Friday

December 19, 2014

December 19, 2014

Total # Posts: 27,630

**Algebra 2 HELP plz**

t=2 is the correct time. So, plug in t=2 for your formula for h to get the max height. to see when it hits the ground, just set h=0: -16t^2+64t+80 = 0 -16(t+1)(t-5) = 0 so, h=0 at t=5
*November 12, 2014*

**Pre-Calc/Trig...**

2(x+5)/(x-3)
*November 12, 2014*

**College Algebra**

This is just a parabola, with its vertex at h = -b/2a, or in this case h = 3.817/0.14
*November 12, 2014*

**Math**

w(3w-5) = 12 ...
*November 12, 2014*

**College Algebra**

y = a(x+2)(x-8) The vertex is midway between the roots, at x=3, so y(3) = 5 a(5)(-5) = 5 a = -1/5 y = -1/5 (x+2)(x-8) See http://www.wolframalpha.com/input/?i=-1%2F5+%28x%2B2%29%28x-8%29
*November 12, 2014*

**College Algebra**

usually college algebra students haven't learned about derivatives. However, s(t) is just a parabola, with its vertex at t = 112/32 so, plug that into s(t) to find the max height Note that you got your derivative wrong, anyway.
*November 12, 2014*

**algebra**

if x at 2%, then the rest (10000-x) at 5%. So, 10000-x = 3x so solve for x, and the interest earned is .02x + .10(3x) = ?
*November 12, 2014*

**algebra**

y = ab^x a*b^2 = 3 a*b^9 = 33 a = 3/b^2, so a*b^9 = 3/b^2 * b^9 = 3b^7 = 33 b = 11^(1/7) a = 3/11^(2/7) y = 3/11^(2/7) * 11^(x/7) = 3*11^((x-2)/7) y(15) = 3*11^(13/7) = k
*November 12, 2014*

**math olympiad**

there are many such rectangles. 1 x √88 2 x √85 ... √44 x √45
*November 12, 2014*

**intermediate algebra**

the line through (2,3) and (2,0) is x = 2 That is a vertical line. So, the perpendicular line through (2,3) is y = 3
*November 12, 2014*

**Maths**

h/10 = 1.8/1.2
*November 12, 2014*

**St Ans School**

75-x = x-45 ...
*November 12, 2014*

**Math**

How about writing that as 3x - 8y - 4 = 0
*November 12, 2014*

**Math Help**

ln√x-8=5 ln√x=13 √x = e^13 x = e^26 log2x+log2(x+2)=log2(x+6) log2(x(x+2)) = log2(x+6) x(x+2) = x+6 x^2+2x = x+6 x^2+x-6 = 0 (x-2)(x+3) = 0 x = 2 or -3 But, log2(x) is not defined for x <= 0, so x=2 is the only solution check: log2(2)+log2(2+2) = log2(2+6...
*November 12, 2014*

**Math**

if course not "all reals" plug in just some value and see whether the equation balances. In particular, try x=0: log(4) =?= log(2) I don't think so. What you need to realize is that if 2^x = 2^y or log2(x) = log2(y) or 2x = 2y or 2+x = 2+y or in general, for any ...
*November 12, 2014*

**Maths**

1.3/1.8 = h/(7+1.8)
*November 12, 2014*

**Math**

you want e^.12x = 3
*November 12, 2014*

**maths**

a:b = 3:5 = 12:20 b:c = 4:7 = 20:35 a:b:c = 12:20:35
*November 12, 2014*

**math algebra**

Hard to tell, but it appears that you have (x-3)/4 - (y-2)/5 = -1/4 (4-x)/3 + (3-y)/2 = -13/20 which is the same as 5x - 4y = 2 20x + 30y = 209 20x-16y = 8 20x+30y = 209 46y = 201 so, x = 448/115 y = 201/46 Hmmm. Maybe you had x - 3/4 - y - 2/5 = -1/4 4 - x/3 + 3 - y/2 = -13/...
*November 12, 2014*

**Pre-Calculus**

recall that cos 2x = cos^2 x - sin^2 x
*November 12, 2014*

**Pre-Calculus**

7/12 = 1/3 + 1/4 now review your addition formula for tan(x+y)
*November 12, 2014*

**Pre-Calculus**

recall that cos(x-y) = cosxcosy+sinxsiny hence the name cosine -- sine of complementary angle
*November 12, 2014*

**maths**

we need (1+x)(0.9) = 1.08 x = 0.20 = 20% check: on $1.00 cost, mark up to $1.20 discount = $0.12, so the final price is $1.08, or 8% profit
*November 12, 2014*

**Pre-Calculus**

cos(cos^2+sin^2) don't forget your Algebra I now that you're taking trig!
*November 12, 2014*

**Math**

surely the explanation in your text will be more thorough than any offered here. Also, google is your friend...
*November 12, 2014*

**Geometry**

s = rθ, so θ = s/r
*November 11, 2014*

**math**

12 and 15
*November 11, 2014*

**math**

yes, you did
*November 11, 2014*

**Calculus**

just use the power rule: d/dx(x^n) = n x^(n-1) so, f'(x) = 3x^2 + 5/x^2
*November 11, 2014*

**Geometry**

No, it's November 11. I think you need to provide a little more context. p -> q where p=It's snowing today and q=It is December
*November 11, 2014*

**math**

Assuming that's $2.90 each, then 2*2.9 = 5.80 So, now, which has cost more for two packs of batteries?
*November 11, 2014*

**calculus related rates**

PV = k V dP/dt + P dV/dt = 0 dP/dt = -(P/V) dV/dt Now just plug in your numbers. By the way, this is true only if the temperature remains constant. More generally, PV = kT
*November 11, 2014*

**calculus inverted cone**

when the contents have depth y, the radius of the surface of the liquid is y/2 So, v = pi/3 r^2 y = pi/3 (y/2)^2 y = pi/12 y^3 dv/dt = pi/4 r^2 dy/dt Now just plug in your numbers.
*November 11, 2014*

**Math**

e+s = 80 e-17 = 2(2/3 s) Now just solve for e and s, and then get e-s.
*November 11, 2014*

**Math**

looks ok. But that depends on what x represents...
*November 11, 2014*

**Math**

9^2+35
*November 11, 2014*

**Math**

5 * 50/(5+8+7) = 12.5 m
*November 11, 2014*

**College Algebra**

there are lots of good graphing sites. Try here: http://www.wolframalpha.com/input/?i=plot+y%3D1%2F2x-3 The two equations describe the same line.
*November 11, 2014*

**Advanced Algebra**

3x=4(4-2x)=11 I suspect a typo
*November 11, 2014*

**Pre-Calc/Trig**

extrema where 12x^2 - 24x = 0 12x(x-2) = 0 Now apply the 2nd derivative test, and what you know about the general shape of cubics.
*November 11, 2014*

**math**

I suspect that V = XYZ So, just multiply the three values
*November 11, 2014*

**Calculus**

Go to http://www.wolframalpha.com/input/?i=8ln%28sec%28x%29%29+ and scroll down a bit.
*November 11, 2014*

**Derivatives**

Hmmm. You have y = uv, so y' = u'v+uv' = 3x^2(5x-1)^4 + x^3 * 4(5x-1)^3(5) = x^2(5x-1)^3 (3(5x-1) + 20x) = x^2(35x-3)(5x-1)^3
*November 11, 2014*

**Geometry**

if the circle has diameter d, then the square has sides equal to the diameter. area = d^2
*November 11, 2014*

**Pre-Calc/Trig**

x^5 is positive for x>0 and negative for x<0. So, the graph curves down on the left, and up on the right. All you really need to look at is the highest power of x, which dominates everything for large values of x.
*November 11, 2014*

**math - oops**

Oops. That is Nx - My I'll let you make the change and redo the arithmetic.
*November 11, 2014*

**math**

Green's Theorem says that ∮CMdx+Ndy = ∫∫RMx+Ny dx dy Nx = 8xy My = 2xy So, we end up with ∫[0,2]∫[x,2x] 10xy dy dx = ∫[0,2] 5xy^2 [x,2x] dx = ∫[0,2] 5x(4x^2-x^2) dx = ∫[0,2] 15x^3 dx = 5x^4 [0,2] = 80
*November 11, 2014*

**algebra 2**

Just as with normal fractions, you need to find a common denominator. For example, with 2/3 + 3/5, the commond denominator is (3)(5) Here, you will have a common denominator of (x+y)(x-y) So, you will wind up with (x-y)(x-y)/(x+y)(x-y) + (x+y)(x+y)/(x-y)(x+y) = [(x-y)^2 + (x+y...
*November 11, 2014*

**math**

telepathy
*November 11, 2014*

**Algebra please help answer ASAP**

(A) 5km/.5hr = 10.0 km/hr (B) 20km/.8hr = 25.0 km/hr (C) 10.0+25.0 = 35.0 km
*November 11, 2014*

**Algebra please help answer thx**

1/4 of the guitarists play jazz, so 3/4 play classical. So, 3/4 * 5/8 * 64 = 30
*November 11, 2014*

**math**

you are correct, because 6 and23 have no common factors
*November 11, 2014*

**math**

every 12 minutes both change at the same time.
*November 11, 2014*

**Math**

every 35 days the schedules coincide
*November 11, 2014*

**Precalc/Trig**

The first 4 are all ok cot is negative in QII and QIV, and you are right about π±θ. But, that θ is the reference angle, which is always in QI. θ = arccot(5), not -5.
*November 11, 2014*

**Math**

(-x+4)log11 = (3x-2)log5 -log11 x + 4log11 = 3log5 x - 2log5 -(log11+3log5)x = -(2log5 + 4log11) x = (2log5+4log11)/(log11+3log5) and, of course, you could go on and say x = log(5^2*11^4)/log(11*5^3) = log366025/log1375 or even log1375366025
*November 11, 2014*

**Grammar**

more in keeping with the original might be: The medication had a noticeable positive effect on the patient.
*November 11, 2014*

**physics**

Snell's Law says that sinθ/sin30° = 1/1.33
*November 11, 2014*

**math**

3 whites out of 10 total balls makes P(white) = 3/10
*November 10, 2014*

**Calculus**

Not sure why there are two = signs. Also, Phi is usually an angle, unless it is the golden ratio. I've never seen either one raised to such a power. Where did you get the "equation"?
*November 10, 2014*

**algebra**

both domain and range are (-∞,+∞) the domain for all polynomials is the same.
*November 10, 2014*

**Calculus**

so, why toss in the product rule as well? :-)
*November 10, 2014*

**calc/trig**

to find θ, note that tanθ = 84/130
*November 10, 2014*

**trig**

cosθ = 1/9 θ = 83.6° That is your reference angle. Since cosine is positive in QI and QIV, your two angles are 83.6° and (360-83.6)°
*November 10, 2014*

**Math**

(6 1/2)/(5 1/5) = (13/2)/(26/5) = 13/2 * 5/26 = (13*5)/(2*26) = 5/4 = 1 1/4
*November 10, 2014*

**science**

s = 1/2 at^2 v = at = √(2as) So, just plug in the numbers to get v.
*November 10, 2014*

**algebra**

Note that A(8,12)+(-5,-13) = M(3,-1) Since M is halfway to B, we need to add the same amounts to M that got us there from A: M(3,-1)+(-5,-13) = B(-2,-14)
*November 10, 2014*

**Math please help**

Just so you can check your work, you should come up with 1/8
*November 10, 2014*

**math**

5.6930 x 10^4
*November 10, 2014*

**pre calc**

f(x) = 6x^7 + 7x^6 f'(x) = 42x^6 + 42x^5 = 42x^5(x+1) so, now it's easy to see where f'(x) = 0
*November 10, 2014*

**Math Pre Calc Help**

(cos^4x -sin ^4 x)/(sin^2x) (cos^2x-sin^2x)(cos^2x+sin^2x)/sin^2x (cos^2x-sin^2x)(1)/sin^2x cos^2x/sin^2x - sin^2x/sin^2x cot^2x - 1
*November 10, 2014*

**College Math**

v0 t - 16t^2 = 0 when t=8, so 8 v0 - 1024 = 0 v0 = 128 Now just find the vertex of the parabola, at t = -b/2a which in this case is at t = v0/32 = 4 that sounds correct, since it is at half the time to hit. 4 sec up, 4 sec down. So, just find s(4)
*November 10, 2014*

**Pre-Calc/Trig**

y-intercept is where x=0: f(0) = 0 x-intercept is where y=0 y = x(4-x^2) = x(2-x)(2+x) So, the x-intercepts are at -2,0,+2
*November 10, 2014*

**Math 201 ( Algebra)**

If there are x at $40 and y at $80, then 40x+80y = 1000 x = y+80 Now just solve for x and y
*November 10, 2014*

**calculus**

if the square has side s and the circle has radius r, then r+s = 360 pi r^2 = (360-r)^2 (pi-1)r^2 + 720r - 129600 = 0 r = 129.85 ...
*November 10, 2014*

**Calculus**

y = x^(2/3) * ((5/2)-(x) y' = (2/3)x^(-1/3)*(5/2 - x) + x^(2/3)(-1) = 5(1-x) / (3*x^(1/3)) So, y'=0 when x=1 y" = -5(2x+1) / (9x^(4/3)) y"(1) < 0, so f(1) is a maximum See the graph at http://www.wolframalpha.com/input/?i=+%28x^2%29^%281%2F3%29+*+%28%285%...
*November 10, 2014*

**Calculus angle of elevation with elevator**

The two solutions below have exactly the same problem, and each step is shown. Where do you get stuck?
*November 10, 2014*

**Calculus angle of elevation with elevator**

see the related questions below - just change the numbers
*November 10, 2014*

**limerick poem**

oops - bad closing tag...
*November 10, 2014*

**limerick poem**

Just because it has five lines and rhymes, that does not make it a limerick. There is a certain very strict meter which must be followed. Read a bunch of them, then try again. Your meter is way off. The rhythm must be as in There was an old man from Nantucket Who kept all his ...
*November 10, 2014*

**Math! Algebra 1!**

If the still-water speed is s, then since distance = speed*time, then in equal time (say, t=1), (s+1.5) = 2(s-1.5) s+1.5 = 2s-3 s = 4.5 100/4.5 = 22.22 hrs or 22 hr 13 min 20 sec
*November 10, 2014*

**Calculus angle of elevation**

of course there is... when the elevator is x meters up, tanθ = (x-31)/22 So, sec^2θ dθ/dt = 1/22 dx/dt when x=20, tanθ = -11/22, so sec^2θ = 5/4 5/4 dθ/dt = 1/22 (3) dθ/dt = 55/6 m/s Is one of the steps unclear?
*November 10, 2014*

**Calculus angle of elevation**

see the related questions below. The first is the identical problem -- just change the numbers.
*November 10, 2014*

**math**

so, OHE = 3HER Sure you didn't leave something out?
*November 10, 2014*

**mathematics**

2OR = 2x+5 2OR = HE+5 HE = 2OR-5
*November 10, 2014*

**mathematics - incomplete**

can you not tell there is some missing information here?
*November 10, 2014*

**Science**

(8m/s)/(3s) = 8/3 m/s^2 (40m/s)/(8s) = 5 m/s^2 which is greater?
*November 10, 2014*

**Algebra**

the factors of -7 are -1 and 7 1 and -7 can you pick them so they add up to -6? (x+1)(x-7)
*November 10, 2014*

**math**

So, you must want the pair (cosθ,sinθ) which is the point on the unit circle associated with θ. 30: (0.866,0.5) 60: (0.5,0.866)
*November 10, 2014*

**math**

ok - I thought that might be it. So, now, what ordered pair did you have in mind? 30 degrees is just an angle. Is there supposed to be an ordered pair associated with an angle?
*November 10, 2014*

**math**

eh? Not sure what it is you are asking. ff?
*November 10, 2014*

**geometry - incomplete**

so, what is the question, again? You seem to have dropped some data, as well as what it is you are asking.
*November 10, 2014*

**math**

just write the data as given: a+b+c+d = 9000 a+c+d = 4b b+d = 4/5 (a+c) The first says that a+c+d = 9000-b using the second, that means that 9000-b = 4b b = 1800 So, now we have a+c+d = 7200 1800+d = 4/5 (a+c) Now we know that a+d = 7200-d, so 1800+d = 4/5 (7200-d) 1800+d = ...
*November 10, 2014*

**math(fractions)**

no idea. How long is the Blue Trail?
*November 10, 2014*

**calculus**

online, we generally use ^ for powers, and * for multiplication. Why bother changing the way it was posted? And the quotient rule says 5(3x+4)^4(3)(4x+1)^2 - (3x+4)^5(2(4x+1)(4)) ----------------------------------------- (4x+1)^4 (3x+4)^4 (15(4x+1)-8(3x+4...
*November 10, 2014*

**what is the rate of change of the angle of eleva**

see related questions below
*November 10, 2014*

**calculus**

when the elevator is x meters up, tanθ = (x-20)/20 So, sec^2θ dθ/dt = 1/10 dx/dt when x=10, tanθ = -1/2, so sec^2θ = 5/4 5/4 dθ/dt = 1/10 (5) dθ/dt = 2/5 m/s
*November 9, 2014*

**physics**

Look out below!! Was there a question in there somewhere?
*November 9, 2014*

**phyics**

If the tug pulls at speed v on a course of θ, then in x-y coordinates, 15 + 9cos60° + 35 sinθ = 0 19.5 + 35 sinθ = 0 sinθ = -0.5571 θ = N 33.9° W Now you can figure part B.
*November 9, 2014*

**Calculus**

revenue = price * quantity, so r = x(160-2x) = 160x - 2x^2 max revenue when r'=0, or x = 40 So, just plug that in as needed.
*November 9, 2014*

Pages: <<Prev | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | **11** | 12 | 13 | 14 | 15 | 16 | 17 | Next>>