Tuesday

July 7, 2015

July 7, 2015

Total # Posts: 32,149

**maths**

d = (32-11)/(12-5) = 3 a = 11-4*3 = -1 now you can fill in the rest.
*May 9, 2015*

**math**

nth from front is a + (n-1)d nth from l is l - (n-1)d their sum is a+l Not sure what you meant by "a l"
*May 9, 2015*

**precalculus**

the coefficients are 1,3,3,1 so you have 1(2x)^3(-3y)^0 + 3(2x)^2(-3y)^1 + 3(2x)^1(-3y)^2 + 1(2x)^0(-3y)^3 = 8x^3 - 36x^2y + 54xy^2 - 27y^3
*May 9, 2015*

**extreme math help please**

change in x: 2 change in y: 3 y = 3/2 x - 5
*May 9, 2015*

**college precalculus**

check for common difference or ratio: #1: d = 1 #2: r = 2/3 #3: differences: -5/2, +7/2, -9/2, ... ratios: -3/2, +4/3, -5/4, ... neither d nor r is constant, so the sequence is not AP or GP
*May 9, 2015*

**algebra**

y grows by 5 when x grows by 1. So, since 6-1=5, when x=6, y=4+5*5 = 29 or, using the point-slope form, y-4 = 5(6-1)
*May 8, 2015*

**Calculus**

(a) ok (b) ok (c) g" = (x^2-4x+16)/(x-2)^2 Since g" is never negative, g is never concave down (d)correct, but since you know g(3)=4, y-4 = -9(x-3) (e) since the graph is always concave up, any tangent lines must lie below the graph. Doodle around some, and you will ...
*May 8, 2015*

**Physics**

distance = speed * time, so that wold be 3.00*10^5 km/s * 1.28s = 3.84*10^5 km
*May 8, 2015*

**trigonometry**

well, geez. I gave you the information: h(cot35+cot25) = 1650 Now just solve for h!
*May 8, 2015*

**trigonometry**

Drop the altitude from C to P, with height h AP = h cot35 PB = h cot25 AP+PB = 1650
*May 8, 2015*

**calculus**

rather than just sit around waiting, why not actually try it yourself? 1/[x^2(1+x^2)] = 1/x^2 - 1/(1+x^2) and those are both easy, right?
*May 8, 2015*

**math**

no ideas? Recall that L{y} = F(s) L{y'} = s F(s) - f(0) L{y"} = s^2 F(s) - s f(0) - f'(0) then just collect terms, solve for s, and do L^-1{s}
*May 8, 2015*

**math**

5/8 x = 13 x = 13 * 8/5 Strange, since 13*8 is not a multiple of 5.
*May 8, 2015*

**algebra**

Are the clubs mutually exclusive?
*May 8, 2015*

**Physics**

A heat engine is 20% efficient. If it loses 800 J to the cooling system and exhaust during each cycle, the work done by the engine is: 200 J 1000 J 800 J 20 J
*May 8, 2015*

**extreme math help please**

note that y decreases by 1 when x increases by 2. So, start with y = -1/2 x But, -1/2 (2) = -1, and y(2) = 1. So, y = -1/2 x + 2
*May 8, 2015*

**Physics**

How many kg of ice needs to be added to 1.31 kg of water at 64.9°C to cool the water to 14.9°C when the Latent heat of ice = 80 kcal/kg?
*May 8, 2015*

**Calculus, derivatives**

x^2+2r√y = xy If y is constant, it's just like any other dumber, so 2x dx/dr + 2√y = y dx/dr dx/dr (2x-y) = -2√y dx/dr = 2√y / (y-2x)
*May 8, 2015*

**math**

a quadratic function involves x^2 So, (C) is a quadratic. Surely you could have discovered that information (a) in your text (b) via google
*May 8, 2015*

**math**

when x increases by 2, y increases by 3. So, start with y = 3/2 x But y(0) would be 0, and the table says it is -5. So, y = 3/2 x - 5
*May 8, 2015*

**Calculus**

y = 5x√(121-x^2) This is an odd function, so algebraically, the area is zero -- equal areas above and below the x-axis. By symmetry, the geometric area is a = 2∫[0,11] 5x√(121-x^2) dx If you let u = 121-x^2, du = -2x dx, so a = 2∫[121,0] -5/2 √u ...
*May 8, 2015*

**Calc 2**

well, consider that 11n just gets bigger and bigger without bound. csc(x) has an asymptote at x = π That is, csc(x) -> ∞ as x -> π So, csc^1(x) -> π as x -> ∞ Or, in the case of our sequence, lim {csc^(-1) 11n} = π n->∞
*May 8, 2015*

**math**

.05x = 95
*May 8, 2015*

**Math**

Assuming a sinusoidal function, we have the area is (letting x=0 in September) a(x) = 7.5 cos(pi/6 x) + 10.5 we want a(x) > 15. In other words, cos(pi/6 x) > 0.6 -1.77 < x < 1.77 so, using symmetry, in one complete period (a year), there are 3.54 months with more ...
*May 8, 2015*

**maths**

Let v = victor v+24 = lahai so, adding them up, you have v + v+24 = 60
*May 8, 2015*

**STS**

since s=w+3, w=s-3 3/2 = s/(s-3)
*May 8, 2015*

**Math**

correct
*May 8, 2015*

**STS**

.10(40) + 1.00(x) = .20(40+x)
*May 8, 2015*

**STS**

.12x = 69
*May 8, 2015*

**mathematics**

that would be totalpoints/totalstudents: (35*80 + 15*70)/50
*May 8, 2015*

**Geometric Series**

a = 5(-1/6)^5 Note that (-1/6)^(5k) = ((-1/6)^5)^k, so r = (-1/6)^5 S = a/(1-r) = (5(-1/6)^5)/(1-(-1/6)^5) = -5/7777
*May 7, 2015*

**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*