Wednesday

August 27, 2014

August 27, 2014

Total # Posts: 24,450

**algebra**

If you mean 2y/3 + 2y - 6/15 then that is just 2y/3 + 6y/3 - 6/15 = 8y/3 - 2/5 = (40y-6)/15 If that's not it, then use some parentheses and show your own work.
*June 19, 2014*

**Math**

well, you have ax^2 = -h ax^2 = h In either case, a and h must have the same sign, since x^2 is always positive. Unless x=0, in which case, h=0. x = ±√|h/a| Naturally, if a=0, there is no solution unless h is also zero.
*June 19, 2014*

**math**

looks like you need to review your basic triangles: sin 60°/cos^2 45° - cot 30° + 15cos 90° √3/2 / 1/2 - √3 + 15*0 √3 - √3 0
*June 19, 2014*

**Math**

You have calculated the number of feet in 26 miles. Now divide that by (32/12), the number of feet in one stride. Or, you can multiply 137280 by 12 to get the number of inches in 26 miles, and then divide that by 32, the number of inches in one stride. Same difference.
*June 19, 2014*

**Trigonometry**

secx - √2 = 0 secx = √2 x = π/4 If this gave you trouble, it looks like you need to review your basic trig functions and common triangles.
*June 19, 2014*

**algebra**

x+y = 15000 .08x + .07y = 1100 Now just solve for x at 8% and y at 7%
*June 18, 2014*

**Physics**

(30,0) + (9.64,-11.50) = (39.64,-11.50) √(39.64^2 + 11.50^2) = 41.27 tanθ = -11.50/39.64
*June 18, 2014*

**Math**

750000 * 0.8^3 at the start of the 4th year. 750000 * 0.8^4 at the end of the 4th year you will have to decide when "during" the year you want its value.
*June 18, 2014*

**Math**

sinθ = 10/15
*June 18, 2014*

**Math**

(1*1 + 5*2 + 3*5 + 1*10 + 1*100) ------------------------------------ (1+5+3+1+1)
*June 18, 2014*

**finite math**

(# left field) / (# hits) = 6/59
*June 18, 2014*

**math**

Clearly, attendance on M-W is 33,34,35 So, attendance on Th-Sat is 36,37,38 with average=37
*June 18, 2014*

**algebra**

think 3-4-5 right triangle. or, do the math: √(50^2 - 30^2)
*June 18, 2014*

**Math**

481,924,821,947,217,481,274,832,748,971,298
*June 18, 2014*

**math - ouch**

Dang. Don't know what I was thinking.
*June 18, 2014*

**math**

what's the trouble? Just start working in the clues. The number is 2.abcdefg... c = a+b b = c-a (well, duh) b = c-1 It appears that the number is just 2.abc c = a+c-1 a = 2c+1 Now, since a < 10, c < 5 So, if a = 0,1,2,3,4 then c = 1,3,5,7,9 and b = 1,2,3,4,5 So, we ...
*June 18, 2014*

**Algebra 1**

the slope of the line is positive, so maximum y will occur where the maximum x occurs. So, y <= 5 Since lines extend to infinity, the range is thus (-∞,5]
*June 18, 2014*

**math**

The 48 is just noise. The cost (c) of the tickets for p pupils is c = 150p 150*43 = 6,450 So, the figure of 6,350 is incorrect. Whether or not it is wrong for the teacher to say otherwise is a moral question ...
*June 18, 2014*

**math**

1/(1+√2+√3) * 1/(1-(√2+√3))/(1-(√2+√3)) = (1-√2-√3)/(1 - (√2+√3)^2) = (1-√2-√3)/(-4-2√6) = (1-√2-√3)/(-2(2+√6)) = (1-√2-√3)/(-2(2+√6)) * (2-√6)/(2-√6...
*June 18, 2014*

**Math**

yes each value of x produces a distinct value of y.
*June 18, 2014*

**Math**

yes no maybe
*June 18, 2014*

**math**

If those "x" signs mean multiplication, then you surely know that 0x anything is zero.
*June 18, 2014*

**math**

n(n-1)/2 = 171 Since 342 = 18*19, there were 19 teachers.
*June 18, 2014*

**physics**

g = GM/r^2 at r+h, g1 = GM/(r+h)^2 at r-x, g2 = GM/(r-x)62 ∆g = GM(1/(r-x)^2 - 1/(r+h)^2) = GM/((r-x)^2(r+h)^2) ((r+h)^2-(r-x)^2) = GM/((r-x)^2(r+h)^2) (2rh+h^2 - 2rx+x^2) Since r-x and r+h are both approximately r, we have ∆g = GM/r^4 (2r(h-x) + h^2+r^2) = 2GM/r^3...
*June 18, 2014*

**finite mate**

anythingP0 = 1 There is only 1 way to choose nothing.
*June 17, 2014*

**Algebra**

You can factor out a^2 b, giving a^2 b (a^2 + b^2) Not sure that's any simpler.
*June 17, 2014*

**Math**

multiply RS top and bottom by cos2x and you have 2/(1-cos2x) = 2/(1-(cos^2 x - sin^2 x)) = 2/(1-cos^2 x + sin^2x) = 2/(2sin^2 x) = 1/sin^2 x = csc^2 x
*June 17, 2014*

**Algebra**

you want two factors of 45 which add to 18. Since 45 = 15*3, you get (x+15)(z+3)
*June 17, 2014*

**math**

y = 6x+4 Since you know the number of games, and the cost depends on that, x is the independent variable.
*June 17, 2014*

**math**

Arrange the coefficients into a 3x3 matrix, filling in zeros for missing values. A matrix is in row echelon form if (1) All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (if any). (2) The leading coefficient of a nonzero row is always ...
*June 17, 2014*

**Algebra/please can some one help I forgot**

Just look up the formula for monthly payment of loan amortization, and plug in your numbers.
*June 17, 2014*

**MATH**

no problem. Literally -- no problem. All you have done is give us the definition of some set of BOYs.
*June 17, 2014*

**geometry**

C = 3.14 * 60 = 188.4 cm 33.3 rpm means the bicycle travels 33.3 * 188.4 cm in one minute.
*June 17, 2014*

**geometry**

360 degrees is just one revolution. So, in 48 minutes, it turns 48 times. By the way, most LPs rotate at 33 1/3 rpm, not just 1 rpm. Also, they don't play for any 48 minutes either. Stupid question.
*June 17, 2014*

**geometry - oops**

I misread 1008 as 1080. Go with Reiny.
*June 17, 2014*

**geometry**

720 = 2^4 * 3^2 * 5 1080 = 2^3 * 3^3 * 5 672 = 2^5 * 3 * 7 So, the largest number that evenly divides all of those is 2^3 * 3 = 24
*June 17, 2014*

**algebra**

x+y=15 8x+5y=96 now just solve for x and y
*June 17, 2014*

**Math**

150 is less than the mean for Natalia's class. So, if that score is removed, the mean rises. Similarly, since 150 is greater than the mean for Eric's class, if that score is added, the mean rises.
*June 17, 2014*

**Math**

y = x^2 + 3x + 1/4 = x^2 + 3x + (3/2)^2 + 1/4 - (3/2)^2 = (x + 3/2)^2 - 2 That should answer most of your questions.
*June 17, 2014*

**Math**

30x+20y+10z = 800000 x+y+z = 40000 x = y/2 x=10,000 y=20,000 z=10,000
*June 17, 2014*

**algebra 1**

If we model the situation as y = mx+b then we have 120m+b = 18 300m+b = 6 Subtract to get -180m = 12 m = -1/15 So, b = 26 y = -1/15 x + 26 y=0 when x = 390
*June 17, 2014*

**finite math**

Hmm. Looks like 0.438
*June 17, 2014*

**finite math - PS**

That's the experimental probability. The theoretical probability is still 1/2.
*June 17, 2014*

**finite math (probability**

I'd say 57/100
*June 17, 2014*

**math**

400m+b = 2150 300m+b = 1650 subtract to get 100m = 500 m = 5 so, b = 150 y = 5x+150
*June 17, 2014*

**physics**

given the drawing as shown below, I calculate 3.62 km at 46°
*June 17, 2014*

**geometry**

Assuming the chords are parallel, think of 8-15-17 right triangles.
*June 17, 2014*

**Science**

Seems to me that the only possible reason can be that ice provides less friction, whatever the underlying properties. Friction translates into a restraining force, reducing the speed.
*June 17, 2014*

**science**

F = ma, so 20 = 4m m = 5 kg
*June 17, 2014*

**math**

what a lot of noise. Distance traveled: 128100 ft = 24.26 mi Average speed: 335.92 mi/hr so, time falling is 24.26mi / 335.92mi/hr = 0.0722hr = 4.3 minutes
*June 17, 2014*

**chemistry**

1000cm^3 * 13.6g/cm^3 * 1mole/200.59g = 67.8 moles
*June 17, 2014*

**chemistry**

.09g * 1mole/18g * 6.02*10^23mol/mole = 3.01*10^21 molecules
*June 17, 2014*

**chemistry**

just plug in the mol wts of the compounds in 0.78*N2 + 0.21*O2 + 0.009*Ar * 0.001*CO2
*June 17, 2014*

**algebra**

the ratio is 6:6, making 12 total. 48 = 12*4, so 24:24 will satisfy everyone.
*June 17, 2014*

**Geometry**

figure the length and slope of AB and CD. If they match, it's a ||-ogram.
*June 16, 2014*

**Geometry**

well, what are the diagonals? AC and BD midpoint of AC = (-1/2,-2) midpoint of BD = (-1/2,-2) Looks like the diagonals bisect each other.
*June 16, 2014*

**projectile motion**

we know that the path follows x(t) = v cosθ t y(t) = 160 + v sinθ t - 4.9t^2 I will assume that the cliff and the tower are the same, otherwise I'm confused. x(20) = 1800 y(20) = 0 So, 20v cosθ = 1800 20v sinθ = 1800 Clearly θ = π/4 and v = ...
*June 16, 2014*

**Algebra**

no solution the lines are parallel, and so do not intersect.
*June 16, 2014*

**Algebra**

see the solution at http://www.wolframalpha.com/input/?i=solve+4y%3Dx%2B4%2C+x%3D2%2F3y%2B8%2F3+
*June 16, 2014*

**Physics**

150km/2.5hr = 60km/hr 150km/2.0hr = 75km/hr velocities is the same, but directed. 300km/4.5hr = 66.67km/hr avg velocity is 0, since no net distance was traveled.
*June 16, 2014*

**Physics**

4 hr * 300km/hr = 1200km 2.5 hr * 400km/hr = 1000km 2200km/6.5hr = 338.46km/hr
*June 16, 2014*

**Math: Trig**

since cos pi/3 = 1/2, sec pi/3 = 2 But, arcsec has principal values from 0 to pi, so arcsec(-2) = pi - pi/3 = 2pi/3. Note: Some authors define the range of arcsecant to be ( 0 ≤ x < π/2 or π ≤ x < 3π/2 ), because the tangent function is ...
*June 16, 2014*

**MATH**

No idea. Do it the way I did, but you have to use integration by parts. If you get stuck, show how far you got. You should wind up with -(x+1)e^-x for x<0 (x-1)e^x for x>=0
*June 16, 2014*

**MATH**

since |x| = -x for x < 0. ∫[-∞,x] e^-|t| dt = ∫[-∞,x] e^t dt if x < 0 = e^t So, for x>=0, ∫[-∞,x] e^-|t| dt = ∫[-∞,0] e^t dt + ∫[0,x] e^-t dt = 1 + (1-e^-x) = 2 - e^-x
*June 16, 2014*

**maths**

32 is 2^5 so, 1/32 = 1/2^5 = 2^-5
*June 16, 2014*

**Math - incomplete**

did you want some particular value for x?
*June 16, 2014*

**Calculus**

Note that you have ∫1/(x^2 − 2x + 8)^(3/2) dx = ∫1/((x-1)^2 + 7)^(3/2) dx If x-1 = √7 tanu (x-1)^2 + 7 = 7 sec^2 u dx = √7 sec^2 u du and you have ∫1/(√7 secu)^3 * √7 sec^2 u du = ∫ 1/7 cosu du = 1/7 sinu = 1/(7 csc u) = 1...
*June 16, 2014*

**pseudocode app**

this kind of code is about as simple as it gets. I don't know what language you plan to use, but do a little google to see how others have performed the operations you desire. There are lots of code snippets on line.
*June 16, 2014*

**pseudocode app**

read price tax = price * 0.08 print price,tax
*June 16, 2014*

**Dela ray Mkhatshwa j.s.s**

Let AC and BD intersect at E. Note that AE = 3 Draw a vertical line which will contain AC. Mark point E by intersecting an arc of radius 3 from A to E. Construct the perpendicular at E to AC. That will contain BD. Draw an arc centered at A of radius 6, which will intersect the...
*June 16, 2014*

**Arithmetic**

since the ratio between terms is constant, (k+6)/(2k+3) = k/(k+6) (k+6)^2 = k(2k+3) k^2+12k+36 = 2k^2+3k k^2 - 9k - 36 = 0 (k-12)(k+3) = 0 k = 12 and the sequence is 27,18,12,... so the ratio is 2/3 and S = 27/(1/3) = 81
*June 16, 2014*

**Science**

you need 4m = 600
*June 16, 2014*

**sci**

ever been on a roller coaster? ever stuck your finger in a wall outlet?
*June 16, 2014*

**algebra**

If the larger monitor has width w, then its length is w+5. So, we have w(w+5)-80 = 70 w^2 + 5w - 150 = 0 (w+15)(w-10) = 0 ...
*June 16, 2014*

**Math**

or, you can avoid b altogether and go straight to the point-slope form of the line with slope 1/2: y-12 = 1/2 (x-10)
*June 16, 2014*

**physics**

total distance is 30m + 40m = 60m = 6000cm total time is 5s + 20*60s = 1205s avg speed is thus 6000/1205 = 4.98cm/s
*June 16, 2014*

**science**

Look, I'm Furiously Engaged!
*June 16, 2014*

**Chemistry**

Since the reaction is 2HNO3 + Ba(OH)2 -> Ba(NO3)2 + 2H2O It takes 2 moles of HNO3 to neutralize 1 mole of Ba(OH2) 2.5g Ba(OH)2 is 0.0146 moles So, you need 0.0292 moles of HNO3 0.0292M / 0.109M/L = 0.268L of acid.
*June 16, 2014*

**maths**

since θ is in QII and Ø is in QIII, tanθ = 3/-4 secØ = √5/-2 now go figure.
*June 16, 2014*

**physics please help!!!!!!**

You want to find where 30t - 4.9t^2 = 60 - 4.9(t-.5)^2 t = 2.34 now plug that in to find either height. Assuming you meant 60m above where the first mass started. Things are different if the second mass is dropped when the first mass is 60m below it. But I'm sure you can ...
*June 16, 2014*

**Calculus**

Your function becomes less unwieldy if you note that if you divide top and bottom by cosx, you have y = (1/2 sin2x)/(sinx+cosx) All those nasty sec^2 and stuff go away and you have 2nd: wolframalpha.com is your friend. Enter 2nd derivative sinx/(1+tanx) and it pops right up. ...
*June 16, 2014*

**Calculus**

From what I can decipher from your cryptic use use of arcana (like A, II DC, Nr, etc.) your reasoning appears sound. I have a bit of trouble distinguishing between II DC and second DC. If they are the same, how can you check the sign of second DC without calculating it? If y&#...
*June 16, 2014*

**physics**

depends on the size of the umbrella.
*June 16, 2014*

**math**

The height y(t) is given by y(t) = h + vt - 16t^2 h + v - 16 = 148 h + 2v - 64 = 272 So, h=-8 v=172 y(t) = -8 + 172t - 16t^2 y(6) = 448
*June 16, 2014*

**algebra**

Since the population grew by a factor of 1.64 over 8 years, we have p = 4.2 * 1.64^(t/8) Since 1.64 = e^0.49 (call it e^.5), we have p = 4.2 * e^(t/16)
*June 15, 2014*

**math**

23% of wages is 14204.80 SS tax is 7658.24 FICA tax is 1791.04 so, subtract all those from her gross. Or, note that the total taxes add up to 38.3%, so just subtract that amount. Or not. You never really did ask a question.
*June 15, 2014*

**Precalc**

An equation is not a function. However, the definition of a function is usually written as an equation, as in f(x) = 2e^x - 7 But, that's really just a definition. An equation is usually something you have to solve for a particular value. For example, you might have to ...
*June 15, 2014*

**calculus**

Oops. The w(t) given above is the y-value of Janice's position. You also need to use the x-value, which I'm sure you can do.
*June 15, 2014*

**calculus**

If Janice is at (0,0) when t=0, then the wheel obeys w(t) = 20(1-cos(π/8 t)) The ball follows the parametric equations x(t) = -70 + 60cos60° t y(t) = 60sin60° t - 16t^2 I have no idea where point A is, but having the equations should make things amenable to solution.
*June 15, 2014*

**maths**

(x+y)/2 = 44 If x=12, then (12+y)/2 = 44 12+y = 88 y = 76
*June 15, 2014*

**Algebra**

we have x^2-2 where x <= 1 2x where x < 1 Not quite clear where you want to go with this. I suspect you really meant 2x where x < 1 x^2-2 where x >= 1 If so, then the graph is a line which stops at x=1, and a parabola from there on. There is a jump discontinuity at...
*June 15, 2014*

**bearing and distance**

Use the law of cosines. After t hours, the distance d between the vehicles is d^2 = (8t)^2 + (5t)^2 - 2(8t)(5t)cos70° because the angle between the bearings if 25+45 = 70° So, just plug in t=3 and solve for d. As for the bearing from q to p, just find the slope of the ...
*June 14, 2014*

**math,bearing and distance**

not sure what bearing "s30 degree of north" is.
*June 14, 2014*

**Math**

you can easily see that K = 18-C So, put that into the 2nd equation and you have 2(18-C)+4C = 50 36-2C+4C = 50 2C = 14 C = 7 So, K = 11
*June 14, 2014*

**math**

x^3+3x = x(x^2+3) = (a^2/3-a^-2/3)((a^2/3-a^-2/3)^2+3) = (a^2/3-a^-2/3)(a^4/3 + 1 + a^-4/3) Note we have the factorization of the difference of two cubes: = a^2 - a^-2
*June 14, 2014*

**Another this math tho**

ding ding -- we have a winner!
*June 13, 2014*

**this math tho**

well, does 3(-10)+54 = 18?
*June 13, 2014*

**3 questions Math**

Oops. The synthetic division site was http://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.php
*June 13, 2014*

**3 questions Math**

#18 - did you try this at the syntehtic division site I gave you? You can see polynomial long division at calc101.com; click on the "long division" button. #19 If you meant 1/(2x + 14) - 9/(x + 7) = -6 then that's just 1/(2x+14) - 18/(2x+14) = -6 -17/(2x+14) = -6...
*June 13, 2014*

**Algebra ll**

take a visit to http://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.php and enter your coefficients. It will show the calculation, and even explain it if needed. It's hard to do it here, due to browser formatting of html.
*June 13, 2014*

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