Monday

May 25, 2015

May 25, 2015

Total # Posts: 31,368

**trig**

find the angle between the pole and the wire using the law of sines: 35/sin30 = 25/sinθ Now use the law of sines again to find x, since you can now figure the angle opposite x.
*May 10, 2015*

**Trig**

or, having found B, you can use the law of cosines: x^2 = 35.4^2 + 10.2^2 - 2(35.4)(10.2)cosB
*May 10, 2015*

**Trig**

If the ladder makes an angle A with the ground and B with the embankment, then using the law of sines, 35.4/sin 117.5 = 10.2/sinB = x/sinA Now you can find B, and A = 180-(117.5+B) and you can find x.
*May 10, 2015*

**binomial expansion**

since the powers add up to 12, that would be the 10th term: 12C9 (√2)^9y^3 = 12C3 (16√2)y^3 = 3520√2 y^3
*May 9, 2015*

**binomial theorem**

That would be 22C19 a^3 b^19 = 22C3 a^3 b^19 = 1540a^3b^19
*May 9, 2015*

**trig**

note that h cot38° - h cot75° = 65
*May 9, 2015*

**Pre-Calculus**

cos x cot^2 x= cos x cos x cot^2 x - cos x = 0 cos x (cot^2 x - 1) = 0 cosx = 0 cotx = ±1 Now you can just find x, right?
*May 9, 2015*

**math**

4 pink out of 14 total, so P(pink) = 4/14 = 2/7
*May 9, 2015*

**Common Core Math II A-CR**

recall that the height h is given by h(t) = 128t - 16t^2 That's just your good old friend, the parabola. Find its vertex -- that is the maximum height.
*May 9, 2015*

**precalculus**

suppose the terms were -3,-2,-1,0,... would that be hard tto figure out? Well, it's the same here: (t-2)-(t-3) = 1 (t-1)-(t-2) = 1 The common ratios for the 2nd one are found by dividing each term by the previous one: 1,-3/2,2,-5/2 (-3/2)/(1) = -3/2 2/(-3/2) = -4/3 (-5/2...
*May 9, 2015*

**precalculus**

see related questions below
*May 9, 2015*

**Trig**

all you need here is the perpendicular distance from Nancy's plane to a N-S line. If that distance is x, then x/400 = sin 23° x = 156.29 you are correct. Nasty how they sneak in unrelated problems, eh?
*May 9, 2015*

**algebra**

1.22x = 3200000 x = 2622950.82
*May 9, 2015*

**math**

5x+7 = 7x-21
*May 9, 2015*

**math**

well, A+B=180, so solve for x. Also, A+D = 180
*May 9, 2015*

**math**

2*15/√3
*May 9, 2015*

**math**

the two angles are equal, so just set them equal and solve for x Then figure MRQ
*May 9, 2015*

**Math**

one statement is by definition independent.
*May 9, 2015*

**algebra**

If x at 40%, the rest (20-x) is 85%. So, .40x + .85(20-x) = .60(20)
*May 9, 2015*

**Maths**

2560/58.0 = x/(9.50*8.75)
*May 9, 2015*

**maths**

clearly a = √7 r = √3 An = √7*√3^(n-1)
*May 9, 2015*

**english**

1st step is learning to write it. As in Can you help me in my debate topic, 'Knowing English well is equal to having a college degree'?
*May 9, 2015*

**physics**

consider that T^2 = kr^3 That means that T^2/r^3 is constant T^2/r^3 = (2T)^2/(xr)^3 = 4T^2/x^3r^3 So, 4/x^3 = 1, and x = ∛4 That is, our small planet's orbital radius is ∛4 times earth's.
*May 9, 2015*

**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*