2015*

**GEOMETRY**

you'd better describe the figure. Some of your copy/paste is garbled, and we have no idea of the relative locations of the points.
*May 6, 2015*

**math**

y changes by +1 when x changes by 1. So, the slope is +1/1 = 1.
*May 6, 2015*

**math**

total water: (4*.750 + 2*1.5)L at 1L/min, how long is that?
*May 5, 2015*

**science**

(5000+750)/(180) servings I'll let you decide how much is left over
*May 5, 2015*

**Calc**

h(t) = 200 + 40t - 16t^2 now work your magic on (b) and (c)
*May 5, 2015*

**Math**

both wrong. see earlier post.
*May 5, 2015*

**Math**

if the roots are a and b, the function is (x-a)(x-b) = 0 So, plug in a=(2+√5)/3 b=(2-√5)/3
*May 5, 2015*

**336**

Bzzzt, but thanks for playing! since the number of messages is an integer, it is clearly discrete since time can take on any value, it is continuous
*May 5, 2015*

**math**

repeated multiplication x^2 = x * x x^3 = x * x * x and so on
*May 5, 2015*

**Math**

reflection in the y-axis takes (x,y) -> (-x,y) Just fold the paper along the y-axis and see what the points do.
*May 5, 2015*

**Math**

you have y = (x-a)^2 - b so, (x-a)^2 = b x = a±√b = (k/2)±√((k-2)^2/4) = (k/2)±(k-2)/2 = k/2 + k/2 - 1 = k-1 or k/2 - k/2 + 1 = 1 check x=1: (1-(k/2))^2 - (k-2)^2/4 (2-k)^2/4 - (k-2)^2/4 0
*May 5, 2015*

**math**

guess you missed the deadline, eh? 5 kinds, choose 3. Sound familiar?
*May 5, 2015*

**Probability**

clearly, based on the 69 draws, P(pink) = 36/69 P(brown) = 33/69
*May 5, 2015*

**Math**

That's the same as the number of ways to select 6 items all at once, then read them once a week. 20P6 Unless you are allowed to choose the same book more than once. Then the number is 20^6
*May 5, 2015*

**Math**

There are 20 books on a summer reading list. In how many ways can you choose 1 per week for 6 weeks.
*May 5, 2015*

**Math**

well, how many perfect squares or odds are there in 1..8?
*May 5, 2015*

**Math**

You spin a spinner that has 8 equal sections numbered 1 to 8.Find p(perfect square or odd).
*May 5, 2015*

**math**

given your matrix A, we need to solve det(nI-A) = 0 so we need to solve |n-1 -1 0 0| |-1 n-1 0 0| |0 0 n 0| |0 0 0 n| = 0 That is n^3(n-2) = 0 As you can see, n = 0,0,0,2
*May 5, 2015*

**Math**

6 6/7 is near 7 4 2/7 is near 4 1 3/5 is near 1 3/4 = 7/4 4 * 7/4 = 7 Sounds good to me as an estimate
*May 5, 2015*

**trigonometry**

I AM WRONG >
*May 5, 2015*

**trigonometry**

How about some parentheses, so we can tell what's what? As it stands, it means 1 + tan2A = cosA + tanA - sinA which is clearly false
*May 5, 2015*

**functions**

you want R where k/R^2 = 10(k/r^2) R^2/r^2 = 1/10 R/r = 1/√10
*May 5, 2015*

**functions**

v = x(10-2x)(20-2x)
*May 5, 2015*

**Math**

so, now you can figure how many miles the light travels in 7200 seconds, right?
*May 4, 2015*

**Math**

well, how many seconds in 2 hours?
*May 4, 2015*

**Math**

42
*May 4, 2015*

**math**

you need 2/x < x/22 44 < x^2 so, 6 < x Now, you need x/22 < 0.33 x < 7.32 So, we have x = 7 2/7 < 7/22 < .33 Do the other in like wise.
*May 4, 2015*

**Algebra**

what's the problem? Just start working out the values: a1 = -2 a2 = 2(a1)^2 = 2(-2)^2 = 8 a3 = 2(a2)^2 = 2(8)^2 = 128 ... a1 = ln(e^2) = 2 a2 = ln(e^4) = 4 ... b0 = 1 b1 = 2 b2 = 2(2)-1 = 3 b3 = 2(3)-1 = 5 ...
*May 4, 2015*

**math**

it will be a line sloping upward, passing through (0,-4)
*May 4, 2015*

**Math**

the intersection(s) will be where x^2 + (x+k)^2 - 25 has one solution. That is, where the discriminant is zero. x^2 + x^2 + 2kx + k^2-25 = 0 2x^2 + 2kx + (k^2-25) = 0 The discriminant is (2k)^2 - 4(2)(k^2-25) 4k^2 - 8k^2 + 200 = 0 k^2 = 50 k = ±√50 So, check out ...
*May 4, 2015*

**Math**

y = 4(x^2+6x)-5 = 4(x^2+6x+9) - 5 - 4*9 = 4(x+3)^2 - 41 The vertex is at (-3,-41), and the parabola opens upward, y cannot be less than -41. Did you actually try plugging in, say, 7=700 to see whether there was a solution there?
*May 4, 2015*

**Math**

correct
*May 4, 2015*

**Algebra**

do this just like the ellipse in your earlier post. That is, complete the squares, then review your text about hyperbolas. 36x^2-24x - (y^2-6y) = 41 36(x - 1/3)^2 - (y-3)^2 = 41 + 36/9 - 9 (x-1/3)^2 - (y-3)^2/36 = 1
*May 4, 2015*

**Math**

Switching dircetion is just a shorthand way of moving stuff from one side to the other: -4k > -36 add 4k to both sides: 0 > 4k-36 add 36 to both sides: 36 > 4k or, as you saw above, 4k < 36 k < 9 multiplying and dividing by a negative value change the direction...
*May 4, 2015*

**Algebra**

for this ellipse, the major axis is vertical. a = 3 b = 2 The vertices are at (-3,1±3) The co-vertices are at (-3±2,1) Looks like it's time to review your ellipses.
*May 4, 2015*

**Algebra**

rearrange stuff and complete the squares: 9x^2+54x + 4y^2-8y = -49 9(x^2+6x) + 4(y^2-2y) = -49 9(x^2+6x+9) + 4(y^2-2y+1) = -49 + 9*9 + 4*1 9(x+3)^2 + 4(y-1)^2 = 36 (x+3)^2/4 + (y-1)^2/9 = 1 Now answer the questions.
*May 4, 2015*

**geometry - eh?**

what is the "lip"? and where is your punctuation?
*May 4, 2015*

**Math**

F = ma
*May 4, 2015*

**Math**

so, if you have the answer, why are you posting the question? Check out the Pythagorean Theorem. x^2+y^2 = 1600 x+y+40 = 96.22
*May 4, 2015*

**Math**

2r+h = 25 a = πr(r+2h) = πr(r+25-2r) = π(25r-3r^2) da/dr = 0 when r = 25/3 d = 2r = 50/3 h = 25 - 50/3 = 25/3
*May 4, 2015*

**Math**

x+y = 37 x^2+y^2 = 765.5 x^2+(37-x)^2 = 765.5 2x^2 - 74x + 603.5 x^2 - 37x + 301.75 = 0 x = (37±9√2)/2 x = 12.1, 24.8 Hmmm. I don't get their answers either, but I'm a lot closer than you. What did you do?
*May 4, 2015*

**geometry**

the diameters are in the ratio 5:6 the larger is 12, so the smaller is 10 the smaller cone has height h=3, so the larger cone has height 3*(6/5) = 18/5
*May 4, 2015*

**Math**

Laid out on the x-y plane, the track's line can be described by y = 400 - 2/3 x So, if the rectangle has one corner on the line and the opposite corner at (0,0), its area is a = xy = x(400 - 2/3 x) = 400x - 800/3 x^2 This is just a parabola. Find its vertex, and that is ...
*May 4, 2015*

**Calculus**

all of them? f is concave down with a max at x = -3
*May 4, 2015*

**math**

divide the total amount by the amount needed for each batch, giving the number of batches: 6/(2/3) = 6 * 3/2 = 9
*May 4, 2015*

**geometry**

Recall your rotation matrix. (x,y) -> (x',y') where x' = xcosθ - ysinθ y' = xsinθ + ycosθ So, plug in your θ. You can easily check your work by noting that the point will rotate down to the x-axis.
*May 4, 2015*

**math**

300*1.04^5
*May 4, 2015*

**alg**

no ideas? Review your conic sections and show us whatcha got.
*May 4, 2015*

**math**

since the dice are independent events P(odd,odd) = 1/2 * 1/2
*May 4, 2015*

**math**

$5200 / $6.50/m = 800m of fence maximum area for a given perimeter is a square, so we have an isosceles right triangle with both legs 400m long.
*May 4, 2015*

**math**

#1 clearly the circle is of radius 5. So, since x^2+y^2 = 5^2, (0,5) lies on the circle. the center is the midpoint of AC: (0,0) so, the radius is 13. 0^2+13^2 = 13^2 7^2 + (2√30)^2 = 49+120 = 13^2 so, all vertices lie on the circle.
*May 4, 2015*

**critical points**

critical values are where f' = 0 or undefined. f'(x) = 18x^2-18 = 18(x^2-1) So, where is f' = 0?
*May 4, 2015*

**math**

a = s^2 da = 2s ds
*May 4, 2015*

**trigonometry**

You can expand both binomials, and watch things cancel out, or note that 2sin^2(45+A) = 1-cos(90+2A) 2sin^2(45-A) = 1-cos(90-2A) cos(90+x) = -sinx cos(90-x) = sinx Now just add it up.
*May 4, 2015*

**math**

v = (12-2x)(16-2x)x dv/dx = 4(3x^2-28x+48) set dv/dx=0 and solve for x.
*May 4, 2015*

**math**

it reverses direction when v changes from + to - v(t) = 3t^2-18t+24 = 3(t^2-6t+8) = 3(t-2)(t-4) So, where does v(t) cross the axis and become negative? Now use that value to find s(t) and a(t)
*May 4, 2015*

**Physics**

1/48 + 1/di = 1/110 M = 110/(110-48)
*May 3, 2015*

**calc (please help)**

I'm assuming that the angle between the planes is situated on the edge of the cylinder, so that the two planes intersect the cylinder in a circle and an ellipse. So, the ellipse has major axis of length 2r secβ, and the far side of the major axis is a distance 2r tan...
*May 3, 2015*

**Math--ONE QUESTION**

well, -14 = -9-5 0 = 8-8 So, what do you think?
*May 3, 2015*

**Calculus max/min**

If you walk directly to a point x feet from the bus stop, your walking distance is park: √(600^2+(2000-x)^2) sidewalk: x so, your time taken (distance/speed) is t(x,d) = √(600^2+(2000-x)^2)/(3.5+d) + x/(5.5+d) t(x) = √(600^2+(2000-x)^2)/4.267 + x/7.267 so, ...
*May 3, 2015*

**Math Trigonometric Identities**

if you set the left side over a common denominator of sinx, you have 1/sinx - sin^2x/sinx (1-sin^2 x)/sinx If that does not get you started, I don't know what will...
*May 3, 2015*

**Math Trigonometric Identities**

that is correct, but it would have been simpler when you got the LS expanded, to see that you had LS = sin^2x - 2sinxcosx + cos^2x = sin^2x + cos^2x -2sinxcosx = 1 - 2sinxcosx
*May 3, 2015*

**Earth Science**

the magnification of a telescope is changed by changing what? my answer is the eyepiece
*May 3, 2015*

**Math**

well, just plug in 180 = pi
*May 3, 2015*

**Math**

Since you are actually performing the action, I'd have to go with Experimental. If asked just to calculate the probability, then it would be theoretical. The actual observations might or might not agree with the calculation. In fact, for any finite number of trials, the ...
*May 3, 2015*